Quantitative Chemistry 1 GCSE

Quantitative Chemistry

GCSE Workbook Quantitative Chemistry Part 1

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GCSE Chemistry 8462. GCSE exams June 2018 onwards. Version 1.0 21 April 2016 Quantitative Chemistry

4.3.1 Chemical measurements, conservation of mass and the quantitative interpretation of chemical equations 4.3.1.1 Conservation of mass and balanced chemical equations Content

Key opportunities for skills development

The law of conservation of mass states that no atoms are lost or made during a chemical reaction so the mass of the products equals the mass of the reactants.

WS 1.2

This means that chemical reactions can be represented by symbol equations which are balanced in terms of the numbers of atoms of each element involved on both sides of the equation.

Students should understand the use of the multipliers in equations in normal script before a formula and in subscript within a formula.

4.3.1.2 Relative formula mass Content

Key opportunities for skills development

The relative formula mass (Mr) of a compound is the sum of the relative atomic masses of the atoms in the numbers shown in the formula.

In a balanced chemical equation, the sum of the relative formula masses of the reactants in the quantities shown equals the sum of the relative formula masses of the products in the quantities shown.

4.3.1.3 Mass changes when a reactant or product is a gas Content

Key opportunities for skills development

Some reactions may appear to involve a change in mass but this can usually be explained because a reactant or product is a gas and its mass has not been taken into account. For example: when a metal reacts with oxygen the mass of the oxide produced is greater than the mass of the metal or in thermal decompositions of metal carbonates carbon dioxide is produced and escapes into the atmosphere leaving the metal oxide as the only solid product.

AT 1, 2,6 Opportunities within investigation of mass changes using various apparatus.

Students should be able to explain any observed changes in mass in non-enclosed systems during a chemical reaction given the balanced symbol equation for the reaction and explain these changes in terms of the particle model.

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Quantitative Chemistry

4.3.1.4 Chemical measurements Content

Key opportunities for skills development

Whenever a measurement is made there is always some uncertainty about the result obtained.

WS 3.4

Students should be able to: • represent the distribution of results and make estimations of uncertainty • use the range of a set of measurements about the mean as a measure of uncertainty.

4.3.2 Use of amount of substance in relation to masses of pure substances 4.3.2.1 Moles (HT only) Content

Key opportunities for skills development

Chemical amounts are measured in moles. The symbol for the unit mole is mol.

WS 4.1, 4.2, 4.3, 4.5, 4.6

MS 1a The mass of one mole of a substance in grams is numerically equal Recognise and use to its relative formula mass. expressions in decimal One mole of a substance contains the same number of the stated form. particles, atoms, molecules or ions as one mole of any other MS 1b substance.

Recognise and use The number of atoms, molecules or ions in a mole of a given expressions in standard substance is the Avogadro constant. The value of the Avogadro form. constant is 6.02 x 1023 per mole.

MS 2a Students should understand that the measurement of amounts in moles can apply to atoms, molecules, ions, electrons, formulae and Use an appropriate number equations, for example that in one mole of carbon (C) the number of of significant figures. atoms is the same as the number of molecules in one mole of MS 3a carbon dioxide (CO2).

Understand and use the symbols: =, <, <<, >>, >, â&#x2C6;?, ~ MS 3b Change the subject of an equation.

Students should be able to use the relative formula mass of a substance to calculate the number of moles in a given mass of that substance and vice versa.

MS 1c

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GCSE Chemistry 8462. GCSE exams June 2018 onwards. Version 1.0 21 April 2016 Quantitative Chemistry

4.3.2.2 Amounts of substances in equations (HT only) Content

Key opportunities for skills development

The masses of reactants and products can be calculated from balanced symbol equations.

MS 1a

Chemical equations can be interpreted in terms of moles. For example:

Recognise and use expressions in decimal form.

Mg + 2HCI

MS 1c

MgCI2 + H2

shows that one mole of magnesium reacts with two moles of hydrochloric acid to produce one mole of magnesium chloride and one mole of hydrogen gas.

Use ratios, fractions and percentages.

MS 3b

Students should be able to:

Change the subject of an â&#x20AC;˘ calculate the masses of substances shown in a balanced symbol equation. equation MS 3c â&#x20AC;˘ calculate the masses of reactants and products from the Substitute numerical values balanced symbol equation and the mass of a given reactant or into algebraic equations product. using appropriate units for physical quantities.

4.3.2.3 Using moles to balance equations (HT only) Content

Key opportunities for skills development

The balancing numbers in a symbol equation can be calculated from the masses of reactants and products by converting the masses in grams to amounts in moles and converting the numbers of moles to simple whole number ratios.

MS 3b

Students should be able to balance an equation given the masses of reactants and products.

Students should be able to change the subject of a mathematical equation.

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Change the subject of an equation.

MS 3c Substitute numerical values into algebraic equations using appropriate units for physical quantities.

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Quantitative Chemistry

4.3.2.4 Limiting reactants (HT only) Content

Key opportunities for skills development

In a chemical reaction involving two reactants, it is common to use an excess of one of the reactants to ensure that all of the other reactant is used. The reactant that is completely used up is called the limiting reactant because it limits the amount of products.

WS 4.1

Students should be able to explain the effect of a limiting quantity of a reactant on the amount of products it is possible to obtain in terms of amounts in moles or masses in grams.

4.3.2.5 Concentration of solutions Content

Key opportunities for skills development

Many chemical reactions take place in solutions. The concentration of a solution can be measured in mass per given volume of solution, eg grams per dm3 (g/dm3).

MS 1c

Students should be able to: â&#x20AC;˘ calculate the mass of solute in a given volume of solution of known concentration in terms of mass per given volume of solution â&#x20AC;˘ (HT only) explain how the mass of a solute and the volume of a solution is related to the concentration of the solution.

Use ratios, fractions and percentages.

MS 3b Change the subject of an equation.

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Quantitative Chemistry

Balancing Chemical Equations In a chemical reaction, no atoms are ....................... or ............................. The same atoms that were in the ...............................(s) must all appear in the .....................................(s).

More specifically, the number of atoms of each element on one side of a chemical equation must equal the number of atoms of that element on the ................................ side.

When this condition is met, the equation is BALANCED. Understanding the numbers in chemical equations: • Small numbers to the lower right of chemical symbols in a formula are called ..............................; this number indicates the specific number of atoms of that element found in one molecule of the compound.

H2O Subscript

**Note: a molecule contains two or more atoms and is the smallest particle of a substance (element or compound) that retains all the properties of the substance.

For example, 1 water molecule contains 2 hydrogen atoms and 1 oxygen atom. If you remove any of these components, the substance is no longer water.

• Big numbers in front of chemical formulas are called ......................................; this number applies to every element within the compound.

6CO2 Coefficient The coefficient indicates that there are ______ carbons atoms and ______ oxygen atoms in 6 molecules of carbon dioxide.

•

Subscripts of 1 are not written but understood (you don’t write H2O1; instead you write H2O).

Practice Problem: 9/5 8/30

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Quantitative Chemistry

Is the following equation balanced? Yes or no.

C3H6

+

3O2

3CO2

à

+

C =

C =

H =

H =

O =

O =

3H2O

Rules for balancing equations: 1. Balance the elements one at a time by changing the coefficients in front of reactants and products. When no coefficient is written, it is assumed to be 1.

2. You may NOT change the subscripts in the chemical formula of a substance when trying to balance an equation. *Note: Pick an element to balance first and work through the elements in the equation one by one.

It is best to begin with an element other than hydrogen or oxygen. These two elements often occur more than twice in an equation.

3. Finally, your coefficients must be in their smallest whole number ratio for the equation to be properly balanced.

Example: The numbers 2, 2, 4, and 6 are not in their smallest whole number ratio.

The numbers 1, 1, 2, and 3 are.

Practice: 1. P4

+

O2

à

P4O10

2. Na

+

O2

à

Na2O

3. C5H12

+

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O2

à

CO2

Mr Singh

+

H2O

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Quantitative Chemistry

Mass Changes when a reactant or product is a gas Some reactions appear to involve a change in mass.

This is usually explained because a reactant or product is a .......... and so has not been considered when calculating.

Example 1: Mass reduction during reaction between marble chips and acid

Example 2: Mass increased when Magnesium is heated in air.

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Isotopes An element is a substance that is made of only one kind of atom. There are over 105 different known elements of which 90 are naturally occurring. Every element has its own …………… and has a unique number of ………………...

The number of protons an atom has is called the …………………….. Number or the Proton Number. In a neutral atom it is also the number of ………………… that the atom has. The protons are located in the ………………… of the atom along with the neutrons and the electrons orbit around the nucleus.

The Mass number of an atom tells us how massive an atom is and because electrons are relatively mass less, the mass number tells us the number of ……………….. and neutrons.

The number of protons is always the same for every element but atoms of elements can have different numbers of Neutrons. Two atoms of the same element with different numbers of neutrons are called …………………. They exist naturally in different relative abundances.

Isotopes are atoms of elements with the same number of protons but different number of neutrons.

Examples: Carbon has three Isotopes

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3rd Form Notes

Quantitative Chemistry

Hydrogen also has three isotopes:

Helium has two naturally occurring isotopes:

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Relative Atomic Mass A single atom is very very light, about 0.000,000,000,000,000,000,000,002g. As these sorts of numbers are very awkward to use, we use ………………. mass units. (amu) The standard to which everything is compared has been decided to be 1/12th the mass of a Carbon-12 atom. This is because is almost the size of an Hydrogen atom, the smallest atom in the universe. The masses of all other atoms are therefore compared to this. We called this it’s Relative Atomic Mass.

