Essential Maths Skills
for AS/A-level
Chemistry
Answers
Nora Henry
1 Arithmetic and numerical computation
Appropriate units in calculations
Guided question (p.8)
1 203 K − 273 = −70 °C
Practice question (p.8)
2
a
b
c
d
e
f
246 K − 273 = −27 °C
−35 °C + 273 = 238 K
344 K − 273 = 71 °C
−19 °C + 273 = 254 K
45 K − 273 = −228 °C
192 K − 273 = −81 °C
Guided questions (p.10)
12000 cm 3
= 12dm 3
1000
12 dm 3
Step 2:
= 0.012 m 3
1000
2 1.2 kg × 1000 = 1200 g
1 Step 1:
Practice question (p.11)
3 a To convert from dm3 to cm3 multiply by 1000.
1.2 dm3 × 1000 = 1200 cm3
b To convert from cm3 to dm3 divide by 1000.
420 cm 3
= 0.42 dm 3
1000
c To convert from cm3 to dm3 divide by 1000 and then to convert to m3 divide by
1000 (or divide by 106 if you wish to do it in one step).
3452 cm 3
= 0.003452 m 3
6
10
d To convert from tonnes to grams, first convert to kg by multiplying by 1000 and
then convert to grams by multiplying by 1000.
1.4 tonnes × 1000 × 1000 = 1 400 000 g
Essential Maths Skills for AS/A-level Chemistry
1
e
f
To convert from kg to g multiply by 1000.
4 kg × 1000 = 4000 g
To convert from kPa to Pa multiply by 1000.
101 kPa × 1000 = 101 000 Pa
Guided question (p.12)
1 Step 1: 3.312 kg × 1000 = 3312 g
Step 2: (1 × 207.2) + (2 × 14.0) + (6 × 16.0) = 331.2
Step 3: amount (in moles) =
mass (g) 3312
=
=10
331.2
Mr
Practice questions (p.12)
2 The mass of ammonia must be converted from kg to g by multiplying by 103.
Mass of ammonia in grams = 17 kg × 1000 = 17 000 g
amount (in moles) =
mass (g) 17 000 g
=1000 mol
=
Mr
17.0
3 Mass of iron(III) oxide in grams = 2.1 tonnes × 1000 × 1000 = 2 100 000 g
amount (in moles) =
mass (g) 2 100 000 g
=
=13 149.7 mol
Mr
159.7
Note that in questions 3 and 4 the answer is given to one decimal place. Decimal
places are used when large numbers are involved as the precision of the answer would
be limited if three or four significant figures were used.
4 Mass of magnesium nitrate in grams = 2.42 kg × 1000 = 2420 g
amount (in moles) =
mass (g) 2420 g
=
=16.3 mol
Mr
148.3
5 Mass of calcium carbonate in grams = 3.2 kg × 1000 = 3200 g
amount (in moles) =
mass (g) 3200 g
=
= 32.0 mol
100.1
Mr
Guided question (p.13)
1 Step 1:
p = 100 kPa × 1000 = 100 000 Pa
R = 8.31 J K–1 mol–1
T = 25 °C + 273 = 298 K
n = 0.324 mol
Essential Maths Skills for AS/A-level Chemistry
2
Step 2:
pV = nRT
pV
= nRT
p
p
This simplifies to:
V = nRT
p
Step 3: Now substitute all the values, in the correct units.
0.324 × 8.31 × 298 = 8.02 × 10 –3 m 3
100 000
Step 4: 0.00802 m3 × 1000 = 8.02 d m3
Practice questions (p.14)
2 p = 104 kPa. The pressure must be in Pa. To convert from kPa to Pa you must
multiply by 1000. p = 104 kPa × 1000 = 104 000 Pa.
V = 1822 cm3. The volume must be in m3. To convert from cm3 to m3 divide by
106 (1000 × 1000).
V =
1822 cm 3
= 0.001822 m 3
106
R = 8.31 J K–1 mol–1
n = 0.136 mol
At this point, all the quantities are in the correct units.
To find T, you must change the subject of the equation.
pV = nRT
pV
=T
nR
Now substitute all the values, in the correct units.
104 000 × 0.001822
=167.7 K
8.31 × 0.136
3 p = 100 kPa. The pressure must be in Pa. To convert from kPa to Pa you must
multiply by 1000. p = 100 kPa × 1000 = 100 000 Pa.
T=
R = 8.31 J K–1 mol–1
n = 0.502 mol
T = 323 K
At this point all the quantities are in the correct units.
To find V, change the subject of the equation:
pV = nRT
Essential Maths Skills for AS/A-level Chemistry
3
This simplifies to:
V = nRT
p
Now substitute all the values, in the correct units.
V =
0.502 × 8.31 × 323
= 0.0135m 3
100 000
The question asked for the volume in dm3. To convert from m3 to dm3 multiply by 1000.
V = 0.0135 m3 × 1000 = 13.5 dm3
4 p = 100 kPa. The pressure must be in Pa. To convert from kPa to Pa you must
multiply by 1000.
p = 100 kPa × 1000 = 100 000 Pa
V = 0.0320 m3
R = 8.31 J K–1 mol–1
n = 0.658 mol
At this point all the quantities are in the correct units.
To find T, you must change the subject of the equation.
pV = nRT
pV
=T
nR
Now substitute all the values, in the correct units.
T=
100 000 × 0.0320
= 585.2 K
8.31 × 0.658
5 p = 100 kPa. The pressure must be in Pa. To convert from kPa to Pa you must
multiply by 1000.
p = 100 kPa × 1000 = 100 000 Pa
V = 1.63 × 10−2 m3
R = 8.31 J K–1 mol–1
T = 37 °C. The temperatures must be in kelvin. To convert from Celsius to kelvin add 273.
Therefore T = 37 °C + 273 = 310 K.
At this point all the quantities are in the correct units.
To find n, you must change the subject of the equation.
pV = nRT
n=
pV 100 000 × 1.63 × 10 –2
=
= 0.63 mol
RT
8.31 × 310
Guided question (p.15)
1 Step 1: list all of the quantities given and convert them to the correct units.
ΔH = −46.2 kJ mol−1. The units are correct.
