Quantitative Chemistry 2 GCSE

GCSE Workbook Quantitative Part 2

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

4.3.3 Yield and atom economy of chemical reactions (chemistry only) 4.3.3.1 Percentage yield Content

Key opportunities for skills development

Even though no atoms are gained or lost in a chemical reaction, it is WS 4.2, 4.6 not always possible to obtain the calculated amount of a product MS 1a because: Recognise and use • the reaction may not go to completion because it is reversible expressions in decimal • some of the product may be lost when it is separated from the form. reaction mixture MS 1c • some of the reactants may react in ways different to the expected reaction.

Use ratios, fractions and percentages.

The amount of a product obtained is known as the yield. When compared with the maximum theoretical amount as a percentage, it MS 2a is called the percentage yield.

Use an appropriate number M ass o f product actuall y made % Y ield = M aximum theoretical mass o f product × 100 of significant figures.

Students should be able to:

MS 3b

• calculate the percentage yield of a product from the actual yield of a reaction • (HT only) calculate the theoretical mass of a product from a given mass of reactant and the balanced equation for the reaction.

Change the subject of an equation.

4.3.3.2 Atom economy Content

Key opportunities for skills development

The atom economy (atom utilisation) is a measure of the amount of starting materials that end up as useful products. It is important for sustainable development and for economic reasons to use reactions with high atom economy.

WS 4.2, 4.6 MS 1a

Recognise and use expressions in decimal The percentage atom economy of a reaction is calculated using the form. balanced equation for the reaction as follows: MS 1c Relative f ormula mass o f desired product f rom equation Sum o f relative f ormula masses o f all reactants f rom equation × 100 Use ratios, fractions and Students should be able to: percentages.

• calculate the atom economy of a reaction to form a desired MS 3b product from the balanced equation Change the subject of an • (HT only) explain why a particular reaction pathway is chosen to equation. produce a specified product given appropriate data such as atom economy (if not calculated), yield, rate, equilibrium position and usefulness of by-products.

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Quantitative Chemistry

4.3.4 Using concentrations of solutions in mol/dm3 (chemistry only) (HT only) Content

Key opportunities for skills development

The concentration of a solution can be measured in mol/dm3.

WS 4.2, 4.3, 4.6

The amount in moles of solute or the mass in grams of solute in a given volume of solution can be calculated from its concentration in mol/dm3.

MS 1a

Recognise and use expressions in decimal If the volumes of two solutions that react completely are known and form. the concentration of one solution is known, the concentration of the MS 1c other solution can be calculated.

Use ratios, fractions and Students should be able to explain how the concentration of a percentages. solution in mol/dm3 is related to the mass of the solute and the volume of the solution.

MS 3b Change the subject of an equation.

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

AT 1, 3, 8 Opportunities within titrations including to determine concentrations of strong acids and alkalis.

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

4.3.5 Use of amount of substance in relation to volumes of gases (chemistry only) (HT only) Content

Key opportunities for skills development

Equal amounts in moles of gases occupy the same volume under the same conditions of temperature and pressure.

WS 1.2, 4.1, 4.2, 4.3, 4.6

The volume of one mole of any gas at room temperature and pressure (20oC and 1 atmosphere pressure) is 24 dm3.

The volumes of gaseous reactants and products can be calculated from the balanced equation for the reaction.

Students should be able to: • calculate the volume of a gas at room temperature and pressure from its mass and relative formula mass • calculate volumes of gaseous reactants and products from a balanced equation and a given volume of a gaseous reactant or product • change the subject of a mathematical equation.

MS 1a Recognise and use expressions in decimal form.

MS 1c Use ratios, fractions and percentages.

MS 3b 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

Percentage Yield Calculations The Yield is the amount of product you obtain from a reaction. The greater the yield, the better the reaction!

The Yield is defined as the mass that was actually made in the reaction as a percentage of the mass one could theoretically make from the amount of reactant used. !"#$"%&'(" !"#$% =

!"#$%& !"## !" !"#$%&' !"#$%&'( ×100 !ℎ!"#!$%&'( !"## !" !"#$%&' !"#!$#"%&'

Or my succinctly % !"#$% =

!"!"#$ !"## ×100 !ℎ!"#!$%&'( !"##

For example, a certain student calculates that she should obtain 7.0g from an experiment but her product only weighs 6.3g. What is her percentage yield? % !"#$% =

6.3! !"#$!% !"## ×100 = ×100 = 90% 7.0! !ℎ!"#!$%&!" !"##

More often we have to calculate the theoretical yield ourselves from the mass of reactant and using the equation.