Element Hydrogen Carbon Nitrogen Oxygen Sodium Magnesium Sulfur

Symbol H C N O Na Mg S

RAM 1 12 14 16 23 24 32

Element Chlorine Potassium Calcium Iron Copper Zinc Iodine

Symbol Cl K Ca Fe Cu Zn I

RAM 35.5 39 40 56 64 65 127

Some atoms have isotopes so the ……………………. Atomic Mass is the ‘mean’ or average mass. The isotopes are also in different natural abundances so it is also the ‘weighted’ or proportioned average mass.

The Relative Atomic Mass is the weighted mean mass of an atom of an element compared to the mass of 1/12th the mass of an atom of Carbon-12.

We can calculate the Relative Atomic Mass (RAM or Ar) by summing all the Mass numbers for each isotope multiplied by its relative abundance.

!! =

[ !"## !"#$%& !" !"#$#%& 1 ! !"#$%&'" !"#$%&$'( + (!"## !"#$%& !" !"#$#%& 2 ! !"#$%&'" !"#$%&$'() + ⋯ ]

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3rd Form Notes

Quantitative Chemistry

e.g Chlorine.

Chlorine has two isotopes of 35 and 37 amu. Their natural abundance is 75% and 25% respectively. !"#$%&'" !"#$%& !"#! (!! ) = [35 Ă&#x2014;

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!" !""

] + [37 Ă&#x2014;

!" !""

] = 35.5 amu

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Quantitative Chemistry

Relative Formula Mass The world is made up of compounds and so itâ&#x20AC;&#x2122;s helpful to know how massive a molecule or unit of a compound weighs.

Since the Relative Atomic Masses is the mass of an atom we simple add together the Relative Atomic Masses for every atom in the molecule.

The Mass of all the atoms in the formula of a compound is called the Relative Formula Mass (RFM) sometimes known as the Relative Molecular Mass (RMM or Mr). e.g. Water, H2O In the molecule there are 1 x H and 1 x O H has an Ar of 1, O has an Ar of 16 Therefore RFM = [(2 x 1) + (1 x 16)] = 18 amu Substance

Formula

Nitrogen Ammonia

N2 NH3

Sodium Chloride

NaCl

Magnesium Nitrate

Mg NO!

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!

Atoms in the formula 2xN 1xN 3xH

RAM of atoms

Formula Mass

N = 14 N = 14 H=1

1 x Na 1 x Cl

Na = 23 Cl = 35.5

1 x Mg 2xN 6xO

Mg = 24 N = 14 O = 16

2 x 14 = 28 1 x 14 = 14 3x1 = 3 Total = 17 1 x 23 1 x 35.5 Total = 58.5 1 x 24 = 24 2 x 14 = 28 6 x 16 = 96 Total = 148

Mr Singh

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Quantitative Chemistry

Experimental Uncertainty In an experiment, when you make a measurement of whatever kind, you cannot be sure just how close it is to the ...................... value, that is, how ........................ it is. There is an experimental ............................. (often called 'experimental error').

Experimental uncertainty arises because of: • • •

Limits in the how exact the measuring apparatus is. This is the precision of the apparatus.

Imperfections in experimental procedures.

Judgements made by the operator.

When can my results be said to be precise?

If you repeat a measurement several times and obtain values that are close together, your results are said to be ............................

If the same person obtains these close values, then the experimental procedure is ..................................

If a number of different people carry out the same measuring procedure and the values are close the procedure is .......................................

What is a systematic error?

A .............................. error is one that is repeated in each measurement taken. If this is realised after the experimental work is done, it can be taken into account in any calculations.

What are random errors?

Even the most careful and experienced operator cannot avoid ....................... errors.

However, their effect can be reduced by carrying out a measurement many times (if the opportunity exists) and working out an average value.

Let's look in more detail at 'built-in' uncertainty of some laboratory equipment...

Some measurement uncertainties are given below: Equipment

Measurement to the nearest:

Balance (1 decimal place)

±0.08 g

Balance (2 decimal place)

±0.008 g

Balance (3 decimal place)

±0.0008 g 3

Measuring Cylinder (25 cm )

±0.5 cm

3

3

Graduated Pipette (25 cm , Grade B) ±0.04 cm 3

Burette (50 cm , Grade B)

±0.08 cm

3

3

Volumetric Flask (250 cm , Grade B) ±0.2 cm Stopwatch (digital)

Source: www.avogadro.co.uk

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3 3

±0.01 s

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Quantitative Chemistry

Interquartile range & Outliers & Boxplots The IQR is calculated as follows : IQR = UQ – LQ. The UQ is found ¾ of the way through the data i.e. at position

3 n 4

1 .

The LQ is found ¼ of the way through the data i.e. at position

1 n 4

1 .

To find an outlier we work out 1.5 times the IQR and subtract/add to the LQ/UQ respectively. If an item is outside this range, it is considered an outlier.

This data can also be shown on a box plot.

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Quantitative Chemistry

3rd Form Notes

The Mole There are certain names which we give to describe a number. For instance:

A Pair (2) of Shoes

A Couple (2) of people

A Dozen (12) eggs

A Century (100) of runs

Atoms are far too small to weigh out individually but if we know the relative atomic mass in atomic mass units and we weigh out that substanceâ&#x20AC;&#x2122;s relative atomic mass in grams, we know how many atoms we have.

For instance ..........g of Carbon-..... will always contain 602,000,000,000,000,000,000,000 carbon atoms or in standard form 6.02 X 1023 carbon atoms. Amedeo Avagadro discovered this.

We give this number of anything a name â&#x20AC;&#x201C; A ...................... So in chemistry we describe atoms or molecules (or anything) using moles.

A Mole of atoms of any element will always equal its Relative Atomic Mass in .....................

A Mole of molecules of any compound will always equal its Relative Formula Mass in .......................

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Quantitative Chemistry

3rd Form Notes

Sodium is made up of sodium atoms. It has an RAM of 23

Iodine is a molecule I2 and has a Formula Mass of 254

This means that 23 grams of sodium contain 6.02 X 1023 sodium atoms, or a Mole of sodium atoms.

Therefore 254g of Iodine will contain 6.02 X 1023 iodine molecules or a Mole of Iodine molecules

Water is made up of water molecules with the formula H2O. It has a formula mass of 18

Therefore a beaker containing 18ng of Water contains 6.02 X 1023 water molecules or a Mole of Water molecules.

This is how we can work out the mass of a mole of molecules of a compound.

The Mole is the method by which we count by mass.

Atoms or Molecules are far far too small to count them so we need to count they by weighing a very large amount of them. We can do this because we know the mass of a single atom of every element.

(Relative Atomic Mass), just like we can do with coins.

When we need to know the number of atoms we have, we can weigh the compound and calculate it using the formula below.

This is useful because chemicals react not in equal masses but in simple, whole number ............................. of atoms.

For example the reaction between Sodium and Chlorine. !!" + !!! â&#x2020;&#x2019; !!"#$ + In this reaction 2 atoms of sodium reaction with 1 molecule of Chlorine to give us one Sodium Chloride. Simple whole number ratios of atoms or molecules, not masses.

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Quantitative Chemistry

3rd Form Notes

We can also say that 2 moles of atoms of sodium will react with 1 mole of molecules of Chlorine as well. But using masses we can count how many moles and therefore atoms were involved in the reaction.

The relationship between mass, moles and relative formula mass is very simple and very very important.

For elements, !"#$% =

!"## !"#$%&'" !"#$%& !"##

For compounds, !"#$% =

!"## !"#$%&'" !"#$%&' !"##

This can be rearranged to give any of the two other forms: !"## = !"#$% Ă&#x2014;!"#$%&'" !"#$%&' !"## Or, !"#$%&'" !"#$%&' !"## =

!"## !"#$%

The triangle above shows that neatly.

It is often abbreviated to:

! != !!

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Quantitative Chemistry

Amount of Substances in equations

Moles to Masses

The Masses of reactants and products is can be calculated from balanced symbol equations 2Mg(s) + O2(g) --> 2MgO(s) Mg(s) + 2HCl(aq) --> MgCl2(aq) + H2(g) The reaction coefficients show the number of ............... that are reacting together.

THIS IS NOT THE SAME THING AS THE MASS. Using the equations moles = mass/relative formula mass can then calculate the mass of each reactant or product from the .....................

Masses to Moles We can also work out the moles of each substance in a reaction from the masses if each.

See later notes.

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Quantitative Chemistry

Using Balanced Equations to work out reacting masses To work out what mass of which compounds reacts with a given mass of another compound we need a couple of key skills.

1. Working out the Relative Formula Mass (RFM) Work out the RFM for the following: (Add up all the Ar of each atom)

a) Ca(OH)2

b) K2SO4

c) NH4NO3

d) Ca(NO3)2

e) Al2(SO4)3

2. Converting Mass into moles and moles into mass

Mole = Mass/RFM a) How many moles of CCl are there in 56 g?

4

b) How much does 2.50 mol of H SO weigh?

2

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Quantitative Chemistry

c) How much does 0.25 mol of Fe O weigh?

2

3

d) How many moles are there in 52 g of CO?

3. Balancing equations and working out ratios.

Balance these and write underneath the ratios by which all the compounds react

a)

Ca +

H2O à Ca(OH)2 + H2

b)

CO + O2 à

c)

Ca + O2 à

d)

Fe2O3 + HCl à FeCl3 + H2O

e)

NH3 + H2SO4 à (NH4)2SO4

ratio: CO2

ratio:

CaO

ratio:

ratio:

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Quantitative Chemistry

ratio:

Finally we’re ready!

1)

What mass of hydrogen is produced when 192 g of magnesium is reacted with hydrochloric acid?