Essential Maths Skills for AS/A-level Chemistry
4
∆S = −99.5 J K−1 mol−1. This must be converted to kJ K−1 mol−1 by dividing by 1000.
–1
–1
∆S = –99.5 J K mol = –0.0995kJ K –1 mol–1
1000
Step 2:
∆G = ∆H − T∆S
0 = −46.2 – T(−0.0995)
0 = −46.2 + 0.0995T
T = 464.3 K
Practice questions (p.16)
2 List all of the quantities given and convert them to the correct units.
ΔH = −198 kJ mol−1. The units are correct.
∆S = 180 J K−1 mol−1. This must be converted to kJ K−1 mol−1 by dividing by 1000.
Therefore ΔS = 0.180 kJ K−1 mol−1
T = 30 °C. This must be converted to kelvin by adding 273.
Therefore T = 30 °C + 273 = 303 K.
At this point all the quantities are in the correct units.
Substitute the values into the equation.
∆G = ∆H – T ∆S
= –198 – (303 × 0.180)
In calculations, multiply out the brackets, before taking away.
∆G = –198 – 54.54
= – 253 kJ mol–1
3 List all of the quantities given and convert them to the correct units.
ΔH = +323 kJ mol−1. The units are correct.
∆S = +153.7 J K−1 mol−1. This must be converted to kJ K−1 mol−1 by dividing by 1000.
ΔS = +0.1537 kJ K−1 mol−1
At this point all the quantities are in the correct units.
Substitute the values into the equation.
∆G = ∆H – T ∆S
0 = +323 – T (0.1537)
0 = +323 – 0.1537 T
–323 = – 0.1537T
323 = 0.1537T
T = 323 = 2101K
0.1537
To convert into degrees Celsius subtract 273.
2101 K − 273 = 1828 °C
Essential Maths Skills for AS/A-level Chemistry
5
Guided questions (p.18)
1 Step 1: mol dm−3 s−1 = k (mol dm−3)2 × (mol dm−3)
s−1 = k(mol dm−3)2
Step 2: k(mol dm−3)2 = s−1
k(moldm –3 )2
s –1
=
–3 2
(moldm )
(moldm –3 )2
k=
s –1
(moldm –3 )2
k=
s –1
= mol–2 dm 6 s –1
2
–6
(mol dm )
2 Step 1: the square brackets mean concentration so substitute concentration units into
the expression.
Kc =
[CO 2 ][H 2 ]4
[CH 4 ][H 2O]2
(moldm –3 )(moldm –3 )4
(moldm –3 )(moldm –3 )2
Step 2: cancel the mol dm−3 on the top and bottom of the fraction.
Kc =
=
(moldm –3 )(moldm –3 )4 (moldm –3 )4
=
(moldm –3 )(moldm –3 )2 (moldm –3 )2
Step 3 and 4:
Kc =
(moldm –3 )4 2
= mol2 dm –6
–3 2
(moldm )
3 Step 1:
Kc =
(moldm –3 )
1
(moldm –3 )(moldm –3 ) 2
Step 2: units of K c =
Step 3:
Kc =
1
1
–3 2
(moldm )
Step 4: K c = mol
–1
2
(moldm –3 )
1
(moldm –3 )(moldm –3 ) 2
=
1
2
1
mol dm
– 23
3
dm 2
Essential Maths Skills for AS/A-level Chemistry
6
Practice questions (p.19)
4 a mol dm−3 × mol dm−3 = mol2 dm−6
b mol dm−3 × (mol dm−3)2 = mol dm−3 × mol2 dm−6 = mol(1+2) dm(−3−6) = mol3 dm−9
c (mol dm−3)2 × (mol dm−3)2 = (mol dm−3)(2+2) = (mol dm−3)4 = mol4 dm−12
d
(moldm –3 )2
1
=
= mol–1 dm 3
–3 3
–3
(moldm )
moldm
e
(moldm –3 )4 2
= (mol dm –3 )2 = mol2 dm –6
–3 2
(moldm )
(moldm –3 )2
1
=
= mol–1 dm 3
(moldm –3 ) × (moldm –3 )2 moldm –3
1
= (moldm –3 ) –2 = mol–2 dm 6
g
–3 2
(moldm )
f
5
(moldm –3 )(moldm –3 )3 (moldm –3 )(moldm –3 ) 3 2
=
= (mol dm –3 )2 = mol2 dm –6
–3
–3
–3
–3
(moldm )(moldm )
(moldm )(moldm )
6 a mol dm−3 s−1
b mol dm−3 s−1 = k mol dm−3
The units of k are s−1.
c mol dm−3 s−1 = k × mol dm−3 × (mol dm−3)2
s−1 = k (mol dm−3)2
k=
7 K c =
s –1
= s –1 (moldm –3 ) –2 = s –1 mol–2 dm 6
(moldm –3 )2
(moldm –3 )2
(moldm –3 ) × (moldm –3 )
The units cancel top and bottom, so there are no units for Kc in this case.
8 K c =
[NH 3 ]2
1
2
[N 2 ] [H 2 ]
3
2
=
(moldm −3 )2
3
1
(moldm −3 ) 2 (moldm −3 ) 2
Cancel mol dm−3 on the top and bottom.
Kc =
(moldm −3 )2
1
–3 2
31
−3 2 2
(moldm ) (moldm )
=
moldm −3
= no units
moldm −3
Expressions in decimal and ordinary form
Guided questions (p.21)
1 Step 1: 3.418 g
tep 2: the number after the underlined number is 8 (above 5) so the rule ‘if the next
S
number is 5 or more, round up’ is followed.
The answer is 3.42 g (to 2 d.p.).
2 a Step 1: subtract the mass of the evaporating basin from the mass of the
evaporating basin + anhydrous solid.
27.799 − 26.250 = 1.549 g
Essential Maths Skills for AS/A-level Chemistry
7
Step 2: underline the numbers up to two numbers after the decimal point.
1.549
Step 3: 1.549 g
The next number is 9, which is greater than 5, so round up. The answer is 1.55 g.
b Step 1 and 2:
28.465 − 27.799 = 0.666 g
Step 3: the next number is 6 (greater than 5) so round up to 0.67 g.