For example, when 6.4g of Copper were heated in air, 7.6g of copper(II) oxide, CuO, were obtained.

2!! ! + !! ! → 2!"#(!)

Calculate the mass of copper(II) oxide formed and from this the percentage yield.

1. Turn the mass of copper into moles of copper !=

! !!

, Ar (Cu) = 64

6.4! = 0.1 !"#$% 64!/!"# 2. From the equation we can see that copper and copper oxide react in a ration of 1 to 1. !=

1 mole of Copper gives 1 mole of copper(II) oxide Therefore 0.1 moles of copper give 0.1 moles of copper(II) oxide 3. Using this we can then work out the mass of copper(II) oxide ! = !!! , Mr (CuO) = 64 + 16 = 80g/mol ! = 0.1 !"# ×80!/!"# = 8!

4. Calculate the Percentage Yield % !"#$% =

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7.6! !"#$!% = ×100 = 95% 8.0! !ℎ!"#!$%&'(

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Quantitative Chemistry

Atom Economy A reaction sometimes not only produces the products we desire, but also products we don’t want, called by-products.

They are often considered waste and need to be disposed of which is costly and posses potential environmental problems.

Sometimes by-products may be sold or used elsewhere in the chemical plant in an attempt to use all the resources wisely, minimise waste and therefore make the process more cost effective.

The atom economy of a reaction is a measure of the extent to which the atoms in the starting materials end up in the desired products. đ?‘Žđ?‘Ąđ?‘œđ?‘š đ?‘’đ?‘?đ?‘œđ?‘›đ?‘œđ?‘šđ?‘Ś =

đ?‘&#x;đ?‘’đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘Łđ?‘’ đ?‘“đ?‘œđ?‘&#x;đ?‘šđ?‘˘đ?‘™đ?‘Ž đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘œđ?‘“ đ?‘Ąâ„Žđ?‘’ đ?’…đ?’†đ?’”đ?’Šđ?’“đ?’†đ?’… đ?‘?đ?‘&#x;đ?‘œđ?‘‘đ?‘˘đ?‘?đ?‘Ą đ?‘–đ?‘› đ?‘Ąâ„Žđ?‘’ đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘› Ă—100 đ?‘ đ?‘˘đ?‘š đ?‘œđ?‘“ đ?‘&#x;đ?‘’đ?‘™đ?‘Žđ?‘Ąđ?‘–đ?‘Łđ?‘’ đ?‘“đ?‘œđ?‘&#x;đ?‘šđ?‘˘đ?‘™đ?‘Ž đ?‘šđ?‘Žđ?‘ đ?‘ đ?‘’đ?‘ đ?‘œđ?‘“ đ?’•đ?’‰đ?’† đ?’“đ?’†đ?’‚đ?’„đ?’•đ?’‚đ?’?đ?’•đ?’” đ?‘“đ?‘&#x;đ?‘œđ?‘š đ?‘’đ?‘žđ?‘˘đ?‘Žđ?‘Ąđ?‘–đ?‘œđ?‘›

Atom economy describes the efficiency of a reaction in terms of all the atoms involved. A reaction with a high atom economy uses atoms with minimal waste.

The type of reaction used for chemical processes is a major factor in achieving a higher atom economy. Addition reactions all have an atom economy of 100%. Reactions involving elimination or substitution have an atom economy of less than 100%.

To improve the atom economy we must find a use for all the products of a reaction For example, what is the atom economy for making hydrogen by reacting coal with steam?