Mg + 2 HCl → MgCl2 + H2

(3)

Compound Mass/g RFM/g/mol Mole/mol Ratio

2)

What mass of oxygen is needed to react with 8.5 g of hydrogen sulphide (H2S)?

2 H2S + 3 O2 à 2 SO2 + 2 H2O

(3)

Compound Mass/g RFM/g/mol Mole/mol Ratio

3)

What mass of potassium oxide is formed when 7.8 g of potassium is burned in oxygen?

4 K + O 2 à 2 K 2O

(3)

Compound Mass/g RFM/g/mol Mole/mol Ratio

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Quantitative Chemistry

4)

Railway lines are welded together by the Thermite reaction, which produces molten iron.

What mass of iron is formed from 1 kg of iron oxide?

Fe2O3 + 2 Al Ă 2 Fe + Al2O3

5)

What mass of oxygen is required to oxidise 10 g of ammonia to NO?

4 NH3 + 5 O2 â&#x2020;&#x2019; 4 NO + 6 H2O

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Quantitative Chemistry

From Masses to Balanced Equations 1.

Working out balanced Chemical Equations The Formula of a compound tells us a lot of information. For instance CO2 1 Carbon atom combines with 2 Oxygen atoms Therefore, 1 Mole of Carbon Atoms combines with 2 Moles of Oxygen Atoms Moles can be changed into Masses so therefore from the Relative Atomic Masses we can say 12g of Carbon combines with 32g of Oxygen Therefore, 6g of Carbon will combine with 16g of Oxygen and so on.

From the formula we can tell: i) How many moles of the different atoms combine ii) How many grams of the different element combine.

From the Masses that react to form a compound we can work out the simplest whole number ratio the moles react in.

1. Start with the number of grams for each reactant or product

2. Change the grams to moles of atoms.

3. Divide all mole quantities by the smallest mole quantity

4. This results in the simplest ratio or reaction coefficients.

For Example, 32g of Sulfur combine with 32g of Oxygen to form the compound Sulfur Dioxide. Write the balanced equation.

1. đ?&#x2018;&#x2020; đ?&#x2018; + đ?&#x2018;&#x201A;& đ?&#x2018;&#x201D; â&#x2020;&#x2019; đ?&#x2018;&#x2020;đ?&#x2018;&#x201A;& (đ?&#x2018;&#x201D;) Masses into Moles RAM (S) = 32 RAM (O) = 16 Therefore using the formula đ?&#x2018;&#x203A; = 0&1 0&1/-34 0&1 0&1/-34

./

= 1 Mole of S = 1 Moles of O2

2. Divide each mole quantity by the smallest mole quantity 1 mole is smallest so we divide everything by 1.

3. This results in the coefficients for each element

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Quantitative Chemistry

đ?&#x2018;&#x2020; đ?&#x2018; + đ?&#x2018;&#x201A;& â&#x2020;&#x2019; đ?&#x2018;&#x2020;đ?&#x2018;&#x201A;& (đ?&#x2018;&#x201D;) For Example, 20 grams of Calcium react completely with 19g of Fluorine, to form Calcium Fluoride.

What is the balanced chemical equation?

Substances

Ca

F2

CaF2

Mass/g

20

19

(20+19) = 39

Write out the balanced equation:

2. Limiting Reactants When you carry out a reaction in experiments exact amounts according to the ........................... equation are not used. One reactant is always in ....................

The reason the reaction stops is that the other reactant has run out. We say that it is the limiting ........................

The reactant that gets used up first is the limiting reactants.

1.

2.

3.

4.

Determine the balanced chemical equation for the chemical reaction Convert all given information into moles Calculate the molar ratio form the given information.

Compare calculated ratio to the actual ratio in the equation. Whichever reactant is not larger, is the limiting reactant.

5. Use the amount of limiting reactant to do further calculations.

Alternatively.

1.

2.

3.

4.

5.

Determine the balanced chemical equation Convert all given information into moles Using the equation, calculate the mass of product produced for each reactant.

Whichever produces the least amount of product is the limiting reactant.

Use this is do any further calculations.

EXAMPLE 1: PHOTOSYNTHESIS

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Quantitative Chemistry

Consider respiration, one of the most common chemical reactions on earth. đ??ś7 đ??ť9& đ?&#x2018;&#x201A;7 + 6đ?&#x2018;&#x201A;& â&#x2020;&#x2019; 6đ??śđ?&#x2018;&#x201A;& + 6đ??ť& đ?&#x2018;&#x201A; + đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x201D;đ?&#x2018;Ś What mass of carbon dioxide forms in the reaction of 25 grams of glucose with 40 grams of oxygen?

SOLUTION When approaching this problem, observe that every 1 mole of glucose (C6H12O6C6H12O6) requires 6 moles of oxygen to obtain 6 moles of carbon dioxide and 6 moles of water.

Step 1: Determine the balanced chemical equation for the chemical reaction.

The balanced chemical equation is already given.

Step 2: Convert all given information into moles (most likely, through the use of molar mass as a conversion factor).

25đ?&#x2018;&#x201D;Ă&#x2014;

1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; = 0.1388đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ??ś7 đ??ť9& đ?&#x2018;&#x201A;7 180.06đ?&#x2018;&#x201D; 40đ?&#x2018;&#x201D;Ă&#x2014;

1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; = 1.25đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& 32đ?&#x2018;&#x201D;

Step 3: Calculate the mole ratio from the given information. Compare the calculated ratio to the actual ratio. a. If all of the 1.25 moles of oxygen were to be used up, there would need to be 1.25Ă&#x2014;161.25Ă&#x2014;16 or 0.208 moles of glucose. There is only 0.1388 moles of glucose available which makes it the limiting reactant.

1.25đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& Ă&#x2014;

1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ??ś7 đ??ť9& đ?&#x2018;&#x201A;7 = 0.208đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ??ś7 đ??ť9& đ?&#x2018;&#x201A;7 6 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;&

b. If all of the 0.1388 moles of glucose were used up, there would need to be 0.1388 x 6 or 0.8328 moles of oxygen. Because there is an excess of oxygen, the glucose amount is used to calculate the amount of the products in the reaction.

0.1388đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ??ś7 đ??ť9& đ?&#x2018;&#x201A;7 Ă&#x2014;(

6 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& ) = 0.8328đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& 1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ??ś7 đ??ť9& đ?&#x2018;&#x201A;7

If more than 6 moles of O2 are available per mole of C6H12O6, the oxygen is in excess and glucose is the limiting reactant. If less than 6 moles of oxygen are available per mole of glucose, oxygen is the limiting reactant. The ratio is 6 mole oxygen per 1 mole glucose, OR 1 mole oxygen per 1/6 mole glucose. This means: 6 mol O2 / 1 mol C6H12O6 .

Therefore, the mole ratio is: (0.8328 mol O2)/(0.208 mol C6H12O6) This gives a 4.004 ratio of O2 to C6H12O6.

Step 4: Use the amount of limiting reactant to calculate the amount of CO2 or H2O produced.

For carbon dioxide produced: 0.1388molesglucoseĂ&#x2014;61=0.8328molescarbondioxide0.1388molesglucoseĂ&#x2014;61=0.8328mole scarbondioxide.

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Quantitative Chemistry

Step 5: If necessary, calculate how much is left in excess.

1.25 mol - 0.8328 mol = 0.4172 moles of oxygen left over

EXAMPLE 2. OXIDATION OF MAGNESIUM Calculate the mass of magnesium oxide possible if 2.40 g MgMg reacts with 10.0 g O2O2 đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; + đ?&#x2018;&#x201A;& â&#x2020;&#x2019; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; SOLUTION Step 1: Balance equation 2đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; + đ?&#x2018;&#x201A;& â&#x2020;&#x2019; 2đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; Step 2 and Step 3: Converting mass to moles and stoichiometry 2.40đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; Ă&#x2014;

1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; 2.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 40.31đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; Ă&#x2014; Ă&#x2014; = 3.98đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 24.31đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; 2.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; 1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A;

10.0đ?&#x2018;&#x201D; đ?&#x2018;&#x201A;& Ă&#x2014;

1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& 2đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 40.31đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; Ă&#x2014; Ă&#x2014; = 25.2đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& 1đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 32.0đ?&#x2018;&#x201D; đ?&#x2018;&#x201A;&

Step 4: The reactant that produces a smaller amount of product is the limiting reagent Mg produces less MgO than does O2 (3.98 g MgO vs. 25.2 g MgO), therefore Mg is the limiting reagent in this reaction.

Step 5: The reactant that produces a larger amount of product is the excess reagent O2 produces more amount of MgO than Mg (25.2g MgO vs. 3.98 MgO), therefore O2 is the excess reagent in this reaction.

Step 6: Find the amount of remaining excess reactant by subtracting the mass of the excess reagent consumed from the total mass of excess reagent given.

Mass of excess reagent calculated using the limiting reagent:

2.40đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; Ă&#x2014;

1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& 32.0đ?&#x2018;&#x201D; đ?&#x2018;&#x201A;& 1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; Ă&#x2014; Ă&#x2014; = 1.58đ?&#x2018;&#x201D; đ?&#x2018;&#x201A;& 2.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D; 1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;& 24.31đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;

OR Mass of excess reagent calculated using the mass of the product:

3.98đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; Ă&#x2014;

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1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;2 32.0đ?&#x2018;&#x201D; đ?&#x2018;&#x201A;& 1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; Ă&#x2014; Ă&#x2014; = 1.58đ?&#x2018;&#x201D; đ?&#x2018;&#x201A;& 40.31đ?&#x2018;&#x201D; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 2.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;đ?&#x2018;&#x201D;đ?&#x2018;&#x201A; 1.00đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x201A;&

Mr Singh

27

Quantitative Chemistry

Mass of total excess reagent given â&#x20AC;&#x201C; mass of excess reagent consumed in the reaction 10.0g â&#x20AC;&#x201C; 1.58g = 8.42g O2 is in excess.

