IGCSE chemistry
calculations extension exercises
1. To find the % of aluminium in an alloy which contains only aluminium and copper, the following experiment
was carried out.
7.00g of the alloy was added to excess HCl. When the reaction was complete traces (small amounts) of
copper were left and 1.32 dm3 of hydrogen, at RTP, was collected.
2Al + 6HCl  2AlCl3 + 3H2
Calculate the percentage of aluminium in the alloy by completing the following:
(a) number of moles of H2 formed : 1.32/24 = 0.055 moles
(b) number of moles of Al reacted ratio = H2: Al is 3:2 so (0.055/2) x 3 = 0.037 moles
(c) mass of aluminium reacted 0.037 x 27 = 0.99g
(d) percentage of aluminium in the alloy. 0.99/7.00 x 100 = 14.1%
2. Copper (II) sulphate-5-water was prepared by the following reactions
CuO + H2SO4  CuSO4 + H2O
CuSO4 + H2O  CuSO4 .5H2O
In an experiment, 25 cm3 of 2.0 mol dm-3 sulfuric acid was neutralized with an excess of CuO. The yield
of crystals, CuSO4 .5H2O, was 7.3 g. Complete the following to calculate the percentage yield.
(a) Number of moles of H2SO4 in 25 cm3 of 2.0 mol dm-3 solution 2 x 0.025 = 0.05 moles
(b) Maximum number of moles of CuSO4 .5H2O that could be formed 0.05
(c) Maximum mass of crystals, CuSO4 .5H2O, that could be formed 12.5g
(d) The percentage yield 58.4%
3. The lead compound used by Midgley can be represented by the formula Pb(C2H5)n. It contains 64% by
mass of lead.
(a) Calculate the composition by mass of 100g of Pb(C2H5)n by the following steps.
Mass of lead in 100g of the compound =
Mass of (C2H5)n in 100g of the compound =
(b) The number of moles of Pb in 100g of Pb(C2H5)n =
(c) The mass of one mole of C2H5 =
64 g
36 g
0.31
29 g
(d) The number of moles of C2H5 in 100g of the compounds =
1.24
(e) The mole ratio Pb: C2H5 is 0.31 moles : 1.24 moles simplified = 1 : 4
(f) The value of n is
4
Chemistry calculations extensions
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4.
Green vitriol is hydrated (II) sulphate, FeSO4 .xH2O. Two pupils were asked to investigate the action of
heat on this mineral. They carried out the following experiments. The solid was gently heated to
constant mass to drive off all the water leaving anhydrous iron (II) sulphate.
Mass of hydrated salt FeSO4 .XH2O = 27.8 g
Mass of anhydrous salt FeSO4 = 15.2 g
(a) Calculate the mass of water driven off in experiment A. 12.6g
(b) The mass of 1 mole of H2O is 18g. How many moles of water were driven off? 12.6/18 = 0.7 moles
(c) Calculate the number of moles of the anhydrous salt, FeSO4. The mass of one mole of FeSO4 is
152g. 15.2g /152g = 0.1 mole
(d) What is the value of x? 0.1: 0.7 = 1 : 7 so x = 7
5. Copper (II) nitrate contains water of crystallization and the full formula of the salt is of the type
Cu(NO3)2. xH2O:
1 mole of Cu(NO3)2. xH2O will produce 1 mol of CuO
The following experiment was carried out to find “x” .
A 5.92 g sample of the salt Cu(NO3)2. xH2O was heated to leave 1.60 g of copper (II) oxide.
(a)
What is meant by the phrase one mole of a substance? 6 x 10x23 particles
(b)
How many moles of CuO there in 1.60g of copper (II) oxide? 1.60/80 = 0.02 moles
(c)
How many moles of Cu(NO3)2. xH2O are there in 5.92 g of hydrated copper (II) nitrate? 0.02 as
ratio to CuO is 1: 1
(d)
Calculate the mass of one mole of Cu(NO3)2. xH2O. 5.92/0.02 = 296
(e)
The mass of one mole of Cu(NO3)2 is 188g. Calculate x. mass of water = 296-188=108/18 = 6
6. The Mr of oxalic acid is 90 and its composition by mass is: carbon: 26.7 %
hydrogen 2.2 % . What is the molecular formula of the acid? C2O4H2
oxygen 71.1 %
7. An excess of HCl was added to 1.23 g of impure barium carbonate. The volume of carbon dioxide
collected at rtp was 0.120 dm3. The impurities did not react with the acid. Calculate the percentage
purity of the barium carbonate.