3 Step 1: pH = −log 0.2 = − (−0.698970004) = 0.698970004
Step 2: pH = 0.698970004
tep 3: the number after the second decimal place is 8 so round up.
S
The answer is 0.70.
Practice questions (p.22)
4 a 1.72 g which is 1.7 g to 1 decimal place.
b 9.69 g which is 9.7 g to 1 decimal place.
c 2.11 g which is 2.1 g to 1 decimal place.
5 Table A.1
Mass/g
29.883
0.046
32.6789
13.999
0.0894
19 992.456
6 a
b
c
d
e
Mass recorded to two decimal places/g
29.88
0.05
32.68
14.00
0.09
19 992.46
0.5
1.3
−0.3
2.3
1.7
Guided question (p.24)
1 Step 1: Table A.2
Temperature/°C
10.2
10
10.3
11
Number of decimal places
1
0
1
0
Essential Maths Skills for AS/A-level Chemistry
8
Step 2: the answer should be recorded to 0 decimal places.
Step 3: average =
10.2 +10 +10.3 +11
= 10.375 = 10
4
Practice questions (p.24)
2 Table A.3
Mass of evaporating
basin/g
34.567
29.934
25.49
18.456
Mass of solid
Mass of evaporating basin (to appropriate number of
and solid/g
decimal places)/g
23.4
11.2 (1 d.p.)
25.66
4.27 (2 d.p.)
22.1
3.4 (1 d.p.)
11.9
6.6 (1 d.p.)
3 a 53.667 g is the total. The number with the least decimal places is 43.2 g with one
decimal place hence the answer must be rounded to one decimal place i.e. 53.7 g.
b 13.128 g is the total. The number with the least decimal places is 2.49 g with two
decimal places hence the answer must be rounded to two decimal places i.e.
13.13 g.
c 7.5439 g is the total. The number with the least decimal places is 3.23 or 3.97 with
two decimal places hence the answer must be rounded to two decimal places
i.e. 7.54 g.
Guided questions (p.26)
1 Step 1: write the non-zero digits with a decimal place after the first number and then
write ‘× 10n’ after it.
Step 2: the decimal place has moved three places to the right (or 6.45 divided by 10
three times to get 0.00645) so n is −3.
6.45 × 10−3
2 Step 1:
mass in g
= 2.3 = 0.1
Mr
23.0
Step 2: 0.1 × 6.02 × 1023 = 6.02 × 1022
Practice questions (p.26)
3 a
b
c
d
e
f
g
1.1345 × 104
3.234567 × 106
1 × 10−2
3.45 × 10−3
8.7 × 10−4
1.110343 × 103
9.8760089 × 103
Essential Maths Skills for AS/A-level Chemistry
9
4 a
b
c
d
e
f
g
3 200 000
845.6
0.0056765
50 000
0.000004655
934 000
0.000238
5 a
b
c
d
1600
0.0000000015
6380
0.002040
6 a 1.3244 × 1012
b 1.1333 × 10−5
c 1.5238 × 1026
d4
7 1.6734 × 10−24
8
8.505
= 0.05
170.1
0.05 × 6.02 × 1023 = 3.01 × 1022
9 1 mole S8 contains 8 atoms
1 mole contains 6 × 1023 × 8 atoms
0.0012 mol = 0.0012 × 6 × 1023 × 8 = 5.76 × 1021 atoms
10 2.08 kg = 2080 g
2080 g
= 17.52 moles
118.7
Number of atoms = 17.52 × 6 × 1023 = 1.05 × 1025
Moles tin =
11 [H+]2 = 1.0 × 10−14 mol2 dm−6
[H + ]= 1.0 × 10 –14 =1.0 × 10 –7 moldm –3
pH = −log[H+] = −log(1.0 × 10−7) = 7.0
The pH of pure water at 25 °C is 7.0.
12 [H+]2 = 2.92 × 10−14 mol2 dm−6
[H+] = 2.92 × 10−14 = 1.7 × 10−7 mol dm−3
pH = − log[H+] = −log(1.7 × 10−7) = 6.77
Essential Maths Skills for AS/A-level Chemistry
10
Ratios, fractions and percentages
Guided question (p.28)
1 Step 1:
Mr = 40.1 + (14.0 × 2) + (16.0 × 6) = 164.1
Step 2:
Mass N = 14.0 × 2 = 28.0
Steps 3 and 4:
28
× 100 = 17.1%
164.1
Practice questions (p.29)
2 × 1.0 × 100 = 2.7%
74.1
2 × 39.1 × 100 = 26.6%
b
294.2
2 a
c 2 × 14.0 × 100 = 21.2%
132.1
5 × 18.0 × 100 = 36.1%
d
249.6
e 13 × 16.0 × 100 = 72.7%
286.0
35 × 2.3 = 0.8 g
3
100
4
4.1 × 100 = 61.2%
6.7
Guided question (p.30)
1 Step 1:
P
:
O
0.050
:
0.125
Step 2:
0.050
0.050
:
0.125
0.050
1
:
2.5
1 × 2
:
2.5 × 2
2
:
5
P2O5
Essential Maths Skills for AS/A-level Chemistry
11
Practice questions (p.30)
2 a C4H5N2O
b Na2S2O3
c CH2O
d P2O5
3 a Al(NO3)3.9H2O
b Pb3O4
c Cl2O7
Guided questions (p.32)
1 Step 1:
N2 :
H2
1 :
3
Step 2: there is three times as much H2 as N2, so divide H2 moles by 3.
0.4 = 0.133
3
0.133 moles
2 a Step 1:
P4O10
:
H 2O
1 :
6
Step 2: there is six times as much H2O as P4O10, so multiply P4O10 moles by 6.