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Quantitative Chemistry

Concentration Calculations A solution is made up of two components, a solid â&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Ś. and a liquid â&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Śâ&#x20AC;Ś that have totally dissolved in one another.

Concentration is the amount of solute, in grams or moles, that is dissolved in 1 dm3 of solution.

Amount per unit volume measured in mol/dm3 or g/dm3 Concentration of solution, amount of solute and volume of solvent are simply related by a very important equation. đ??´đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;˘đ?&#x2018;&#x203A;đ?&#x2018;Ą đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x2020;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;&#x2019;(đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;) =

đ??śđ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x2020;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A;(đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x2018;đ?&#x2018;&#x161;56 ) Ă&#x2014; đ?&#x2018;&#x2030;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x161;đ?&#x2018;&#x2019; đ?&#x2018;&#x153;đ?&#x2018;&#x201C; đ?&#x2018;&#x2020;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;Łđ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ą(đ?&#x2018;?đ?&#x2018;&#x161;6 ) 1000

More simply the equation is đ?&#x2018;&#x20AC;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;đ?&#x2018; =

đ??śđ?&#x2018;&#x153;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2019;đ?&#x2018;&#x203A;đ?&#x2018;Ąđ?&#x2018;&#x;đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; Ă&#x2014; đ?&#x2018;&#x2030;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x161;đ?&#x2018;&#x2019; 1000

Abbreviated to

đ?&#x2018;&#x203A;=

đ??śđ?&#x2018;&#x2030; 1000

Amount is always measured in moles but volume is always measured in decimetres cubed.

This is the scientific name for a litre.

In exam questions and during practicals volume will be measured in centimetres cubed so we add into the equation a conversion factor of 1000 .

This can be rearranged depending on what information we know and what information we are trying to find

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Quantitative Chemistry

đ??ś=

>???Ă&#x2014;@ A

and

đ?&#x2018;&#x2030;=

>???Ă&#x2014;@ B

We use this formula for calculating the number of moles in a solution in a given volume.

For example, calculate the number of moles of solute present in 75cm3 of a solution of hydrochloric acid which has a concentration of 2.0 mol/dm3. đ?&#x2018;&#x203A;=

đ??śđ?&#x2018;&#x2030; 2.0 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;&#x2018;đ?&#x2018;&#x161; 56 Ă&#x2014; 75đ?&#x2018;?đ?&#x2018;&#x161; 6 = = 0.15đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; 1000 1000

We can also use it to calculate the concentration of a solution when we make it.

For example, 100g of Sodium Hydroxide was dissolved to make 2.0 litres of solution. What is the concentration?

1. Calculate the number of moles of sodium hydroxide Mr (NaOH) = 23 + 16 + 1 = 40g/mol đ?&#x2018;&#x203A;=

đ?&#x2018;&#x161; 100đ?&#x2018;&#x201D; = = 2.5đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x20AC;G 40đ?&#x2018;&#x201D;/đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;

2. Calculate the concentration of Sodium Hydroxide đ?&#x2018;&#x203A; 2.5 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ??ś = = = 1.25 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; đ?&#x2018;&#x2018;đ?&#x2018;&#x161; 56 đ?&#x2018;&#x2030; 2.0 đ?&#x2018;&#x2018;đ?&#x2018;&#x161; 6 Or it could be used to work out how much of a solute is needed to make a concentration of a specific strength.

For example, what mass of sodium carbonate must be dissolved in 1 l of solvent to give a solution of concentration 1.5 mol/dm3?

1. Amount of sodium carbonate required đ?&#x2018;&#x203A; = đ??śđ?&#x2018;&#x2030; = 1.5 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;&#x2018;đ?&#x2018;&#x161; 56 Ă&#x2014; 1 đ?&#x2018;&#x2018;đ?&#x2018;&#x161; 6 = 1.5 đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; 2. Mass of sodium carbonate required Mr(Na2CO3) = (23 x 2) + 12 + (16 x3) = 106g/mol đ?&#x2018;&#x161; = đ?&#x2018;&#x203A; Ă&#x2014; đ?&#x2018;&#x20AC;G = 1.5đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; Ă&#x2014; 106đ?&#x2018;&#x201D;đ?&#x2018;&#x161;đ?&#x2018;&#x153;đ?&#x2018;&#x2122; 5> = 159đ?&#x2018;&#x201D;

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Atomic Structure

Atomic Structure

1. Copy the table below into your book, then complete it.

Element hydrogen lithium

Symbol 1 1H 7 3 Li 16 8O

Atomic number 1

Atomic mass 1

13

27

1. Copy the table below into your book, then complete it.

Number of: protons neutrons electrons 1 0 1

argon 19

22 20

13

27

Number of: protons neutrons electrons 1 0 1

19 53

Explain why.

22 20

127 238 92

U

2. a) Neutral atoms must have the same number of electrons as protons.

53

U

2. a) Neutral atoms must have the same number of electrons as protons.

Explain why.

b) Atomic mass is the number of protons added to the number of neutrons. For example, beryllium has 4 protons and 5 neutrons, and so its atomic mass is 9.

Explain why the number of electrons is not included in the atomic mass.

3. You can use the symbol of an element to work out the numbers of protons, neutrons and electrons there are in its atom.

Draw, or write, a revision summary about how you do this.

Use the lithium atom 73 Li as your example.

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Symbol 1 1H 7 3 Li 16 8O

Atomic mass 1

argon

127 238 92

Element hydrogen lithium

Atomic number 1

Mr Singh

b) Atomic mass is the number of protons added to the number of neutrons. For example, beryllium has 4 protons and 5 neutrons, and so its atomic mass is 9.

Explain why the number of electrons is not included in the atomic mass.

3. You can use the symbol of an element to work out the numbers of protons, neutrons and electrons there are in its atom.

Draw, or write, a revision summary about how you do this.

Use the lithium atom 73 Li as your example.

40 31

N-m08-05 (Š Nigel Saunders 2000)

Quantitative Chemistry

Moles 1 â&#x20AC;&#x201C; Formula Mass 1)

2)

3)

4)

Calculate the relative molecular mass (Mr) of: a) H2

g) Ca(OH)2

b) Ne

h) K2SO4

c) NH3

i) NH4NO3

d) CH4

j) Ca(NO3)2

e) MgBr2

k) Al2(SO4)3

f) S8

l) H2C2O4.2H2O

(12)

Calculate the percentage by mass of the elements shown in the following compounds (you have worked out the Mr's of (a) to (g) in question 1). a) C in CH4

e) N in Ca(NO3)2

b) Br in MgBr2

f) O in Ca(NO3)2

c) S in K2SO4

g) O in Ca(OH)2

d) N in NH4NO3

h) O in Fe(NO3)3

(9)

Calculate the relative molecular mass (Mr) of: a) sodium oxide

c) copper hydroxide

b) calcium carbonate

d) zinc nitrate

(8)

Calculate the percentage by mass of the elements shown in the following compounds. a) Cl in calcium chloride

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b) O in iron (III) oxide

Mr Singh

(6)

34 32

Quantitative Chemistry

410W

Molar Mass Calculations 2 Calculate the Molar Mass of the following using the Relative Atomic Masses on the Periodic Table. Beware of brackets. Write out the calculation as shown in the example below: Na2CO3 = (Na x 2) + (C x 1) + (O x 3) = (23 x 2) + (12 x 1) + (16 x 3) = 46 + 12 + 48

= 106 g mol-1

1. H2O

21. CaCl2

41. CuSO4

2. CO2

22. Ca(NO3)2

42. ZnCl2

3. NH3

23. Ca(OH)2

43. AgNO3

4. C2H5OH

24. CaSO4

44. NH4Cl

5. C2H4

25. BaCl2

45. (NH4)2SO4

6. SO2

26. AlCl3

46. NH4VO3

7. SO3

27. Al(NO3)3

47. KClO3

8. HBr

28. Al2(SO3)3

48. KIO3

9. H2SO4

29. FeSO4

49. NaClO

10. HNO3

30. FeCl2

50. NaNO2

11. NaCl

31. FeCl3

51. CuSO4·5H2O

12. NaNO3

32. Fe2(SO4)3

52. FeSO4·7H2O

13. Na2CO3

33. PbO

53. CH3CH2(OH)CH3

14. NaOH

34. PbO2

54. Na2S2O3·5H2O

15. Na2SO4

35. Pb3O4

55. (COOH)2·2H2O

16. KMnO4

36. Pb(NO3)2

56. MgSO4.7H2O

17. K2Cr2O7

37. PbCl2

57. Cu(NH3)4SO4·2H2O

18. KHCO3

38. PbSO4

58. CH3CO2H

19. KI

39. CuCl

59. CH3COCH3

20. CaNO3

40. CuCl2

60. C6H5CO2H

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Mr Singh

41 33

Quantitative Chemistry

Ar, Mr and percentage composition calculations Information needed to answer the questions Element Al Cl Cu Fe

Ar 27 35.5 64 56

Relative formula mass (Mr) calculations 1.

2.

3.

4.

Ar 1 14 16 32

Work out the Mr of the following compounds:

Iron(II) sulphide, FeS Copper(II) sulphate, CuSO4 Ammonium chloride, NH4Cl Aluminium sulphate, Al2(SO4)3

Percentage composition calculations 5.