Write the balanced equation: đ??ś(đ?‘ ) + 2đ??ť) đ?‘‚(đ?‘”) → đ??śđ?‘‚) (đ?‘”) + 2đ??ť) (đ?‘”)

Write out the Ar and Mr values underneath: đ??ś(đ?‘ ) + 2đ??ť) đ?‘‚(đ?‘”) → đ??śđ?‘‚) (đ?‘”) + 2đ??ť) (đ?‘”) 12

2 Ă— 18

44

2Ă—2

Remember that the Ar or Mr in grams is one mole, so: total mass of products = 44 + 4 = 48g (note that this is the same as the reactants: 12 + 36 = 48g) mass of desired product (H2) = 4g

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Quantitative Chemistry

% atom economy = 4â „48 Ă— 100 = 8.3% This means the majority of the starting material becomes waste even if the process at a 100% yield.

Percentage yield and atom economy are different Percentage yield tell you the efficiency of converting reactants into products Atoms economy tells you the proportion of desired products compared with all the products formed.

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Quantitative Chemistry

Concentration Calculations A solution is made up of two components, a solid ………………………. and a liquid …………………………… 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. !"#$%& !" !"#$%&(!"#) =

!"#$%#&'(&)"# !" !"#$%&"' !"# !! !! ×!"#$%& !" !"#$%&'(!! ! ) 1000

More simply the equation is !"#$% =

!"#$%#&'(&)"# ×!"#$%& 1000

Abbreviated to

!=

!"

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

Calculations involving chemicals in solution These are often referred to as Volumetric Analysis. The name should not worry you, the basis of the calculations is the same as all the rest ie moles and equations.

Many reactions take place in solution involving solutions of known concentration.

Concentration in solution is generally measured as moles per 1000 cm3 of solution. For example the sodium chloride on the bench may be labelled as 1M NaCl. This means that each 1000 cm3 of the solution contains 1 Mole of NaCl (58.5 g).

It does not mean that 58.5 g of NaCl have been added to 1000 cm3 of water.

The solution will have been made up by measuring out 58.5 g of the solid, dissolving it in about 500 cm3 of water and then adding water to make the total volume of the mixture up to 1000 cm3. (1 dm3) Concentration in mol dm-3 is called molarity.

molarity =

concentration in grams per 1000 cm 3 M r for the material dissolved

molarity × volume (cm 3 ) number of moles of material in a given volume = 1000

mass of material in a given volume of solution =

molarity × volume (cm 3 ) × M r 1000

In reactions in solution it is often more convenient to use molarity rather than g dm–3.

There are two ways you can approach calculations involving solutions. The first method (A) detailed below is really a short cut way of using the more detailed method B. Most of the straight forward calculations you will meet at the start of your course and the ones in this booklet can be carried through using method A.

UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) 8/30Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue Mr3Singh – October 2004

97 8

Quantitative Chemistry

Method A Consider the following reaction between two solutions aA(aq)

+

→

bB(aq)

cC(aq)

+

dD(aq)

In this reaction a moles of substance A react with b moles of substance B Let us suppose that Va cm3 of the solution of A react with Vb cm3 of the solution of B. If this is an acid/alkali reaction we could find these volumes out using an indicator.

∴ Number of moles of A in Va cm3 of solution

=

Va M a 1000

=

a

∴ Number of moles of B in Vb cm3 of solution

=

Vb M b 1000

=

b

∴ If we divide equation (i) by equation (ii) we get

Va M a a = Vb M b b

This relationship will hold good for any reaction between two solutions.

Examples 1

2NaOH(aq)

+

H2SO4(aq)

→

Na2SO4(aq)

+

2H2O(l)

→

BaSO4(s)

+

2HCl(aq)

M NaOH × VNaOH 2 = M H 2SO 4 × VH 2SO 4 1

2

BaCl2(aq)

+

H2SO4(aq)

M BaCl2 × VBaCl2 M H 2SO 4 × VH 2SO 4

3

MnO4-(aq)

+ 5Fe2+(aq)

M Fe 2 + × VFe 2 +

1 1

+ 8H+(aq)

M MnO − × VMnO − 4

=

4

=

→ Mn2+(aq)

+ 5Fe3+(aq)

+ 4H2O(l)

1 5

UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) 8/30Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue Mr3Singh – October 2004

99 9

Quantitative Chemistry

Calculation examples 1

What is the molarity of a solution of NaOH which contains 4 g of NaOH in 250 cm3 of solution?