0.25 :
0.25 × 6 = 1.5
1.5 moles
b Step 1:
H2O :
H3PO4
6
:
4
H2O
:
H3PO4
6 :
4
3 :
2
0.3
:
0.2
Step 2:
Practice questions (p.32)
3 a Cu(NO3)2
2
:
:
O2
1
4
:
2
2 moles
Essential Maths Skills for AS/A-level Chemistry
12
b Cu(NO3)2 :
2
:
NO2
4
1
:
2
0.6
:
0.6 × 2 = 0.12
4 a CaO
1
:
:
C
3
0.33
:
0.33 × 3 = 0.99
b C
:
CO
3
:
3.2
:
1
3.2 × 1 = 1.1
3
0.12 moles
0.99 moles
1.1 moles
5 a Pb
3
:
:
1
:
0.66
:
O2
2
2
3
0.66 × 2 = 0.44
3
0.44 moles
b O2
2
:
:
2.2
:
1.1 moles
c Pb
3
0.33
Pb3O4
1
2.2 × 1 = 1.1
2
:
:
Pb3O4
1
:
0.33 × 1 = 0.11
3
Essential Maths Skills for AS/A-level Chemistry
13
Estimating results
Guided questions (p.34)
1 Step 1: ∆H is positive so the reaction is endothermic.
Step 2: an increase in temperature moves the position of equilibrium from left to right
to oppose the change.
Step 3: there will be a greater concentration of products and so the numerator
increases and the denominator decreases, meaning that Kc increases.
2 Step 1: as the temperature increases, Kc decreases.
Step 2: the denominator is bigger and the concentration of reactants is greater.
Step 3 the equilbrium moves left on increasing the temperature and the reaction is
exothermic.
Practice questions (p.35)
3 The reaction is endothermic due to the positive enthalpy change. An increase in
temperature moves the position of equilibrium from the left to right.
This means there is a greater concentration of products [HI], so the numerator of the
Kc expression is larger and so Kc increases.
4 The reaction is exothermic, due to the negative enthalpy change so increasing the
temperature means the position of equilibrium moves from right to left. There is a
greater concentration of reactants and so the denominator, [SO2]2[O2], is bigger and Kc
is smaller.
5 Kc decreases means that there is a greater concentration of reactants and the
denominator is bigger, making Kc smaller. Hence the equilibrium position has moved
left when temperature increases. Hence the forward reaction is exothermic.
6 The reaction is exothermic, due to the negative enthalpy change, so increasing the
temperature means the position of equilibrium moves from right to left. There is a
greater concentration of reactants and so the denominator is bigger and Kc is smaller.
Essential Maths Skills for AS/A-level Chemistry
14
2 Handling data
Significant figures
Guided question (p.37)
1 Step 1:
3478906
Step 2:
34
Step 3 and 4: the number after 7 is 8, which is greater than 5, so round up and make
all the remaining digits zeros.
3 480 000
Practice questions (p.37)
2 aFour
b Not possible to say as could be two, three or four significant figures.
cFive
dFour
eTwo
3 a 35 560
b 5.28
c400
d 442.5
e 0.000045
Guided question (p.38)
1 Step 1: Table A.4
Measurement
20.5 cm3
0.25 mol dm−3
1.2 mol dm−3
Number of significant figures
Three
Two
Two
Step 2: the least accurate measurement is to two significant figures and only two
significant figures should be given in your final answer.
v × c = 20.5 × 0.25
= 0.005125
Step 3: moles of NaOH =
1000
1000
Essential Maths Skills for AS/A-level Chemistry
15
Therefore, moles of HCl = 0.005125.
Step 4: moles × 1000 = 0.005125 × 1000
conc.
1.2
Step 5: 4.3 cm3 to 2 s.f.
Practice questions (p.39)
2 The mass has four significant figures and the volume has two significant figures,
hence the density should be given to two significant figures.
d = m
v = 7.945098039 = 7.9 (2 s.f.)
2.2
3 moles = mass
Mr = 40.1 = 0.05486
In the original data the mass was only given to two significant figures, hence the
answer must be given to two significant figures.
0.05486 = 0.055 (2 s.f.)
4 Table A.5
Measurement
26.5 cm3
0.200 mol dm−3
0.300 mol dm−3
The least accurate measurement is to three significant figures and so three significant
figures should be given in your final answer.
moles of NaOH = v × c = 26.5 × 0.200 = 0.00530
1000
1000
moles of HCl = 0.00530
v = 0.00530 × 1000 = 17.6666 = 17.7 cm 3 to 3 s.f.
0.300
5 K c =
Number of significant figures
Three
Three
Three
2.9 =
[C]
[A][B]2
0.175
0.2 × [B]2
(2.9 × 0.2)[B]2 = 0.175
[B]2 = 0.175 = 0.3017
0.58
0.58 × [B]2 = 0.175
[B] = 0.5493 = 0.5 mol dm−3 to 1 s.f.
6 Since the density is 1.00 and d = m
v then the mass and volume are equal,
3
25.00 g = 25.00 cm .
moles of acid = 25.00 × 1.00 = 0.0250 mol
1000
moles of alkali = 25.00 × 1.00 = 0.0250 mol
1000
Essential Maths Skills for AS/A-level Chemistry
16
1 mole of H2O is produced for each mole of NaOH and HCl.
So the moles of water formed = 0.0250
Volume of solution = 50.0 cm3
As density is assumed to be 1.00 g cm−3, the mass of solution = 50.0 g.
q = mcΔT = 50.00 × 4.18 × 9.2 = 1922.8 J for 0.0250 moles of water
1922.8
Energy change per mol of water formed =
= 76 912 J mol−1
0.0250
In this calculation the temperature is given to the least accuracy — two significant
figures, hence the answer must be given to two significant figures.
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
76 912 = 77 000 J mol−1
7 Table A.6
Measurement
0.120
25.0
0.400
4.00
4.8
Number of significant figures
Three
Three
Three
Three
Two
The precision of the answer should be to two significant figures as this is the number
of significant figures in the least accurate measurement (4.8).
Guided questions (p.41)
1 Step 1: there are five significant figures.
Step 2: 2.3040 × 10n
Step 3: 2.3040 × 103
2 Step 1 and 2: Table A.7
Number
0.0060
50.08
30.0 g
0.04070
Number of significant
figures
Two
Four
Three
Four
Number in standard
form
6.0 × 10−3
5.008 × 101
3.00 × 101
4.070 × 10−2
Practice questions (p.42)
3 a 5.0 × 10−2
b 1.2 × 10−1
c 1.230010 × 106
d 1.4050 × 104
e 3.003 × 101
Essential Maths Skills for AS/A-level Chemistry
17
4 q = mcΔT = 100.0 × 4.18 × 12.5 = 5225 J
Answer must be given to three significant figures as this is the least accurate reading.