6.

7.

8.

Element H N O S

Calculate the percentage composition of the following:

Fe in FeS Cu in CuSO4 H in NH4Cl O in Al2(SO4)3

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Mr Singh

34

Quantitative Chemistry

410W

Molar Mass Calculations 1 A. Relative Atomic Mass – RAM: The mass of one atom of an element compared to the mass of one atom of carbon-12, with a nominal mass of 12. What is the RAM of the following elements RAM H RAM C RAM O

RAM Cl RAM Na RAM Mg

RAM S RAM F RAM I

B. Relative Molecular Mass – RMM: The mass of a molecule of an element or compound. The sum of the RAM of all the atoms in the molecule. Calculate the Molar Mass of the following using the Relative

Atomic Masses on the Periodic Table. Beware of brackets. Write out the calculation as shown in the example below: H2O = (H x 2) + (O x 1) = ( 1 x 2) + (16 x 1) = 2 + 16 = 18 g mol-1 RMM RMM RMM RMM RMM

H 2O CO2 H2 N2 CH4

RMM RMM RMM RMM RMM

SO2 SO3 I2 Cl2 NH3

C. Relative Formula Mass – RFM: For compounds which don’t have discrete units i.e. compounds which contain a metal and a non-metal (ionic compounds). The sum of the RAM for all the atoms in one formula unit. Calculate the RFM of the following RFM RFM RFM RFM RFM RFM RFM

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NaCl NaOH Ca(OH)2 CuCO3 NH4NO3 ZnS Na2S

RFM RFM RFM RFM RFM RFM RFM

HCl(aq) H2SO4 HNO3 Al2O3 RbBr K 2O ZnCO3

Mr Singh

RFM RFM RFM RFM RFM RFM RFM

Fe2O3 PbBr2 AgF Ca2(PO4)2 LiOH CuSO4·5H2O Mg(HCO3)2

44 35

Quantitative Chemistry

1105WS

Mole Calculations Write out the questions and answer them in your exercise book 1. What is the mass of: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j)

1 mole of silver atoms 2 moles of carbon atoms 1 mole of osmium 0.5 mole of helium atoms 0.25 mole of sulphur atoms 0.1 mole of neon atoms 10 mole nitrogen atoms 0.05 mole mercury 0.05 mole calcium atoms 0.01 mole gold atoms

3. Consider the pairs below and decide in each case which one contains the greater number of atoms. HINT: Calculate the number of moles of each first (a) (b) (c) (d) (e)

40 g of calcium or 50 g of iron 100 g of sulphur or 150 g iron 80 g of bromine or 40 g of argon 28 g of nitrogen or 30 g of oxygen 127 g of iodine or 65 g of zinc

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2. Write down the number of moles of the following (a) 56 g iron (b) 2.4 g magnesium (c) 6 g carbon (d) 2 g neon (e) 64 g sulphur (f) 2.07 g lead (g) 1 g mercury (h) 2 g oxygen (i) 10 g bromine (j) 39.4 g gold

4. Write down the mass of the first named element that has the same number of atoms as there are in the mass of the second named element a) b) c) d) e) f) g) h)

Mr Singh

Carbon – 32 g of oxygen Sulphur – 28 g of nitrogen Lead – 190 g of osmium Helium – 12 g carbon Chlorine – 12.7 g iodine Aluminium – 2.3 g sodium Calcium – 0.32 g sulphur Argon – 0.2 g neon

43 36

Quantitative Chemistry

Simple Mole Calculations 1. How many moles of Na are in 42 g of Na?

2. How many moles of O are in 8.25 g of O?

3. How much does 2.18 mol of Cu weigh?

4. What is the mass of 0.28 mol of iron?

5. How many atoms are in 7.2 mol of chlorine?

6. How many atoms are in 36 g of bromine?

7. How many moles are in 1.0 x 109 atoms?

8. What is the mass of 1.20 x 1025 atoms of sulfur?

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37

Quantitative Chemistry

9. How many moles of CO molecules are in 52 g of CO?

10. How many moles of C2H6 are in 124 g?

11. How many moles of CCl4 are there in 56 g?

12. How much does 2.50 mol of H2SO4 weigh?

13. How much does 0.25 mol of Fe2O3 weigh?

14. How many molecules are there in 52 g of CO?

15. How many formula units are in 22.4 g SnO2?

16. How many molecules are in 116 g CCl4?

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Mr Singh

38

Quantitative Chemistry

17. What is the mass of 3.01 x 1023 formula units of Fe2O3?

18. What is the mass of 1.2 x 1025 molecules of CO?

19. How many O atoms are in 1.25 mol of SO2?

20. How many moles of O atoms do you have when you have 1.20 x 1025 N2O5 molecules?

21. How many formula units are in 5.33 mol of CuCl2?

22. How many copper atoms are in 5.33 mol of CuCl2?

23. How many moles of Cl atoms are in 5.33 mol of CuCl2?

24. How many moles of CuCl2 contain 1.2 x 1023 atoms of Cl?

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Mr Singh

39

Quantitative Chemistry

25. How many O atoms are in 3.15 mol of SnO2?

26. How many H atoms are in 17.5 g (NH4)2C2O4?

27. What is the mass of 1 mole of lithium hydroxide?

28. A saltshaker filled with 17 grams of Sodium chloride contains how many moles of Sodium chloride?

29. Which of the following has the greatest mass? 4.2 mol of carbon or 8.34*1024 atoms of CaCO3 or 12.6 g of Al2(NO3)3

30. In 60 g of N2O there are how many molecules?

31. Find the atomic mass or formula mass for each of the following: a) Mn b) Se c) BaBr2

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Mr Singh

40

Quantitative Chemistry

d) NaOH 32. Find the number of moles of each of the following: a) 25 g of Ar

b) 38.6 g of Mg

c) 76.5 g of SO3

d) 229 g of NaNO3

33. Find the number of atoms in each of the following: a) 0.28 mole Ni

b) 1.84 mole S

c) 1.32 mole F2

d) 3.15 mole K3PO4

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Mr Singh

41

Quantitative Chemistry

Reacting Mass calculations To work out what mass of which compounds reacts with a given mass of another compound we need a couple of key skills.

1. Working out the Relative Formula Mass (RFM) Work out the RFM for the following: (Add up all the Ar of each atom)

a) Ca(OH)2

b) K2SO4

c) NH4NO3

d) Ca(NO3)2

e) Al2(SO4)3

2. Converting Mass into moles and moles into mass

Mole = Mass/RFM a) How many moles of CCl are there in 56 g?

4

b) How much does 2.50 mol of H SO weigh?

2

9/5 8/30

4

Mr Singh

47 42

Quantitative Chemistry

c) How much does 0.25 mol of Fe O weigh?

2

3

d) How many moles are there in 52 g of CO?

3. Balancing equations and working out ratios.

Balance these and write underneath the ratios by which all the compounds react

a)

Ca +

H2O à Ca(OH)2 + H2

b)

CO + O2 à

c)

Ca + O2 à

d)

Fe2O3 + HCl à FeCl3 + H2O

e)

NH3 + H2SO4 à (NH4)2SO4

ratio: CO2

ratio:

CaO

ratio:

ratio:

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Mr Singh

48 43

Quantitative Chemistry

ratio:

Finally we’re ready!

1)

What mass of hydrogen is produced when 192 g of magnesium is reacted with hydrochloric acid?

Mg + 2 HCl → MgCl2 + H2

(3)

Compound Mass/g RFM/g/mol Mole/mol Ratio

2)

What mass of oxygen is needed to react with 8.5 g of hydrogen sulphide (H2S)?

2 H2S + 3 O2 à 2 SO2 + 2 H2O

(3)

Compound Mass/g RFM/g/mol Mole/mol Ratio

3)

What mass of potassium oxide is formed when 7.8 g of potassium is burned in oxygen?

4 K + O 2 à 2 K 2O

(3)

Compound Mass/g RFM/g/mol Mole/mol Ratio

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49 44

Quantitative Chemistry

4)

Railway lines are welded together by the Thermite reaction, which produces molten iron.

What mass of iron is formed from 1 kg of iron oxide?

Fe2O3 + 2 Al Ă 2 Fe + Al2O3

5)

What mass of oxygen is required to oxidise 10 g of ammonia to NO?

4 NH3 + 5 O2 â&#x2020;&#x2019; 4 NO + 6 H2O

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(3)

Mr Singh

(3)

50 45

Quantitative Chemistry

Moles 2 – Reacting mass calculations 1)

What mass of hydrogen is produced when 192 g of magnesium is reacted with hydrochloric acid?

Mg + 2 HCl ® MgCl2 + H2

2)

3)

4)

(3)

What mass of oxygen is needed to react with 8.5 g of hydrogen sulphide (H2S)?

2 H2S + 3 O2 ® 2 SO2 + 2 H2O

(3)

What mass of potassium oxide is formed when 7.8 g of potassium is burned in oxygen?

4 K + O2 ® 2 K2O

(3)

Railway lines are welded together by the Thermitt reaction, which produces molten iron. What mass of iron is formed from 1 kg of iron oxide?

Fe2O3 + 2 Al ® 2 Fe + Al2O3

5)

(3)

What mass of oxygen is required to oxidise 10 g of ammonia to NO? 4 NH3 + 5 O2 ® 4 NO + 6 H2O

(3)

6)

What mass of aluminium oxide is produced when 135 g of aluminium is burned in oxygen?

(3) 2 Al + 3 O2 ® Al2O3

7)

What mass of iodine is produced when 7.1 g of chlorine reacts with excess potassium iodide?