Mr NaOH = 40 g mol–1 4 g per 250 cm3 = 16 g per 1000 cm3

∴ molarity

16 40

=

=

0.040 mol dm-3

This can also be written molarity

2

( 4 ×1000) ×1 250 × 40

=

0.040 mol dm-3

=

What mass of KMnO4 would be needed to prepare 250 cm3 of 0.020 mol dm–3 KMnO4 solution? (Mr = 158) 1000 cm3 of 0.020 mol dm–3 KMnO4 will need 158 x 0.02 g

158 × 0.02 × 250 1000

∴ 250 cm3 will need

3

0.79 g

How many moles of H2SO4 will be contained in 25 cm3 of 0.10 mol dm–3 H2SO4?

0.10 × 25 1000

number of moles

4

=

=

0.0025 moles

25 cm3 of 0.10mol dm–3 NaOH react with 50 cm3 of a solution of H2SO4. What is the molarity of the H2SO4?

2NaOH(aq)

+

H2SO4(aq)

→ Na2SO4(aq)

+

2H2O(l)

M NaOH × VNaOH 2 = M H 2SO 4 × VH 2SO 4 1 ∴

∴

0.1× 25 =2 M H 2SO 4 × 50

M H 2SO 4 =

0.1× 25 2 × 50

=

0.025 mol dm-3

NB If you are required to calculate the concentration in g dm–3 at this stage you need to multiply by the Mr of the material. In this case 98 g mol–1 100 8/30

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Quantitative Chemistry

5

What volume of 0.02 mol dm–3 KMnO4 solution will be needed to react with 25 cm3 of 0.1 mol dm–3 Iron(II) ammonium sulphate ?

NB in Iron(II) ammonium sulphate only the iron(II) ions react with the manganate(VII) ions MnO4-(aq)

+

5Fe2+(aq)

+ 8H+(aq)

M MnO − × VMnO − 4

4

M Fe 2 + × VFe 2 + ∴

0.02 × VMnO − 4

0.1× 25

∴

6

VMnO − = 4

=

=

→ Mn2+(aq)

+ 5Fe3+(aq)

+ 4H2O(l)

1 5

1 5

0.1 × 25 = 25 cm 3 0.02 × 5

25 cm3 of a solution of 0.05 mol dm–3 silver nitrate react with 10 cm3 of a solution of NaCl. What is the concentration of NaCl in g dm–3 in the solution?

NaCl(aq)

+

→ NaNO3(aq)

AgNO3(aq)

+

AgCl(s)

M NaCl × VNaCl 1 = M AgNO3 × VAgNO3 1

∴ ∴

∴

10 × M NaCl 1 = 25 × 0.05 1 M NaCl =

25 × 0.05 = 0.125 mol dm −3 10

concentration of NaCl = 0.125 mol dm −3 × 58.5 = 7.31 g dm −3

UA008883 – Workbook for GCE students – Moles, Formulae and Equations Edexcel Advanced GCE in Chemistry (9080) 8/30Edexcel Advanced GCE in Chemistry (Nuffield) (9086) – Issue Mr3Singh – October 2004

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Quantitative Chemistry

7

In the reaction between an acid HxA and 0.1 mol dm–3 NaOH solution. 25 cm3 of a solution of 0.1 mol dm–3 HxA react with 50 cm3 of the 0.1 mol dm–3 NaOH. What is the value of x?

This is not as difficult as it looks. You need to think what the equation for the reaction would be.

HxA(aq)

+

∴

xNaOH(aq)

M H x A × VH x A M NaOH × VNaOH

∴

=

→ NaxA (aq)

+

xH2O(l)

1 x

25 × 0.1 1 = 50 × 0.1 x ∴x=2

Thus the acid is H2A.

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Quantitative Chemistry

Molar Gas Volume The volume of a gas depends on its temperature and pressure. As a gas is heated it expands and when pressurised, tries to take up a larger volume.

The same amount of all gases occupy the same volume when at the same temperature and pressure.

The size of the molecule has no affect on the volume of the gas. So Oxygen , Hydrogen, Carbon Dioxide or Sulfur Dioxide will all take up the same volume under the same conditions.

At Standard Temperature and Pressure (stp) 1 Mole of any gas will take up 24.0 dm3 or 24,000cm3. We call this the Molar Gas Volume. VM

!! = 24 !!!

