8 Formulae, equations
and moles
Answers to end-of-chapter questions
Page 141 Questions
1
Element
Divide by r.a.m.
Divide by smallest
Carbon
17.8/12.0 = 1.48
1.48/0.991 = 1.5
Hydrogen
3.0/1.0 = 3.0
3.0/0.991 = 3.0
Bromine
79.2/79.9 = 0.991
0.991/0.991 = 1.0
The simplest whole number ratio of 1.5:3:1 is obtained by multiplying them all by 2, i.e. 3:6:2.
The empirical formula is C 3 H 6 Br 2 .
[e] After division by the smallest number, you should round the values to 1 decimal place. If the ratio
is not a whole number ratio, try to obtain integers by multiplying all the numbers by 2. If that fails,
multiply by 3.
2 a)
Element
Divide by r.a.m.
Carbon
36.4/12.0 = 3.03
Hydrogen
6.1/1.0 = 6.1
Fluorine
57.5/19.0 = 3.03
The empirical formula is CH 2 F.
Divide by smallest
3.03/3.03 = 1
6.1/3.03 = 2
3.03/3.03 = 1
b) The mass of the empirical formula is (12.0 + 2.0 + 19.0) = 33.0. This is half the molar mass, so
the molecular formula is C 2 H 4 F 2 .
3 a) As there is one manganese ion to each carbonate ion, the charges have the same numerical
value. Carbonate ions are 2–, so the manganese ions in this compound must be 2+.
b) The charge on two vanadium ions must equal the charge on three sulfate ions. Each sulfate is
–2, so three sulfate ions have a total charge of –6. This means that two vanadium ions have a
total charge of +6, so each has a charge of +3.
4 a) molar mass of Ca(OH) 2 = 40.1 + 2 × (16.0 + 1.0) = 74.1 g mol
–1
b) molar mass of Al 2 (SO 4 ) 3 = (2 × 27.0) + 3 × [32.1 + (4 × 16.0)] = 342.3 g mol
–1
–1
c) molar mass of FeSO 4 .7H 2 O = 55.8 + 32.1 + (4 × 16.0) + 7 × (2.0 + 16.0) = 277.9 g mol
[e] In calculating molar mass, make sure that you use the relative atomic masses (the larger
−1
numbers), not the atomic numbers, from the periodic table. Molar mass has the units of g mol .
Relative molecular mass has no units. The numbers are the same.
5 The balanced equations are:
a) 2Mg(NO 3 ) 2 → 2MgO + 4NO 2 + O 2
[e] In part a), balance the oxygen atoms last. Remember that the number of oxygen atoms on the
right-hand side of the equation must be an even number, because Mg(NO 3 ) 2 contains an even
number of oxygen atoms.
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8 Formulae, equations
and moles
3+
–
Answers to end-of-chapter questions
2+
b) 2Fe + 2I → 2Fe + I 2
3+
[e] Make sure that this equation balances for charge. Having one Fe ion on the left-hand side and
2+
one Fe on the right-hand side balances the number of Fe ions, but the charge on the left would not
equal the charge on the right.
c) 2C 8 H 18 + 25O 2 → 16CO 2 + 18H 2 O or C 8 H 18 + 12½O 2 → 8CO 2 + 9H 2 O
[e] The only source of oxygen on the left-hand side of the equation is O 2 , so the number of oxygen
atoms on the right-hand side must be an even number. Therefore, there must be an even number of
H 2 O molecules on the right. This requires two C 8 H 18 molecules on the left.