Cl2 + 2 KI ® 2 KCl + I2 (3)

8)

What mass of hydrogen is needed to react with 32 g of copper oxide?

CuO + H2 ® Cu + H2O

9)

(3)

What mass of oxygen is formed when 735 g of potassium chlorate decomposes?

2 KClO3 ® 2 KCl + 3 O2

10)

11)

12)

13)

14)

(3)

What mass of hydrogen is produced when 195 mg of potassium is added to water?

2 K + 2 H2O ® 2 KOH + H2

(3)

How much calcium oxide is produced by heating 50 g of calcium carbonate?

CaCO3 ® CaO + CO2

(3)

What mass of magnesium oxide is formed when 6 g of magnesium reacts with oxygen?

2 Mg + O2 ® 2 MgO

(3)

What mass of carbon dioxide is produced when 5.6 g of butene (C4H 8) is burnt.

C4H8 + 6 O2 ® 4 CO2 + 4 H2O

(3)

The pollutant sulphur dioxide can removed from the air by reaction with calcium carbonate in the presence of oxygen. What mass of calcium carbonate is needed to remove 1 tonne of sulphur dioxide?

2 CaCO3 + 2 SO2 + O2 ® 2 CaSO4 + 2 CO2

(3)

15)

5.00 g of hydrated sodium sulphate crystals (Na2SO4.nH2O) gave 2.20 g of anhydrous sodium sulphate on heating to constant mass. Work out the relative molecular mass (Mr) of the hydrated sodium sulphate and the value of n.

Na2SO4.nH2O ® Na2SO4 + n H2O (3) (

16)

5.00 g of a mixture of MgSO4.7H2O and CuSO4.5H2O was heated at 120°C until a mixture of the anhydrous salts was formed, which weighed 3.00 g. Calculate the percentage by mass of MgSO4.7H2O in the mixture. (4)

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Quantitative Chemistry

Unit 3 – Lesson 4 Balancing Chemical Equations using Reacting Masses Name: ________________________ Example In a chemical reaction, 72g of magnesium (Mg) was reacted with exactly 48g of oxygen molecules (O2) to produce 120g of magnesium oxide (MgO).

Use the number of moles of reactants and products to write a balanced equation for the reaction.

To gain the marks you MUST show your working, not just the balanced equation.

Ar of Mg = 24 Mr of O2 = 32 Mr of MgO = 40

Amount of Mg = 72/Mr = 72/24 = 3 mol Amount of O2 = 48/Mr = 48/32 = 1.5 mol Amount of MgO = 120/Mr = 120/40 = 3 mol

Divide through by smallest

Mg = 2 mol O2 = 1 mol MgO = 2 mol

Balanced equation

2Mg + O2 à 2MgO

Practice questions 1) In a chemical reaction, 108g of aluminium (Al) reacts with 96g of oxygen molecules (O2) to make 204 g aluminium oxide (Al2O3).

Use the number of moles of reactants and products to write a balanced equation for the reaction.

To gain the marks you MUST show your working, not just the balanced equation.

Ar Al = __ Mr O2 = __ Mr Al2O3 = __

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Quantitative Chemistry

Amount of Al = mass/Mr = __mol Amount of O2 = mass/Mr = __mol Amount of Al2O3 = mass/Mr = __mol

Balanced equation:

2) In a chemical reaction, 26g of benzene (C6H6) reacts with 24g of chlorine molecules (Cl2) to make 37.5g of chlorobenzene (C6H5Cl) and 12g of hydrochloric acid (HCl).

Use the number of moles of reactants and products to write a balanced equation for the reaction.

To gain the marks you MUST show your working, not just the balanced equation.

Mr C6H6 = __ Mr Cl2 = __ Mr C6H5Cl = __ Mr HCl = __

Amount of C6H6 = mass/Mr = __ mol Amount of Cl2 = mass/Mr = __mol Amount of C6H5Cl = mass/Mr = __mol Amount of HCl = mass/Mr = __mol

Divide through by smallest ___ C6H6 = __ mol Cl2 = __ mol C6H5Cl = __ mol HCl = __ mol

Balanced equation:

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Quantitative Chemistry

3) In a chemical reaction, 36g of magnesium (Mg) reacts with 98g of phosphoric acid (H3PO4) to make 131g of magnesium phosphate (Mg3(PO4)2) and 3g of hydrogen (H2).

Use the number of moles of reactants and products to write a balanced equation for the reaction.

To gain the marks you MUST show your working, not just the balanced equation.

Mr Mg = __ Mr H3PO4 = __ Mr Mg3(PO4)2 = __ Mr H2 = __

Amount of Mg = mass/Mr = __ mol Amount of H3PO4 = mass/Mr = __ mol Amount of = mass/Mr = Mg3(PO4)2 __ mol Amount of H2 = mass/Mr = __ mol

Divide through by smallest ___ Mg = __ mol H3PO4 = __ mol Mg3(PO4)2 = __ mol H2 = __ mol

Balanced equation:

3Mg + 2H3PO4 à Mg3(PO4)2 + 3H2

4) In a chemical reaction, 33g of arsenic oxide (As2O3) reacts with 9g of water (H2O) to make 42g of arsenic hydroxide (As(OH)3).

Use the number of moles of reactants and products to write a balanced equation for the reaction.

To gain the marks you MUST show your working, not just the balanced equation.

Try this one without the helping hints…

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Quantitative Chemistry

FINDING REACTING RATIOS In each of the following, find the molar ratio in which the substances react. The first one has been done for you.

1

140 g of nitrogen (N2) reacts with 30 g of hydrogen (H2) to form ammonia.

Moles of N2 =

𝟏𝟒𝟎 𝟐𝟖

= 5.0 mol

Reacting ratio N2 : H2 = 5.0 : 15.0 =

Moles of H2 = 𝟓.𝟎 𝟓.𝟎

:

𝟏𝟓.𝟎 𝟓.𝟎

𝟑𝟎 𝟐

= 15 mol

= 1 : 3

 N2 + 3H2 →

2

5.8 g of butane (C4H10) reacts with 20.8 g of oxygen (O2) during complete combustion. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………………………………………….

3

11.6 g of tungsten oxide (WO3) reacts with 0.30 g of hydrogen (H2) during the extraction of tungsten. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………………………………………….

4

2.56 g of sulfur dioxide (SO2) reacts with 0.64 g of oxygen (O2) to form sulfur trioxide. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………………………………………….

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50 1100 Chemsheets GCSE

Quantitative Chemistry

In each of the following, find the molar ratio of all the substances in the equation to determine the balanced equation for the reaction.

5

0.920 g of silver carbonate (Ag2CO3) decomposes to form 0.773 g of silver oxide (Ag2O) and 0.147 g of carbon dioxide (CO2). …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………………………………………….

6

10.00 g of octadecane (C18H38) decomposes on heating in a cracking reaction to form 2.28 g of butane (C4H10), 3.31 g of propene (C3H6) and 4.41 g of ethene (C2H4). …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………………………………………….

7

4.00 g of potassium hydrogencarbonate (KHCO3) decomposes on heating to form 2.76 g of potassium carbonate (K2CO3), 0.88 g of carbon dioxide (CO2) and some water (H2O). …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. …………………………………………………………………………………………………………………………………………………………. ………………………………………………………………………………………………………………………………………………………….

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51 1100 Chemsheets GCSE

Quantitative Chemistry

LIMITING REAGENTS 1 In each example one of the reactants is in excess. Work out how many moles of the products are formed in each case.

CaO

1

+

H2O

a)

2 mol

3 mol

b)

10 mol

8 mol

c)

0.40 mol

0.50 mol

2Ca

2

+

O2

a)

2 mol

2 mol

b)

10 mol

2 mol

c)

0.50 mol

0.20 mol

2Fe

3

+

3Cl2

a)

3 mol

3 mol

b)

12 mol

15 mol

c)

20 mol

40 mol

TiCl4

4

+

4Na

a)

4 mol

4 mol

b)

2 mol

10 mol

c)

0.5 mol

1 mol

C2H5OH

5

+

3O2

a)

15 mol

30 mol

b)

0.25 mol

1 mol

c)

3 mol

6 mol

N2

6

+

3H2

a)

3 mol

6 mol

b)

0.5 mol

0.9 mol

c)

6 mol

20 mol

4K

7

+

O2

a)

10 mol

2 mol

b)

6 mol

4 mol

c)

0.50 mol

0.20 mol

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Mr Singh 25-October-2016



Ca(OH)2



2CaO



2FeCl3



Ti

+

4NaCl



2CO2

+

3H2O



2NH3



2K2O

52 1101 Chemsheets GCSE

Quantitative Chemistry

LIMITING REAGENTS 2 1

What mass of calcium hydroxide is formed when 10.0 g of calcium oxide reacts with 10.0 g of water?

CaO + H2O  Ca(OH)2

…………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………..

2

What mass of magnesium bromide is formed when 1.00 g of magnesium reacts with 5.00 g of bromine?

Mg + Br2  MgBr2

…………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………..

3

What mass of copper is formed when 2.00 g of copper(II) oxide reacts with 1.00 g of hydrogen?

CuO + H2  Cu + H2O

…………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………..

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53 1102 Chemsheets GCSE

Quantitative Chemistry

4

What mass of sodium fluoride is formed when 2.30 g of sodium reacts with 2.85 g of fluorine?

2Na + F2  2NaF

…………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………..

5

What mass of iron is formed when 8.00 g of iron(III) oxide reacts with 2.16 g of aluminium?

Fe2O3 + 2Al  2Fe + Al2O3

…………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………..

6

What mass of aluminium chloride is formed when 13.5 g of aluminium reacts with 42.6 g of chlorine?