Standard Temperature and Pressure is 298K (25ËšC) 100kPa (1 atm).

Using this information we can calculate how much of a gas there is from knowing its volume

!"#$%& ! = !"#$%& !" !"#$% !" !"# ! Ă—!"#$% !"# !"#$%&(!! )

! = !!!

For example, a reaction produces 100cm3 of hydrogen at stp. How many moles of H2 is this?

!=

!

100!!! = = 0.00417!"# !! 24000!!! /!"#

Or the other way round. What volume does 22g of carbon dioxide occupy at rtp?

Mr(CO2) = 44g/mol

So, 44g = 1 mole

Therefore, 22g = 0.5 moles,

! = !!! Therefore the volume occupied = 0.5 x 24dm3 = 12dm3

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Quantitative Chemistry

Percent Yield Worksheet 1)

!" + 2 !"# à !"#!! + !!

My theoretical yield of beryllium chloride was 10.7 grams. If my actual yield was 4.5 grams, what was my percent yield?

Percentage Yield Equation:

x 100

Actual mass produced:………………………..g Theoretical Mass produced:………………………g

Percentage Yield:………………………………

2)

!"#$ + !"#

à

!"#$

+

!"#

a) I began this reaction with 20 grams of lithium hydroxide. What is my theoretical yield of lithium chloride?

Percentage Yield Equation:

x 100

Theoretical Mass: LiOH

KCl

LiCl

KOH

Mass(g) RMM(g/mol) Moles (mol) Ratio

Theoretical Mass produced:………………………g b) I actually produced 6 grams of lithium chloride. What is my percent yield?

Actual mass produced:………………………..g

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Quantitative Chemistry

Theoretical Mass produced:………………………g Percentage Yield:………………………………

3)

!! !!

+

5 !! à

3 !!!

+

4 !! !

a) If I start with 5 grams of C3H8, what is my theoretical yield of water?

Theoretical Mass: C 3H 8

5O2

3CO2

4H2O

Mass(g) RMM(g/mol) Moles (mol) Ratio

Theoretical Mass produced:

b) I got a percent yield of 75% How many grams of water did I make?

Percentage Yield Equation:

x 100

Mass of water…………………….g

Percent Yield Worksheet 1)

Write the equation for the reaction of iron (III) phosphate with sodium sulfate to make iron (III) sulfate and sodium phosphate.

2 FePO4 + 3 Na2SO4 à 1 Fe2(SO4)3 + 2 Na3PO4 2)

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If I perform this reaction with 25 grams of iron (III) phosphate and an excess of sodium sulfate, how many grams of iron (III) sulfate can I make?

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Quantitative Chemistry

1)

3 H2 + N2 → 2 NH3

Ammonia is made by reacting hydrogen with nitrogen.

a) Calculate the mass of ammonia that can be formed from 12 g of hydrogen. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. b) 20 g of ammonia was formed in this reaction. Calculate the percentage yield.. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. c) Give three potential reasons why the percentage yield was less than 100%.

1 - ??????????????????????????????????????????????????..

2 - ???????????????????????????????????????????????????

3 - ???????????????????????????????????????????????????

2)

Fe2O3 + 3 CO → 2 Fe + 3 CO2

Iron is made by reduction of iron oxide with carbon monoxide. a) Calculate the mass of iron that can be formed from 100 g of iron oxide.

????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. b) 65 g of iron was formed in this reaction. Calculate the percentage yield. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????..

3)

Chlorine can be made by the electrolysis of sodium chloride solution.

2 NaCl + 2 H2O → 2 NaOH + Cl2 + H2

a) Calculate the mass of chlorine that can be formed from 50 g of sodium chloride. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????.. b) 25 g of chlorine was formed in this reaction. Calculate the percentage yield. ????????????????????????????????????????????????????.. ????????????????????????????????????????????????????..

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Chemsheets GCSE 072 16

Quantitative Chemistry

4)

Chromium is a useful metal. It is extracted from chromium oxide by reaction with aluminium.

Cr2O3 + 2 Al → 2 Cr + Al2O3

a) Calculate the mass of chromium that can be formed from 1 kg of chromium oxide.