6 For equations a) and b), the simplest way is to realise that the component ions of the compound
combine to give the precipitate. All other ions are spectators.
a) The precipitate is Cu(OH) 2 , so the ionic equation is:
2+
–
Cu (aq) + 2OH (aq) → Cu(OH) 2 (s)
[e] Do not forget state symbols in all ionic equations.
b) The precipitate is BaSO 4 , so the ionic equation is:
2–
2+
Ba (aq) + SO 4 (aq) → BaSO 4 (s)
–
c) Aqueous alkalis are fully ionised, so in neutralisation reactions OH (aq) ions are in the
+
–
equation. The reaction is between H and OH ions and the ionic equation is:
+
–
H (aq) + OH (aq) → H 2 O(l)
7 For solids:
moles = mass/molar mass
–1
a) amount of CaCO 3 = 1.11 g/[40.1 + 12.0 + (3 × 16.0)] g mol = 0.0111 mol
–1
b) amount of Ba(OH) 2 = 2.22 g/[137.3 + 2 × (16.0 + 1.0) + 8 × (2.0 + 16.0)] g mol = 0.00704 mol
[e] Make sure that you give your answer to the correct number of significant figures. Here, the data
in the question are given to three significant figures, so the answer is also given to three significant
figures.
8 mass = moles × molar mass
–1
a) mass of 0.0100 mol H 2 SO 4 = 0.0100 mol × [2.0 + 32.1 + (4 × 16.0)] g mol = 0.981g
–1
b) mass of 100 mol Na = 100 mol × 23.0 g mol = 2300 g
–1
9 molar mass of the hydrated salt = mass/moles = 4.56 g/0.0185 mol = 246 g mol
–1
MgSO 4 contributes [24.3 + 32.1 + (4 × 16.0)] = 120.4 g mol ;
–1
therefore, 246 – 120.4 = 125.6 g mol is due to x moles of water:
x(2.0 + 16.0) = 125.6
x = 125.6/18.0 = 6.98
© Hodder & Stoughton Limited 2015
8 Formulae, equations
and moles
Answers to end-of-chapter questions
The number of molecules of water of crystallisation must be a whole number, so hydrated
magnesium sulfate contains seven molecules of water of crystallisation.
–1
10 moles of water = mass/molar mass = 1.2 g/18.0 g mol = 0.0667 mol
23
–1
22
number of molecules = 0.0667 mol × 6.02 × 10 mol = 4.0 × 10
[e] Note that the mass data were given to two significant figures.
23
21
11 number of molecules in 0.0100 mol CO 2 = 0.0100 × 6.02 × 10 = 6.02 × 10
There are two oxygen atoms in each molecule of CO 2 , so:
21
number of oxygen atoms in 0.0100 mol CO 2 = 2 × 6.02 × 10 = 1.20 × 10
22
–1
12 moles of Ba(OH) 2 = mass/molar mass = 10.0 g/(137.3 + 34.0) g mol = 0.05838 mol
−
There are two OH ions per formula unit of Ba(OH) 2 , so:
23
22
number of hydroxide ions = 2 × 0.05838 × 6.02 × 10 = 7.03 × 10
[e] Remember that the number of atoms, molecules or ions is always a huge number, usually around
23
10 . The number of moles is usually less than 1.
Page 142 Exam practice questions
1 a) D ()
b) i)
D ()
ii)
moles Ca(OH) 2 = 1/74 = 0.01351
–
23
21
number of OH ions = 2 × 6.02 × 10 × 0.01351 = 8.1 × 10 ()
c) i)
D ()
ii)
moles of Na 2 CO 3 .10H 2 O = 1.22/286 = 0.004266 = moles of CO 2 ()
pV = nRT so V = nRT/p ()
–2
p = 95 kPa = 95000 Pa or 95000 N m ()
–1
–1
–2
V = (0.004266 mol × 8.31 N m K mol × 291 K)/95000 N m ()
–4
3
3
3
= 1.09 × 10 () m () or 0.0109 () dm () or 10.9 () cm ()
d) C ()
3+
2+
[e] Left side of equation is +5 and right side is +6, so 2Fe and 2Fe are needed for the equation to
balance.
−
+
2 a) Cl (aq) + Ag (aq) → AgCl(s) species (), state symbols ()
b) i)
−1
moles AgCl = 2.22 g/143.4 g mol = 0.01548 mol ()
moles of MCl 2 = ½ × 0.01548 = 0.007741 mol ()
ii)
−1
molar mass = mass/moles = 1.23 g/0.007741 mol = 158.9 g mol ()
−1
A r of M = 158.9 – (2 × 35.5) = 87.9 g mol , so M is strontium ()
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