2Al + 3Cl2  2AlCl3

…………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. …………………………………………………………………………………………………………………………………………….. ……………………………………………………………………………………………………………………………………………..

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54 1102 Chemsheets GCSE

Quantitative Chemistry

Concentration Calculations 1.

What is meant by the concentration of a solution?

2.

How many moles of solute are in: a. 500cm3 of solution, concentration 2mol/dm3

3.

4.

b.

200 cm3 of solution, concentration 0.5 mol/dm3

c.

20 cm3 of solution, concentration 0.4 mol/dm3

What is the Molarity of a solution containing: a. 4 moles in 2 dm3 of solution

b.

0.5 moles in 0.1 dm3 of solution

c.

3 moles in 200cm3 of solution

What volume of a. a 4 mol/dm3solution contains 2 moles?

b.

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a 6 mol/dm3 solution contain 0.03 moles?

Mr Singh

55

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3

3

Mr Singh

=

3

volume (dm )

moles

= vol in cm 1000

conc

3

vol (dm3)

3

3

3

3

3

3

5)

24-October-2016

…………………………………………………………………………………………………

c) 50 cm of 0.10 mol/dm H2SO4 ….………………………………………………………..

3

…………………………………………………………………………………………………

b) 25 cm of 1.50 mol/dm KOH ……………………………………………………………..

3

…………………………………………………………………………………………………

a) 100 cm of 0.20 mol/dm HNO3 …………………………………………………………..

3

Calculate the number of moles in the following solutions.

…………………………………………………………………………………………………

c) 0.020 moles of NaOH in 25 cm ………………………………………………………...

3

…………………………………………………………………………………………………

3

56

Chemsheets GCSE 1104

……………………………………………………………………………………………...

b) Calculate the concentration in mol/dm . ………………………………………….…..

3

a) Calculate the concentration in g/dm . ………………………………………………..

3

5.0 g of KNO3 is dissolved in 100 cm of water.

……………………………………………………………………………………………...

3

b) Calculate the concentration in g/dm . ………………………………………….……..

3

a) Calculate the concentration in mol/dm . ……………………………………………..

3

0.20 moles of NaOH is dissolved in 250 cm of water.

………………………………………………………………………………………………

c) 1.50 mol/dm HNO3 ………………………...…………………………………………...

3

………………………………………………………………………………………………

3

b) 0.20 moles of H2SO4 in 100 cm ………………………………………………………...

3

………………………………………………………………………………………………

a) 0.100 mol/dm NaOH …………………………………………………………………...

3

Calculate the concentration of the following solutions in g/dm .

3

3

Concentration = 2 x 98 =196 g/dm

Mr of H2SO4 = 98

Concentration = 2 mol/dm

conc (g/dm ) = conc (mol/dm ) x Mr

2 moles of H2SO4 196 g of H2SO4

1 dm3

b) 0.250 mol/dm CH3COOH ……………………………………………………………...

4)

3)

A simple conversion is:

Concentration can also be measured in grams per cubic 3 decimetre (g/dm ). This is a measure of the number of grams in one cubic decimetre. [remember that mass = Mr x moles]

…………………………………………………………………………………………………

a) 0.10 moles of NaCl in 200 cm ………………………………………………………….

3

3

moles

vol in dm

Calculate the concentration of the following solutions in mol/dm .

© www.CHEMSHEETS.co.uk

2)

1)

concentration (mol/dm3)

The volume must be in dm (there are 1000 cm in 1 dm ).

3

The concentration of a solution is usually measured in moles per cubic decimetre 3 (mol/dm ). This is a measure of the number of moles in one cubic decimetre.

CONCENTRATION OF SOLUTIONS

3

Quantitative Chemistry

Quantitative Chemistry

Taking it further Why not try some of the mole calculations required for A-level by reading about them in an Alevel textbook?

Why not do some internet research into the man who discovered the ‘mole’ concept and find out how his ideas came to him. Why is a chemical concept named after an animal?

Books to read: The Chemistry Maths Book, Erich Steiner Calculations for AS/A-level Chemistry, Jim Clark Maths Skills for A Level Chemistry, Dan McGowan and Emma Poole The Universe and the Teacup, K.C. Coles Life’s Other Secret, Ian Stewart Documentaries to watch The Great Math Mystery, Nova Dangerous Knowledge, BBC Podcasts www.gscepod.com thenakedscientists.com

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Quantitative Chemistry Revision Questions 20 Marks 1.

The symbol equation below shows the reduction of iron (III) oxide by carbon monoxide.

Fe2O3

+

3CO

à

2Fe

+

3CO2

What percentage (%) of iron (III) oxide is iron? ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

Answer ........................................ % (Total 2 marks)

2.

The information on the Data Sheet will be helpful in answering this question. (a)

Calculate the formula mass (Mr) of the compound iron (III) oxide, Fe2O3.

(Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

(3)

(b)

Calculate the mass of iron produced when 32g of iron (III) oxide is completely reduced by aluminium.

The reaction is shown in the symbol equation: Fe2O3

Bradfield College 9/5 8/30

+

2Al

à

Mr Singh

2Fe

+

Al2O3

56 58

1

Quantitative Chemistry

(Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

Answer

=

..................................... grams (3) (Total 6 marks)

3.

Cosmetic powders were widely used in ancient Egypt.

Cosmetic powders that may have been used in face paints have been analysed. These powders contained compounds that are rare in nature. The compounds must have been made by the ancient Egyptians using chemical reactions.

(a)

One of these compounds is called phosgenite.

Analysis of this compound shows that it contains: 76.0% lead (Pb)

13.0% chlorine (Cl)

2.2% carbon (C)

8.8% oxygen (O)

Calculate the empirical formula of this compound.

To gain full marks you must show all your working.

Relative atomic masses: C = 12 ; O = 16 ; Cl = 35.5 ; Pb = 207 ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ................................................................................................................................... ...................................................................................................................................

(4)

(b)

Another compound that the ancient Egyptians used is laurionite.

Bradfield College 9/5 8/30

Mr Singh

57 59

2

Quantitative Chemistry

The reaction used to make laurionite can be represented by this equation: PbO(s) + NaCl(aq) + H2O(1) ÂŽ PbOHCl(s) + NaOH(aq) laurionite

(i)

Explain why the pH of the solution increases as the reaction takes place. .......................................................................................................................... ..........................................................................................................................

(1)

(ii)

How could laurionite be separated from the other product when the reaction is complete? .......................................................................................................................... ..........................................................................................................................

(1) (Total 6 marks)

4.

Aspirin tablets have important medical uses.

A student carried out an experiment to make aspirin. The method is given below.

1.

2.

3.

4.

5.

6.

7.

8.

(a)

Weigh 2.00 g of salicylic acid.

Add 4 cm3 of ethanoic anhydride (an excess).

Add 5 drops of concentrated sulfuric acid.

Warm the mixture for 15 minutes.

Add ice cold water to remove the excess ethanoic anhydride.

Cool the mixture until a precipitate of aspirin is formed.

Collect the precipitate and wash it with cold water.

The precipitate of aspirin is dried and weighed.

The equation for this reaction is shown below.

C 7H 6O 3 + salicylic acid

Bradfield College 9/5 8/30

C 4H 6O 3

â&#x2020;&#x2019;

C 9H 8O 4 + aspirin Mr Singh

CH3COOH

58 60

3

Quantitative Chemistry

Calculate the maximum mass of aspirin that could be made from 2.00 g of salicylic acid.

The relative formula mass (Mr) of salicylic acid, C7H6O3, is 138 The relative formula mass (Mr) of aspirin, C9H8O4, is 180 ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

Maximum mass of aspirin = .............................. g (2)

(b)

The student made 1.10 g of aspirin from 2.00 g of salicylic acid.

Calculate the percentage yield of aspirin for this experiment.

(If you did not answer part (a), assume that the maximum mass of aspirin that can be made from 2.00 g of salicylic acid is 2.50 g. This is not the correct answer to part (a).) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

Percentage yield of aspirin = .............................. % (2)

(c)

Suggest one possible reason why this method does not give the maximum amount of aspirin. ..................................................................................................................................... .....................................................................................................................................

(1)

(d)

Concentrated sulfuric acid is a catalyst in this reaction.

Suggest how the use of a catalyst might reduce costs in the industrial production of aspirin. ..................................................................................................................................... .....................................................................................................................................

(1)

Bradfield College 9/5 8/30

Mr Singh

59 61

4

Quantitative Chemistry

Simple Moles Revision Simple Moles A ‘mole’ can be thought of a just a very big number.

6.02 ×1023 It is also (roughly) the number of hydrogen atoms in a 1g sample of hydrogen. As atoms are so small a tiny mass of a substance contains a vast number of atoms.

1-H

12-C

24-Mg

32-S

Mass Atoms Moles

You can see that by knowing the mass of a sample (and the relative atomic mass) you can ‘count’ of the number of atoms present

Moles = Mass/Relative Atomic Mass Put simply the relative atomic mass (Ar) is the top number of an element as featured in the periodic table.

Calculate the moles in…

9/5 8/30

(i)

2.4g of carbon

(ii)

710g of chlorine

(iii)

84g of Magnesium

Mr Singh

60 62

Quantitative Chemistry

Calculating relative atomic mass Isotopes: â&#x20AC;&#x153;Isotopes are substances with the same proton number (an identical number of protons) but a different number of neutronsâ&#x20AC;?

Relative Atomic Mass (Ar) Obviously we have elements whose atoms may have differing masses. What is useful is if we calculate an average mass of the atoms involved.

Let us consider Chlorine

Relative atomic mass = (Mass of Isotope A x abundance) + (Mass of isotope B x abundance)..