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB.. b) 600 g of chromium was formed in this reaction. Calculate the percentage yield.

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

5)

Titanium is made by the reaction of titanium chloride with sodium.

TiCl4 + 4 Na → Ti + 4 NaCl

a) Calculate the mass of titanium that can be formed from 10 kg of titanium chloride.

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB.. c) 1950 g of titanium was formed in this reaction. Calculate the percentage yield.

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB..

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Chemsheets GCSE 072 17

Quantitative Chemistry

Atom Economy Calculate the atom economy for the following reactions. The teacher will give you the equations.

1. The addition of bromine to ethene

2. Thermal decomposition of calcium carbonate for calcium oxide

3. Reduction of iron oxide by carbon for iron

4. Electroylsis of aluminium oxide for aluminium

5. Chlorination of methane

6. Hydrogenation of propene

7. Aqueous displacement reaction between copper(II) sulfate and zinc metal

8. Esterification of ethanol and propanoic acid

9. Chlorination of butanol

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18

Quantitative Chemistry

Molar Volume Calculations

Molar Gas volume = 24,000cm3/mol 1. The volume in cm3 at rtp of: (a) 2 moles of helium gas atoms, He

(b) 0.01 moles of oxygen gas molecules, O2

(c) 0.75 moles of carbon dioxide gas molecules, CO2.

2. The number of moles at rtp: (a) of ozone gas molecules, O3, in 240 cm3

(b) of hydrogen gas molecules, H2, in 96 cm3

(c)

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of argon gas atoms, Ar, in 480 cm3

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19

Quantitative Chemistry

General Calculations 1 1) a) Calculate the Mr of:

i) P4

ii) Cu(NO3)2

iii) Na2SO4.10H2O

b) Calculate the percentage of oxygen in Cu(NO3)2.

2)

(3)

(1)

Calculate the empirical formula of the compound found to contain 0.04 grams of oxygen and 0.32 grams of copper.

(2)

3) a) Calculate the empirical formula of the compound found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen.

(2) b) Find its molecular formula given that its Mr is 180.

4)

(1)

Calculate the volume of the following gases at room temperature and pressure: a) 0.15 moles of H2

b) 140 g of CO

(2)

5)

What mass of ammonia is produced when 100 g of hydrogen reacts with an excess of nitrogen?

(3) N2(g) + 3 H2(g) → 2 NH3(g)

6)

Iron oxide reacts with carbon monoxide as shown: Fe2O3(s) + 3 CO(g) → 2 Fe(l) + 3 CO2(g) a) What mass of iron is made from 1 kg of iron oxide?

(3)

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

(2) 7)

The fertiliser ammonium nitrate is made by reaction of ammonia with nitric acid. What mass of ammonium nitrate can be made from 1 tonne of ammonia, providing there is an excess of nitric acid?

NH3(aq) + HNO3(aq) → NH4NO3(aq)

8)

(3)

What volume of 2.0 mol/dm3 copper sulphate solution reacts with 1 g of magnesium?

(3) Mg(s) + CuSO4(aq) → MgSO4(aq) + Cu(s)

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20

Quantitative Chemistry

General Calculations 2 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)

C 6H 6

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)

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)

(2)

Calculate the mass of 0.2 mol/dm3 hydrochloric acid.

calcium

carbonate

(3) that

reacts

with

100

cm3

of

2 HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g)(3) 6)

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) → CH3COONa(aq) + H2O(l)

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Quantitative Chemistry

General Calculations 3 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 = P2O5Mr = 284 ii) empirical formula = CH2Mr = 56

(2)

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

(3) 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) → Na2SO4(aq) + 2 H2O(l)

6)

(3)

(3)

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

2 AgNO3(aq) + Cu(s) → Cu(NO3)2(aq) + 2 Ag(s)

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Mr Singh

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Quantitative Chemistry

Solution Calculations 4

Write a balanced chemical equation for the reaction (you are usually given this).

Write out the information given in the question under the equation beneath the appropriate chemical.

You are always given enough information to work out how many moles there are of one reactant, so work it out.

Using the chemical equation, find out how many moles of the other reactant this quantity reacts with.

Use this to then find whatever quantity the question asked you to.