BaCO3
+
2HCl

BaCl2 + CO2
+ H2O
Molar gas volume at rtp = 24 dm3
(a) The number of moles of CO2 collected
0.12 /24 = 0.005
(b) The number of moles of BaCO3reacted
ratio 1 : 1 so 0.005
(c) Mass of one mole of BaCO3
137g
(d) Mass of barium carbonate
137 x 0.005 = 0.685g
(e) Percentage purity of the barium carbonate. 0.685/1.23 x 100 = 55.6%
Chemistry calculations extensions
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8. Insoluble salts are made by precipitation. An equation for the preparation of barium sulphate is:
FeSO4
+
BaCl2

BaSO4 + FeCl2
This reaction can be used in an experiment to find the value of x in the formula for hydrated iron (II)
sulphate crystals:
FeSO4. xH2O
A known mass of hydrated iron (II) sulphate was dissolved in water. Excess barium chloride solution
was added. The precipitate of barium sulphate was filtered and dried. Finally it was weighed.
Mass of hydrated iron (II) sulphate crystals = 1.390 g
Mass of barium sulphate formed
= 1.165 g
Complete the following to find x.
(a) The mass of one mole of BaSO4 is 233g. How many moles of BaSO4 were formed = 0.005
(b) How many moles of the hydrated iron (II) salt were used in the experiment?
(c) Calculate the mass of one mole of FeSO4.xH2O
(d) The mass of one mole of FeSO4 is 152 g. Calculate x.
= 0.005
= 1.390/0.005 = 278
x = 278 – 152 = 126 /18 = 7
9. A 43 g sample of scandium ore, Sc2Si2O7 produced 12g of scandium. Calculate the percentage yield
by completing the following calculation.
The mass of 1 mole of Sc2Si2O7 is 258g.
Number of moles of Sc2Si2O7 in 43g of the ore = 43/258 = 0.167 (= convert data into moles)
One mole of Sc2Si2O7
will give 2 moles of Sc (= use molar ratio)
43g of Sc2Si2O7 (0.167 moles) will produce 0.33 moles of Sc (= use ratio)
43 g of Sc2Si2O7 will produce (0.33x 41 g)= 15 of Sc (= convert moles into grams)
Percentage yield of scandium = 12/15 x 100 = 80%
10. Potassium chlorate, which has a formula of the type KClOn, decomposes to form oxygen. 2.45g of
the chlorate produced 1.49g of potassium chloride and 0.72 dm3 of oxygen at r.t.p. Find the value of n.
KclOn  KCl + n/2 O2
mass of one mole of KCl = 74.5 g
Number of moles of KCl formed = 0.02 (=convert data into moles)
Number of moles of oxygen molecules formed = 0.03 (= use molar ratio)
Number of moles of oxygen atoms = 0.06 (= use molar ratio)
Mole ratio KCl:O is 0.02/0.06
Chemistry calculations extensions
n=3
3
11. A 20 cm3 sample of butyne, C4H6, is burnt in 150 cm3 of oxygen. This is an excess of oxygen.
2C4H6(g)
+ 11O2(g)
 8CO2(g) + 6H2O(l)
(a) What volume of oxygen reacts ? 110 cm3
(=convert data into moles)
(b) What volume of carbon dioxide is produced? 80 cm3 (=use ratio)
(c) What is the total volume of gases left at the end of the reaction? 120 cm3 (40 excess O2 + 80)
(d) Calculate the mass of water formed when 9.0 g of butyne is burnt. The mass of one mole of
butyne is 54g.
From the above equation, 1 mole of butyne forms 3 moles of water
Number of moles of butyne reacted 0.167
Number of moles of water formed 0.5
Mass of water formed 9g
(=convert data into moles)
(=use ratio)
(=convert moles into data)
12. Sodium reacts with sulphur to form sodium sulphide.
2Na + S 
Na2S
An 11.5 g sample of sodium is reacted with 10 g of sulphur. All of the sodium reacted but there was
an excess of sulphur. Calculate the mass of sulphur left unreacted
(a) number of moles of sodium atoms reacted = 0.5 (= data into moles)
(b) number of moles of sulphur atoms that reacted = 0.25 (= use ratio)
(c) mass of sulphur reacted = 8 g (=convert moles into data)
(d) mass of sulphur left unreacted. 2g
13. Tablets for healthy bones contain the same number of moles of CaCO3 and MgCO3. One tablet
reacted with excess hydrochloric acid to produce 0.24 dm3 of carbon dioxide at r.t.p.
CaCO3 + 2 HCl  CaCl2 + CO2 + H2O
MgCO3 + 2 HCl  CaCl2 + CO2 + H2O
(a)
Calculate how many moles of CaCO3 there are in one tablet.
Number of moles CO2 0.01
Number of moles of CaCO3 and MgCO3 0.01 (all CO2 comes from all carbonates)
Number of moles of CaCO3 0.005
(b)
Calculate the volume of HCl, 1.0 mol/ dm3, needed to react with one tablet.
Number of moles of CaCO3 and MgCO3 in one tablet = 0.01 (see above)
Number of moles of HCl needed to react with one tablet = 0.02
Volume of HCl, 1.0 mol /dm3, needed to react with one tablet = 0.02 dm3
Chemistry calculations extensions
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