100

Another example can be seen with the next element in group VII Bromine: Bromine exists as two isotopes. Br-79 (50% abundance) & Br-81 (50% abundance)

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Mr Singh

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Quantitative Chemistry

Questions: Calculating Relative atomic masses.

1. Calculate the RAM (Ar) of Boron Boron-10

Boron-11

19.9%

80.1%

2. Calculate the RAM (Ar) of Silicon

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Silicon-28

Silicon-29

95.23%

4.67%

Mr Singh

66 64

Quantitative Chemistry

Calculating Mr (relative formula mass) This is really easy. If 12-C is twelve times the mass of 1-H then the mass of molecules made from these atoms can be easily calculated.

Carbon dioxide is comprised of 12-C and 16-O Therefore Mr (it is the molecular mass) of CO2 is â&#x20AC;Ś.. (1 x 12C) + (2 x 16O) = 44 Calculate the Mr (relative formula masses of the following: (i)

H2SO4

(ii)

CaCO3

To calculate the number of moles in a given mass you use the equation: Moles = Mass / Mr Calculate the moles of (i)

9g of Water (H2O)

(ii)

110 g of Carbon dioxide (CO2)

(iii) 146.25g of Sodium chloride (NaCl)

(iv) 9.8g of Sulphuric acid (H2SO4)

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Mr Singh

67 65

Quantitative Chemistry

Moles 11 – General Calculations 1)

Calculate the percentage by mass of nitrogen in Al(NO3)3.

(2)

2) a) The molecular formula of some compounds is shown. What is the empirical formula of each one? i)

N2H4

ii) C6H6

iii) C5H12

(3)

b) What is the molecular formula of a compound with an Mr of 42 and the empirical formula CH2?

(1)

c) Work out the empirical formula of the following compound given the information about its composition by mass: O 0.240 g; C 0.060 g; H 0.005 g; K 0.195 g 3)

(2)

What mass of sodium is needed to reduce 1 kg of titanium chloride?

TiCl4(l) + 4 Na(s) ® Ti(s) + 4 NaCl(s)

4)

(3)

Calculate the volume of hydrogen formed when 1 g of sodium reacts with water.

2 Na(s) + 2 H2O(l) ® 2 NaOH(aq) + H2(g)

5)

Calculate the mass of 0.2 mol/dm3 hydrochloric acid.

2 HCl(aq) + CaCO3(s)

6)

calcium

carbonate

(3)

that

® CaCl2(aq) + H2O(l) + CO2(g)

reacts

with

100

cm3

of

(3)

Vinegar contains the weak acid ethanoic acid. Calculate the concentration of ethanoic acid in vinegar given that 25 cm3 of vinegar reacted with 27.3 cm3 of 1.0 mol/dm3 sodium hydroxide solution in a titration.

CH3COOH(aq) + NaOH(aq)

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® CH3COONa(aq) + H2O(l)

Mr Singh

(3)

66

Quantitative Chemistry

Moles 12 – General Calculations 1) a) Calculate the Mr of the following: i) Br2

ii) Zn(OH)2

iii) Na2SO4

b) Calculate the percentage by mass of oxygen in Zn(OH)2.

(3) (1)

2) a) The molecular formula of some compounds is shown. What is the empirical formula of each one? i)

C9H18

ii)

B4H10

iii)

SO3

(3)

b) Work out the molecular formula of the following compounds given the information below? i)

empirical formula = P2O5

ii) empirical formula = CH2

Mr = 284 Mr = 56

(2)

c) Work out the empirical formula of the following compound given the information about its composition by (3) mass: K 44.8%, S 18.4%, O 36.8%.

3)

What mass of potassium oxide is formed when 2 g of potassium is burned in oxygen?

4 K(s) + O2(g) ® 2 K2O(s)

4)

(3)

What volume of hydrogen is produced when 5.4 tonnes of aluminium reacts with excess hydrochloric acid?

2 Al(s) + 6 HCl(aq) ® 2 AlCl3(aq) + 3 H2(g)

5)

Calculate the concentration of sodium hydroxide solution given that 25 cm3 of it reacts with 18.2 cm3 of 0.100 mol/dm3 sulphuric acid.

H2SO4(aq) + 2 NaOH(aq)

6)

(3)

® Na2SO4(aq) + 2 H2O(l)

(3)

What volume of 0.01 mol/dm3 silver nitrate solution reacts with 0.1 g of copper?

2 AgNO3(aq) + Cu(s)

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® Cu(NO3)2(aq) + 2 Ag(s)

Mr Singh

(3)

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Quantitative Chemistry

Moles 15 – General Calculations 1) a) Calculate the Mr of:

i) Cl2

ii) Mg(NO3)2

iii) Na2CO3.10H2O

b) Calculate the percentage of oxygen in sulphuric acid, H2SO4.

(3) (1)

2)

Calculate the empirical formula of the compound found to contain, by mass, 35% nitrogen, 5% hydrogen and (2) 60% oxygen.

3)

1.05 g of iron reacts with oxygen to form 1.50 g of iron oxide. Find the empirical formula for the iron oxide.

(2)

4)

What mass of aluminium oxide is produced when 54 g of aluminium reacts with excess oxygen?

4 Al(s) + 3 O2(g)

5)

® 2 Al2O3(s)

(3)

Iron reacts with excess steam as shown.

3 Fe(s) + 4 H2O(g) ® Fe3O4(s) + 4 H2(g) a) What mass of Fe3O4 is produced when 56 tonnes of iron reacts with excess steam?

(3)

b) What volume of hydrogen (at room temperature and pressure) is produced in this reaction?

(2)

6)

25.0 cm3 of an acid H2A reacts with 23.45 cm3 of 0.1 mol/dm3 sodium hydroxide solution. concentration of H2A. H2A(aq) + 2 NaOH(aq) ® 2 Na2A(aq) + 2 H2O(l)

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Mr Singh

Find the

(3)

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Quantitative Chemistry

Moles 17 – General Calculations 1) a) Calculate the Mr of:

i) Br2

ii) K2CO3

iii) (NH4)2SO4

b) Calculate the percentage of oxygen in K2CO3.

(3) (1)

2) a) Define the terms empirical formula and molecular formula.

(2)

b) A hydrocarbon was found to contain 82.8% by mass of carbon. It has an Mr of 58. Find the empirical and (3) molecular formulas of this compound. c) 1 g of sulphur was burned forming 2.5 g of an oxide. Find the empirical formula of the oxide.

(2)

3)

What mass of calcium oxide is formed from the decomposition of 180 g of calcium carbonate?

CaCO3(s) ® CaO(s) + CO2(g)

4)

(3)

What volume of carbon dioxide is produced when 10 g of ethene (C2H4) is burnt in oxygen?

C2H4(g) + 3 O2(g) ® 2 CO2(g) + 2 H2O(l)

(3)

5)

What mass of magnesium reacts with 50 cm3 of 2 mol/dm3 hydrochloric acid?

(3) Mg(s) + 2 HCl(aq) ® MgCl2(aq) + H2(g)

6)

Find the concentration of a solution of ethanoic acid given that 25.0 cm3 of the acid reacts with 20.5 cm3 of 1.0 mol/dm3 sodium hydroxide.

CH3COOH(aq) + NaOH(aq) ® CH3COONa(aq) + H2O(l)

7)

If 1.2 dm3 of chlorine is produced during the electrolysis of molten sodium chloride, what mass of sodium is formed?

Na+ + e- ® Na

8)

(3)

2 Cl- - 2 e- ® Cl2

(3)

What mass of aluminium is produced from the electrolysis of aluminium oxide if 100 kg of oxygen is formed?

Al3+ + 3 e- ® Al

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2 O2- - 4 e- ® O2

Mr Singh

(3)

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Quantitative Chemistry

Finding the Formula of Magnesium Oxide from Reacting Masses Formulae are written in numbers of moles of each element that have combined.

The number of moles of substance is simply equal to

mass of substance relative atomic mass

Method • Mass a clean crucible and lid together, recording the total mass.

• Cut a strip of magnesium ribbon of between 10 -20 cm length and clean it using sand paper. Loosely coil the ribbon and place it in the crucible.

• Find the mass of ribbon, crucible and lid together.

• Heat on a clay triangle and tripod, avoiding excessive heat.

• Using tongs, lift the lid occasionally to allow air in but try to avoid loss of oxide smoke.

Heat

• When the reaction is complete, allow to cool and find the new total mass.

• If time allows, heat again and re-mass to ensure that the reaction is completed.

Results Calculate the mass of

(i) the magnesium used (ii) the oxide formed (iii) the mass of

oxygen combined and then record your results on the board.

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Mr Singh

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Quantitative Chemistry

Analysis 1. From the class results, plot a graph of the mass of Oxygen (y-axis) which combined with the masses of Magnesium (x-axis).

2. Draw a ‘best fit’ line.

3. From the ‘best fit’ line find the mass of Oxygen that combines with 1.2 g of Mg 4. Divide the mass (1.2g) of Mg by its RAM to find the moles of Mg Divide the mass of oxygen by its RAM to find the moles of oxygen combining.

5. What should be the simplest whole number ratio?

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Mr Singh

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Quantitative Chemistry

Reducing Metal Oxides Using Hydrogen - Demonstration In this demonstration, you will see three reactions involving competition for oxygen. Can hydrogen take the oxygen away from all or any of the metal oxides?

Additionally, it is possible to calculate the empirical formula of copper oxide by weighing the vessel before and after the reaction. Write down the results and do the calculation:

Write down reasons why the empirical formula that you have calculated does not equal the expected formula of CuO.

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Mr Singh

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Quantitative Chemistry

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Mr Singh

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