You will need to know the following key equations: moles mass

=

concentration (mol/dm3) moles

=

volume (dm3)

Mr mass moles x Mr

moles 3 conc x dm

Note that 1 litre = 1 dm3 = 1000 cm3 Concentration is usually measured in mol/dm3 (M or mol/litre) 1)

25.0 cm3 of a solution of sodium hydroxide solution required 21.50 cm3 of 0.100 mol/dm3 sulphuric acid for neutralisation. Find the concentration of the sodium hydroxide solution.

H2SO4(aq)

2)

2 NaOH(aq) →

Na2SO4(aq) +

2 H2O(l)

(3)

Find the volume of 1.0 mol/dm3 hydrochloric acid that reacts with 25.00 cm3 of 1.50 mol/dm3 sodium hydroxide.

HCl(aq)

3)

+

+

NaOH(aq)

→

NaCl(aq)

+ H2O(l)

(3)

25.0 cm3 of 0.100 mol/dm3 sodium hydroxide neutralises 19.0 cm3 of hydrochloric acid. Find the concentration of the acid.

HCl(aq)

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+

NaOH(aq)

→

NaCl(aq)

+ H2O(l)

Mr Singh

(3)

23

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

Example GCSE Mole Calculations 1.

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

+

2Al

→

2Fe

+

Al2O3

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

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

2.

Calculate the formula mass (Mr), of the compound calcium hydroxide, Ca (OH)2.

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

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Quantitative Chemistry

............................................................................................................................................

(Total 3 marks)

3.

The formula for the chemical compound magnesium sulphate is MgSO4.

Calculate the relative formula mass (Mr) of this compound. (Show your working.) ............................................................................................................................................... ............................................................................................................................................... ............................................................................................................................................... ...............................................................................................................................................

(Total 2 marks)

4.

(a)

The formula for the chemical compound magnesium sulphate is MgSO4.

Calculate the relative formula mass (Mr)of this compound. (Show your working.) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

(2)

(b)

Magnesium sulphate can be made from magnesium and dilute sulphuric acid.

This is the equation for the reaction.

Mg

+

H2SO4

→

MgSO4

+

H2

Calculate the mass of magnesium sulphate that would be obtained from 4g of magnesium.

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

Answer..................................... g (2) (Total 4 marks)

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Quantitative Chemistry

5.

In this question you will need to use the following information:

Relative atomic masses: H 1; O 16; Mg 24.

The volume of one mole of any gas is 24 dm3 at room temperature and atmospheric pressure.

The diagram shows a chemical reaction taking place in a conical flask.

The balanced equation for this reaction is: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (a)

Write a balanced ionic equation for this reaction. ....................................................................................................................................

(2)

(b)

Calculate the mass of magnesium required to produce 0.50 g of hydrogen. Show clearly how you work out your final answer and give the unit. ........................................................................................................................... ...........................................................................................................................

Mass = ...............................

(2)

(c)

(i)

Draw a diagram to show how the electrons are arranged in a hydrogen molecule.

(1)

(ii)

What is the name of the type of chemical bond between the hydrogen atoms in a hydrogen molecule? ...........................................................................................................................

(1)

(d)

The chemical formula for hydrogen peroxide is H2O2.

Calculate, to the nearest whole number, the percentage, by mass, of hydrogen in hydrogen peroxide. Show clearly how you work out your answer. .................................................................................................................................... ....................................................................................................................................

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Mr Singh

27

Quantitative Chemistry

Percentage = ................................. % (2) (Total 8 marks)

6.

In this question you will need to use the following information: The diagram shows a chemical reaction taking place in a conical flask.

The balanced equation for this reaction is: Mg(S) + 2HCl(aq) →

MgCl2(aq) +

H2(g)

Calculate the mass of magnesium required to produce 0.50 g of hydrogen. Show clearly how you work out your final answer and give the unit.

(Relative atomic masses: H = 1, O = 16, Mg = 24) ............................................................................................................................................... ...............................................................................................................................................

Mass = ............................

(Total 2 marks)

7.