5230 J
Written in standard form this is:
5.23 × 103 J = B
Arithmetic mean
Guided question (p.43)
1 Step 1: 20.35 is an outlier, as it is distant from the other results. The other three are
within ± 0.10 cm3 of each other.
Step 2: 19.05 + 19.00 + 19.10 = 57.15
Step 3: 57.15 ÷ 3 = 19.05 cm3
Practice questions (p.43)
2 a Table A.8
Rough
Titration 1
Titration 2
Titration 3
Initial burette
reading/cm3
0.00
14.00
0.00
15.30
Final burette
reading/cm3
13.00
26.50
12.45
28.00
Titre/cm3
13.00
12.50
12.45
12.70
Outlier is 12.70.
mean titre = 12.50 + 12.45 = 12.48cm 3
2
b Table A.9
Rough
Titration 1
Titration 2
Titration 3
Final burette
reading/cm3
4.70
8.65
11.85
16.80
Initial burette
reading/cm3
0.20
4.65
7.65
12.85
Titre/cm3
4.50
4.00
4.20
3.95
Titre 2 is an outlier.
Mean titre = 4.00 + 3.95 = 3.98cm 3
2
Essential Maths Skills for AS/A-level Chemistry
18
c Table A.10
Rough
Titration 1 Titration 2 Titration 3 Titration 4
Final burette reading/cm3
23.16
45.40
22.55
45.20
22.50
Initial burette
reading/cm3
0.01
23.15
0.25
22.50
0.30
23.15
22.25
22.30
22.70
22.20
✓
✓
Titre/cm3
Concordant
titres (✓)
✓
Mean titre = 22.25 + 22.30 + 22.20 = 22.25cm 3
3
Guided question (p.45)
1 Step 1:
79 + 10 + 11 = 100
Step 2:
Ar =
(24 × 79) + (25 × 10) + (26 × 11)
= 2432 = 24.32 = 24.3 to 1 d.p.
100
100
Practice questions (p.46)
2 Ar =
=
3 Ar =
=
4 Ar =
Σ (mass of isotope × relative abundance)
Σ relative abundance
(96.9 × 40) + (0.6 × 42) + (0.2 × 43) + (2.3 × 44)
= 40.11
100
Σ (mass of isotope × relative abundance)
Σ relative abundance
(5.85 × 54) + (91.76 × 56) + (2.12 × 57) + (0.28 × 58)
= 55.92
100
Σ (mass of isotope × relative abundance)
Σ relative abundance
(32 × 95.02) + (33 × 0.76) + (34 × 4.22)
= 32.09
100
Essential Maths Skills for AS/A-level Chemistry
19
Identifying uncertainties
in measurements
Guided question (p.48)
1 Step 1: titre value = 22.55 – 0.05 = 22.50 cm3
Step 2: overall uncertainty = 2 × 0.05 = 0.1 cm3
2 × 0.05 × 100 = 0.44%
Step 3: percentage uncertainty =
22.55
Practice questions (p.48)
2 percentage uncertainty =
uncertainty × 100
quantity measured
percentage uncertainty = 0.2 × 100 = 0.08%
250
3 percentage uncertainty =
uncertainty × 100
quantity measured
percentage uncertainty = 0.06 × 100 = 0.24%
25.0
4 percentage uncertainty =
uncertainty × 100
quantity measured
percentage uncertainty = 0.05 × 100 = 0.22%
22.35
5 percentage uncertainty =
uncertainty × 100
quantity measured
percentage uncertainty = 1 × 100 = 4% (measuring cylinder)
25
percentage uncertainty = 0.01 × 100 = 2.5% (balance)
0.4
The measuring cylinder contributed most to the measurement errors.
6 Use the following equation to calculate the percentage uncertainty:
uncertainty × 100
percentage uncertainty =
quantity measured
A percentage uncertainty = 0.1 × 100 = 0.4%
25
B percentage uncertainty = 1 × 100 = 5%
20
C percentage uncertainty = 0.001 × 100 = 0.31%
0.320
D percentage uncertainty = 0.1 × 100 = 0.12%
83.2
Answer D
Essential Maths Skills for AS/A-level Chemistry
20
7 35.3 − 22.3 = 13.0 °C
% error = 0.1 × 100 = 0.77%
13.0
8 percentage uncertainty =
uncertainty × 100
quantity measured
percentage uncertainty = 2 × 0.05 × 100 = 0.81%
12.40
9 Change in mass = 1.14 g
percentage uncertainty =
uncertainty × 100
quantity measured
percentage uncertainty = 2 × 0.001 × 100 = 0.18%
1.14
10 percentage uncertainty =
uncertainty × 100
titre value
percentage uncertainty = 0.15 × 100 = 0.77%
19.45
Essential Maths Skills for AS/A-level Chemistry
21
3 Algebra
Understanding symbols
Practice question (p.50)
1 a False
b True
c True
d True
e True
f False
g True
h False
Changing the subject of an equation
Guided questions (p.51)
1 Step 1: 3x + 6 = y − 3
Step 2: 3x + 6 − 6 = y – 3 – 6
3x = y − 9
y−9
Step 3: x =
3
2 Step 1:
[CH 3COOC2 H 5 ][H 2O]
= Kc
[C2 H 5OH][CH 3COOH]
Step 2:
[CH 3COOC2 H 5 ][H 2O] × [C2 H 5OH][CH 3COOH]
= K c [C2 H 5OH][CH 3COOH]
[C2 H 5OH][CH 3COOH]
