Quantitative revision sheet d) Relative formula masses and molar volumes of gases Students will be assessed on their ability to: 1.16 calculate relative formula masses (Mr) from relative atomic masses (Ar) Eg. To work out the formula mass for NaOH Step 1: Find the Ar for each element (top number of each element in period table) Na(23) O(16) H(1) Step 2: Add the Ar together 23+16+1 = 40 NB If the formula contains a small (subscript) number e.g Na2SO4 it means we must times the Ar for the element in front by that number so 23x2 +32 + 16x4 = 142 NB If the formula contains a bracket and a small (subscript) number e.g. Ca(OH)2 it means we must times the Ar for the elements in the brackets by that number so 40 + 16x2 + 1x2 = 74
1.17 understand the use of the term mole to represent the amount of substance A mole is defined as the quantity of a substance that has the same number of particles as are found in 12.000 grams of carbon-12. This number, Avogadro's number, is 6.022x1023. The mass in grams of one mole of a compound is equal to the molecular weight of the compound in atomic mass units. e.g One mole of water H2O 1x2 + 16 = 18 A thimble full of water is about 1cm3, 1 mole of water = 18g and ~ 18cm3 because the density of water is ~1.0 g/cm3 Therefore in a thimble full of water there are ~6.023 x 1023/18 = ~3.3 x 1022 = 33 000 000 000 000 000 000 000 molecules! = thirty three thousand million million million molecules of water! So, just think how many molecules of water are in your body! Moles allow us to compare ratios, which is why they are important!
1.18 carry out mole calculations using relative atomic mass (Ar) and relative formula mass (Mr) N2(g) + 3H2(g) ==> 2NH3(g) This tells us that one mole of Nitrogen atoms reacts with three moles of hydrogen atoms to produce two moles of ammonia. So 0.05 mol nitrogen combines with 0.15 mol hydrogen to form 0.10 mol of ammonia. (Using 1:3:2 ratio)
e) Chemical formulae and chemical equations Students will be assessed on their ability to: 1.22 calculate empirical and molecular formulae from experimental data 33·6 g of iron was found to react with 14·4 g of oxygen. What is the empirical formula of iron oxide? 1) Find how many moles of iron react with how many moles of oxygen. RAM of Fe = 56, RAM of O = 16. moles = mass ÷ RAM for iron moles = 33·6 ÷ 56 = 0·6 moles. for oxygen moles = 14·4 ÷ 16 = 0·9 moles. 2) The proportion of moles of iron to moles of oxygen is reduced to the lowest whole number. 0·6 moles of Fe to 0·9 moles of O. Divide by 3, multiply by 10. 2 moles of Fe to 3 moles of O. The empirical formula is Fe2O3. Since iron oxide is an ionic compound, the actual formula is the same as the empirical formula. 1.23 calculate reacting masses using experimental data and chemical equations 2Mg + O2 ==> 2MgO 1. (atomic masses Mg =24, O = 16) 2. converting the equation into reacting masses gives ... (2 x 24) + (2 x 16) ==> 2 x (24 + 16) 3. and this gives a basic reacting mass ratio of 48g Mg + 32g O2 ==> 80g MgO 4. The ratio can be used, no matter what the units, to calculate and predict quite a lot! and you don't necessarily have to work out and use all the numbers in the ratio. 5. What you must be able to do is solve a ratio! e.g. 24g Mg will make 40g MgO, why?, 24 is half of 48, so half of 80 is 40.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O (atomic masses Na = 23, O = 16, H = 1, S = 32) mass ratio is: (2 x 40) + (98) ==> (142) + (2 x 18) = (80) + (98) ==> (142) + (36),
but pick the ratio needed to solve the particular problem e.g. reacting mass ratio of 2NaOH : Na2SO4 is 80 (2 x 40) : 142 calculate how much sodium hydroxide is needed to make 5g of sodium sulphate.
1.
from the reacting mass equation: 142g Na2SO4 is formed from 80g of NaOH
2.
5g Na2SO4 is formed from 5g x 80 / 142 = 2.82 g of NaOH by scaling down from 142 => 5
calculate how much water is formed when 10g of sulphuric acid reacts.
3.
from the reacting mass equation: 98g of H2SO4 forms 36g of H2O
10g of H2SO4 forms 10g x 36 / 98 = 3.67g of H2O by scaling from 98 => 10
down
http://www.bbc.co.uk/schools/gcsebitesize/science/add_gateway_p re_2011/chemical/reactingmassesrev1.shtml 1.24 carry out mole calculations using volumes and molar concentrations. What mass of sodium hydroxide (NaOH) is needed to make up 500 cm3 (0.500 dm3) of a 0.500 mol dm-3 (0.5M) solution? [Ar's: Na = 23, O = 16, H = 1] 1 mole of NaOH = 23 + 16 + 1 = 40g molarity = moles / volume, so mol needed = molarity x volume in dm3 mol NaOH needed = 0.500 x 0.500 = 0.250 mol NaOH
therefore mass = mol x formula mass = 0.25 x 40 = 10g NaOH required How many moles of H2SO4 are there in 250cm3 of a 0.800 mol dm-3 (0.8M) sulphuric acid solution? What mass of acid is in this solution? [Ar's: H = 1, S = 32, O = 16] molarity = moles / volume in dm3, rearranging equation for the sulfuric acid mol H2SO4 = molarity H2SO4 x volume of H2SO4 in dm3 mol H2SO4 = 0.800 x 250/1000 = 0.200 mol H2SO4 mass = moles x formula mass formula mass of H2SO4 = 2 + 32 + (4x16) = 98 0.2 mol H2SO4 x 98 = 19.6g of H2SO4
b) Energetics Students will be assessed on their ability to: 4.10 understand that chemical reactions in which heat energy is given out are described as exothermic and those in which heat energy is taken in are endothermic Remember you go out through an exit and go in through an entry. 4.11 describe simple calorimetry experiments for reactions such as combustion, displacement, dissolving and neutralisation in which heat energy changes can be calculated from measured temperature changes Calorimetry, derived from the Latin calor meaning heat, and the Greek metry meaning to measure, is the science of measuring the amount of heat. For combustion we can measure the temperature change of water