‘Iron tablets’ usually contain iron sulphate (FeSO4). This salt can be made by reacting iron with sulphuric acid.

iron + sulphuric acid → iron sulphate + hydrogen (a)

To react, particles from the sulphuric acid must collide with particles of iron. What could you do to the block of iron to make the reaction faster? ..................................................................................................................................... .....................................................................................................................................

(1)

(b)

Calculate the percentage of iron in iron sulphate (FeSO4).

(Relative atomic masses: Fe = 56, O = 16, S = 32) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

Percentage of iron in iron sulphate = ..........................% (3) (Total 4 marks)

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28

Quantitative Chemistry

8.

‘Iron tablets’ usually contain iron sulphate (FeSO4). (a)

This salt can be made by reacting iron with sulphuric acid.

Fe + H2SO4 → FeSO4 + H2 Calculate the mass of iron sulphate that could be obtained from 4 g of iron.

(Relative atomic masses: Fe = 56, H = 1, O = 16, S = 32) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

Mass of iron sulphate = ........................... g (3)

(b)

Under different conditions, another type of iron sulphate may form.

Balance the symbol equation for this reaction.

Fe

+

H2SO4

→

Fe2(SO4)3

+

H2 (1) (Total 4 marks)

9.

Ammonium chloride, NH4Cl, is made up of nitrogen, hydrogen and chlorine atoms. (i)

Complete the table to show the number of atoms of each element present in NH4Cl.

Element

Number of atoms in NH4Cl

nitrogen hydrogen chlorine

1

(1)

(ii)

Calculate the relative formula mass of ammonium chloride, NH4Cl. (Relative atomic masses: H = 1, N = 14, Cl = 35.5) ..................................................................................................................................... ..................................................................................................................................... ..................................................................................................................................... .....................................................................................................................................

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Quantitative Chemistry

1101P GCSE Moles 18 © RWGrime

26/08/16

THE VOLUME OF ONE MOLE OF GAS Experiment 1 - The volume of one mole of oxygen Heating purple crystals of potassium manganate (VII) produces oxygen gas.

2 KMnO4(s) → K2MnO4(s) + MnO2(s) + O2(g) 1) Place a spatula load of potassium manganate (VII) in a boiling tube, and place some glass wool above the potassium manganate (VII).

2) Find the mass of the boiling tube to 3 decimal places.

3

3) Set up apparatus to link the boiling tube to a 100 cm gas syringe. The syringe should read 0 ml.

4) Heat the tube gently with a micro Bunsen until the syringe is over half full of gas.

5) Allow the syringe to cool for 5-10 minutes and then measure the gas volume.

6) Find the mass of the boiling tube at the end of the experiment.

Mass of tube at the start (g)

3

Mass of tube at the end (g)

Volume of O2 (cm )

Calculation a) Calculate the mass of oxygen that is lost from the tube. …………………………………. b) Calculate the moles of oxygen that are lost from the tube. …………………………..….. ……………………………………………………………………………………………………. c) What is the volume of this mass of oxygen? ……………………………………………….. c) Use this to find the volume of 1 mole of oxygen, O2. ………………………………………. ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. volume of 1 mole of O2(g) = …………………………………

Experiment 2 - The volume of one mole of hydrogen Hydrogen gas is produced when magnesium reacts with hydrochloric acid.

Mg(s) + 2 HCl(aq) → MgCl2(aq) + H2(g) 1) Collect a piece of magnesium and find its mass to 3 decimal place accuracy.

3

2) Place 50 cm of hydrochloric acid in a conical flask.

3

3) Set up the apparatus shown with the conical flask connected to a 100 cm gas syringe.

4) To start the experiment, put the magnesium in the acid and place the bung in the conical flask as soon as possible.

5) Wait until the reaction has finished, let the syringe cool for 5 minutes, and then read the gas volume.

hydrogen collects in syringe 3

50 cm hydrochloric acid + magnesium

Mass of magnesium (g)

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3

Volume of H2 (cm )

30

1101P

Quantitative Chemistry

a) How many moles of magnesium are used in this experiment? ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. b) How many moles of hydrogen are formed from this mass of magnesium? ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. c) What is the volume of one mole of hydrogen, H2? ……………………………………………………………………………………………………. ……………………………………………………………………………………………………. volume of 1 mole of H2(g) = …………………………………

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Quantitative Chemistry

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