Step 3: [CH 3COOC2 H 5 ] [H 2O] = K c [C2 H 5OH][CH 3COOH]
[H 2O]
[H 2O]
[CH 3COOC2 H 5 ] =
K c [C2 H 5OH][CH 3COOH]
[H 2O]
Essential Maths Skills for AS/A-level Chemistry
22
3 Step 1: k[NO]2[O2] = rate
k [NO]2 [O 2 ]
= rate
Step 2:
k [O 2 ]
k[O 2 ]
Step 3: [NO] =
rate
k [O 2 ]
Practice questions (p.52)
4 a y − 1 = x
2
b 4 + y = x
3
c 8 − y = x
d
y−c
=x
m
e
3 − 2y
=x
4
f x = −1 − 2y
g x + 1 = 3 ; x = 3 − 1 OR x = 3 − y
y
y
y
y
h 2 = x
z
w
y
i x =
5 a mass = moles × Mr
b vol (dm3) = moles × 24
c T =
d theoretical yield =
e c =
f [C] =
g [B] =
[C]2
[A]2 K c
h [C] =
Rate
k [A][B]
i [ A] =
PV
nR
v × 1000
n
actual yield × 100
percentage yield
[A][B]
[D] K c
Rate
k
j T = ∆H ° – ∆G °
∆S °
Essential Maths Skills for AS/A-level Chemistry
23
Solving algebraic equations
Guided question (p.54)
1 Step 1:
∑∆ H
f
reactants = (2 × –46.0) + (2 × 0.0) = –92 kJ mol−1
Step 2:
∑ ∆H f products = (1 × 82.0) + (3 × –286.0) = –776.0 kJ mol−1
Step 3:
∆H r =∑ ∆H f products –
= −92 – (–776) = −92 + 776 = +684 kJ mol−1
∑ ∆H
f
reactants
Practice questions (p.54)
2 To calculate enthalpy of formation given combustion values use the equation:
∆H f =
∑ ∆H
c
reactants –
∑ ∆H
c
products
The equation is 3C(s) + 4H2(g) → C3H8(g)
The reactants are 3C(s) + 4H2(g)
∑ ∆H
The product is C4H10(g)
∑ ∆H
c
c
reactants =(3 × –394) + (4 × –286) = –2326kJmol−1
products = –2219 kJ mol−1
∆H f =
∑ ∆H
c
reactants –
∑ ∆H
products
c
= −2326 – (–2219)
= −2326 + 2219 = −107 kJ mol−1
3 The reactants are C4H9OH(l) + 6O2(g)
∑ ∆H
The products are 4CO2(g) + 5H2O(g)
∑ ∆H
f
f
reactants = (1 × –327) + (6 × 0.0) = –327 kJ mol–1
products = (4 × –394) + (5 × –286) = –1576 – 1430 = –3006 kJ mol−1
∆H c = ∑ ∆H f products –
∑ ∆H
f
reactants
= −3006 – (−327) = −3006 + 327 = −2679 kJ mol−1
Essential Maths Skills for AS/A-level Chemistry
24
4 To calculate enthalpy of formation given combustion values use the equation:
∆H f =
∑ ∆H
∑ ∆H
reactants –
c
The reactants are 6C(s) + 7H2(g)
∑ ∆H
The product is C6H14(g)
c
products
reactants = (6 × –394) + (7 × –286) = –4366 kJ mol−1
∑ ∆H
∆H f = ∑ ∆H c reactants –
= −4366 − (−4163)
= −203 kJ mol−1
c
c
products = –4163 kJ mol−1
∑ ∆H
c
products
Guided questions (p.56)
v×c =n
1000
Step 2: v × c × 1000 = n × 1000
1000
v × c = n × 1000
v × c = n × 1000
Step 3:
v
v
c = n × 1000
v
0.0034
×
1000
Step 4: c =
= 0.23 moldm –3 (to 2 s. f.)
15.0
1 Step 1:
2 Step 1:
[C]2 [D]
= Kc
[A][B]2
Step 2: [C]2 [D] = K c [A][B]2
Step 3: [C] [D] =
[D]
K c [A][B]2
[D]
[C]2 =
K c [A][B]2
[D]
2
Step 4: [C] =
K c [A][B]2
[D]
Practice questions (p.57)
3 [HI]2 = Kc × [H2] × [I2]
[HI]2 = 44.0 × 0.20 × 0.20
[HI]2 = 1.76
[HI] = 1.76 = 1.3 mol dm−3
4 [C]2 = Kc × [A]2 × [B]
[C]2 = 44.9 × (1.17)2 × 0.34
[C]2 = 20.8976 mol
[C] =
20.8976 = 4.571 = 4.6 mol dm−3 (to 2 s.f.)
Essential Maths Skills for AS/A-level Chemistry
25
5 Temperature change = 7.6 °C
q = mcΔT
q = 25.0 × 4.18 × 7.6 = 794.2 J
n = mass = 6.00 = 0.080
Mr
74.6
794.2
= 9927.5 J mol−1 = 9.9275 kJ mol−1
0.080
The answer should be to three significant figures i.e. 9.93 kJ mol−1.
enthalpy change per mole =
6 rate = k[Y][Z]2
k[Y][Z]2 = rate
k = rate 2
[Y][Z]
–5
k = 7.4 × 10 –6 = 7.56
9.792 × 10
k = 7.56 mol−2 dm6 s−1
moldm –3 s –1
s –1
=
= mol–2 dm 6 s –1
–3
–3 2
–3 2
(moldm
)
(moldm
)(moldm
)
Guided question (p.58)
1 Step 1:
Ea = 115 kJ mol−1. The units must be converted into J mol−1 by multiplying by 1000
Ea = 115 000 J mol−1
T = 298 K. The units are correct.
R = 8.31 J K−1 mol−1. The units are correct.
A = 2.1 × 1012 s−1
Step 2:
Ea
115000
=
= 46.4388
× 298
RT
8.31
Step 3:
k = 2.1 × 1012 × e−46.4388
k = 2.1 × 1012 × 6.790 × 10−21 = 1.43 × 10−8 s−1
Practice questions (p.59)
2 Ea = 50 kJ mol−1. The units must be converted into J mol−1 by multiplying by 1000.
Ea = 50 000 J mol−1
T = 20 °C. The units must be changed to K by adding 273, therefore T = 293 K.
R = 8.31 J K−1 mol−1; the units are correct.
A = 2.1 × 1012 s−1
At this point all the quantities are in the correct units.
Essential Maths Skills for AS/A-level Chemistry
26
Now you need to use the equation:
k = Ae
E
− RTa
Ea
50 000
=
= 20.535
RT 8.31 × 293
E
− a
Substitute into k = Ae RT
k = 2.1 × 1012 × e−20.535
k = 2.1 × 1012 × 1.207 × 10−9 = 2.5 × 103 s−1
3 T = 30 °C. The units must be changed to K by adding 273 = 303 K.
k = Ae
E
− RTa
Ea
50 000
=
=19.858
RT 8.31 × 303
E
− a
Substitute into k = Ae RT
k = 2.1 × 1012 × e−19.858
k = 2.1 × 1012 × 2.376 × 10−9 = 5.0 × 103 s−1
The rate constant increases with increasing temperature.
Using calculators to find logs
Guided questions (p.61)
1 pH = −log(0.60)
= 0.22
2 pKa = −log(1.72 × 10−5)
= 4.76
Practice questions (p.61)
3 a −2.0
b −1.5
c 0.3
d −3.9
e −4.7
f 2.3
4 a 2
b 0.6
c −0.1
d 2.5
e 1.5
Essential Maths Skills for AS/A-level Chemistry
27
5 a For strong monobasic acids, the proticity is 1 (they have one H+)
so [H+] = 1 × [acid].
HCl → H+ + Cl−
If [HCl] = 0.05 mol dm−3 [H+] = 0.05 mol dm−3
pH = −log[H+] = −log(0.05) = 1.3
b For strong diprotic acids, the proticity is 2 so [H+] = 2 × [acid].
H2SO4 → 2H+ + SO 2−
4
If [H2SO4] = 1.0 mol dm−3 then [H+] = 2.0 mol dm−3
pH = −log[H+] = −log(2.0) = −0.3
6 pKa = −logKa = −log 5.26 × 10−7 = 6.3
Guided question (p.62)
1 Step 1:
pH = −log[H+]
0.75 = −log[H+]
−0.75 = log[H+]
Step 2: [H+] = 0.178 mol dm−3
Practice questions (p.63)
2 pKa = −logKa
2.99 = −logKa
−2.99 = logKa
On your calculator type in ‘2ndF’ or ‘shift’ or ‘inv’ then ‘log’ then ‘−2.99’. This
should give the value of Ka of 1.02 × 10−3.
An alternative way of answering this question is by using the equation Ka = 10−pKa
Ka = 10−2.99
3 a pH = −log[H+]
1.2 = −log[H+]
−1.2 = log[H+]
b pH = −log[H+]
0.063
0.2 = −log[H+]
−0.2 = log[H+]
0.63
Essential Maths Skills for AS/A-level Chemistry
28
c pH = −log[H+]
0.7 = −log[H+]
−0.7 = log[H+]
0.2
4 0.8
5 [H+] = 10(−pH) = 10(−0.91) = 0.12 mol dm−3
Nitric acid (HNO3) is a strong monobasic acid so the [acid] = [H+]
6 [H+] = 10(−pH) = 10(−1.1) = 0.08 mol dm−3
Sulfuric acid (H2SO4) is a strong diprotic acid so the
[acid] =
[H + ] 0.08
=
= 0.04 moldm –3
2
2
Essential Maths Skills for AS/A-level Chemistry
29
4 Graphs
Plotting graphs
Practice questions (p.67)
Volume of hydrogen/cm3
1
120
100
80
60
40
20
0
0
20
40
60
100
200
300
80
100
Time/s
Figure A.1
% of C in equilibrium mixture
2
70
60
50
40
30
20
10
0
0
500
400
Temperature/°C
Figure A.2
3a Result at 4.5 minutes
b Mass of flask and contents/g
cTime/mins
d A graph of mass of flask and contents against time for the reaction of calcium
carbonate and acid.
e Yes, as it fits most of the graph paper.
Essential Maths Skills for AS/A-level Chemistry
30
The slope and intercept of a linear graph
Guided question (p.70)
1 a Step 1: choose two points which are far apart on the line, these will form the
hypotenuse of the triangle.
Step 2: complete the triangle as shown on each graph in Figure A.3.
Step 3: find the ∆y (rise) value.
Step 4: find the ∆x (run) value.
Step 5: to find the gradient of each line, use the equation
change in y -axis ∆y
=
gradient (m) =
change in x -axis ∆x
change in y -axis ∆y 6 3
=
= =
Graph A: gradient (m) =
change in x -axis ∆x 4 2
change in y -axis ∆y 4
=
=
change in x -axis ∆x 5
change in y -axis ∆y
4
=
Graph C: gradient (m) =
=3
change in x -axis ∆x
Step 6: decide if it is a positive gradient sloping from lower left to top right or a
negative gradient sloping from top left to lower right.
Graph B: gradient (m) =
Graph A: gradient = −1.5
Graph B: gradient = +0.8
Graph C: gradient = −1.3
b Step 1: write down the number at which the blue line cuts through the y-axis
(at x = 0). This is the intercept, c.
Step 2: substitute the values for m and for c into the equation y = mx + c.
3
Graph A: intercept = −2.5; equation = y = − x − 2.5
2
Graph B: intercept = −1; equation = y = 4 x − 1
5
Graph C: intercept = −0.5; equation = y = − 4 x − 0.5
3
Essential Maths Skills for AS/A-level Chemistry
31
A
B
5
4
3
2
1
C
y
−5 −4 −3 −2−1 O
−1
−2
−3
−4
−5
5
4
3
2
1
1 2 3 4 5x
−5 −4 −3 −2−1 O
−1
−2
−3
−4
−5
y
5
4
3
2
1
1 2 3 4 5x
−5 −4 −3 −2−1 O
−1
−2
−3
−4
−5
y
1 2 3 4 5x
Slope = – 3
Slope = 4
Slope = – 4
y = – 3 x – 2.5
2
y = 4x – 1
5
y = – 4 x – 0.5
3
2
5
3
Figure A.3
Practice questions (p.71)
2 A positive
B negative
C positive
D zero
E negative
F negative
G positive
3 aA 0.6
D −0.75
B – 0.66
C 1.25
E 2.3
F 1.3
( Note: These answers are approximate values. The gradients are determined by
the triangles you use to calculate them.)
bA −2
B −1
C −3
D −3
E −3
F0
c A y = 0.8x – 2
B y = −0.6x − 1
C y = 1.25x − 3
D y = −0.8x − 3
E y = 2.3x − 3
F y = 1.3x
Guided question (p.72)
1 Table A.11
Units on y-axis
Units on x-axis
Units of gradient
mol dm−3
min
moldm –3
= moldm –3 min –1
min
g dm−3
s
g
cm3
g dm –3
= g dm –3 s –1
s
g
= g cm −3
cm 3
Essential Maths Skills for AS/A-level Chemistry
32
Practice question (p.73)
2 a mol dm−3 s−1
b s−1
c g s−1
d cm3 s−1
Calculating the rate of change
Guided question (p.74)
1 a This is a linear straight-line graph, hence the gradient (rate) is constant and does
not change as concentration changes. It is a zero order reaction and the gradient of
the graph represents the rate.
b Step 1: to find the gradient, choose two points far apart on the line, and form a
triangle as shown on the graph below.
[A]/mol dm–3
0.20
0.15
0.10
0.05
0.00
0
20
40
60
Time/s
Figure A.4
change in y -axis ∆y 0.145
=
=
= 0.0029
change in x -axis ∆x
50
Step 3: to find units, use the expression
y -axis units mol dm –3
units of gradient =
=
= mol dm –3s –1
s
x -axis units
Step 2: gradient (m) =
The answer is 0.0029 mol dm−3 s−1.
Essential Maths Skills for AS/A-level Chemistry
33
2
Rate/mol dm–3 s–1
Practice questions (p.76)
0.0125
0.0100
0.0075
0.0050
0.0025
0
0
0.01
0.02
Figure A.5
gradient = 0.0125 = 0.516 s –1
0.05
0.03
0.04
0.05
[X]/mol dm–3
3 a The order is zero.
0.300 = 1.2 × 10 –4 moldm –3 s –1
b rate = gradient =
2500
Tangents and measuring the rate
of change
Guided question (p.79)
1 Step 1 and 2:
5
y
4
3
2
C
Δy = rise = 2 – 1 = 1
A (2, 1.5)
1
B Δ x = run = 3 – 1 = 2
x
0
0
1
2
3
4
Figure A.6
Step 3: gradient (m) =
5
6
change in y -axis ∆y 1
=
=
change in x -axis ∆x 2
Essential Maths Skills for AS/A-level Chemistry
34
2 [Ester]/mol dm–3
Practice questions (p.80)
0.35
0.30
Concentration of
0.200 mol dm–3
0.25
0.20
Tangent to the curve
at 0.200 mol dm–3
0.15
0.240
0.10
0.05
3500
0.0
0
500 1000 1500 2000 2500 3000 3500 4000 4500 5000
Time/s
Figure A.7
At a concentration of 0.200 mol dm−3, the gradient of the tangent is
0.240
= 6.86 × 10−5 mol dm−3 s−1
3500
3a
Volume of carbon dioxide/cm3
70
60
50
40
30
20
10
0
0
20
30
40
50
60
70
80
90 100
Time/s
Figure A.8
Volume of carbon dioxide/cm3
10
70
c
60
b
50
40
30
20
10
0
0
10
20
30
40
50
60
70
80
90 100
Time/s
Figure A.9
b gradient of tangent = 30 1.1 cm3 s−1
28
Rate is about 1.1 cm3 s−1
c gradient of tangent 10 = 0.3 cm3 s−1
34
Rate is about 0.3 cm3 s−1
Essential Maths Skills for AS/A-level Chemistry
35
Guided question (p.83)
1 Step 1 and 2:
Table A.12
278
288
298
308
318
k/mol−1 dm3 s−1
0.013
0.530
1.04
1.67
2.14
ln k
−4.34
T/K
3.60 × 10
1/T K−1
Step 3:
3.47 × 10
0.04
−3
3.36 × 10
0.51
−3
3.25 × 10
0.76
−3
3.14 × 10−3
1.5
In k
−0.63
−3
1
0.5
0
0.0031 0.0032 0.0033
0.0034 0.0035 0.0036 0.0037
–0.5
–1
–1.5
1 × 10–3 /K–1
T
Figure A.10
Step 4: gradient (m) = –
Step 5: Ea = −mR
Ea
; from the graph m = –2.0 −3 = – 588.24
R
3.4 × 10
Step 6: Ea = −588.24 × 8.31 = −4888.3 J mol−1
Step 7: Ea = −4888.3 kJ mol−1
Step 8: The y-axis intercept value is 1.2 and this is equal to ln A.
Step 9: ln A = 1.2
e1.2 = 3.3 = A
Practice questions (p.84)
2 Table A.13
T/°C
k/s−1
T/K
1/T K−1
ln k
327
427
527
0.00034
0.0031
0.029
600
700
800
1.67 × 10−3 1.43 × 10−3 1.25 × 10−3
−8.0
−5.8
−3.5
627
0.100
900
1.11 × 10−3
−2.3
727
0.380
1000
1.0 × 10−3
−1.0
Essential Maths Skills for AS/A-level Chemistry
36
In k
0.50
0.00
0.70
0.90
1.10
1 × 10–3 /K–1
T
1.50
1.70
1.90
1.30
–1.00
–2.00
–3.00
–4.00
–5.00
–6.00
–7.00
–8.00
–9.00
Figure A.11
gradient (m) = –
Ea
R
Rearranging this equation: mR = −Ea; Ea = −mR = −8000 × 8.31 = −66 480 J mol −1
3 Table A.14
` T/°C
k/mol dm s
T/K
1/T /K−1
ln k
In k
−1
3
−1
55
5.7 × 10−8
328
0.0030
−16.7
85
1.6 × 10−4
358
0.0028
−8.7
125
3.9 × 10−2
398
0.0025
−3.2
205
7.8 × 10−1
478
0.0021
−0.2
305
6.3
578
0.0017
1.8
5
0
1
1.5
2
2.5
3
3.5
–5
–10
–15
–20
1 × 10–3 /K–1
T
Figure A.12
gradient (m) =
–25
= –14 285
1.75 × 10 –3
−Ea = −14 285 × 8.31
Ea = 118 708 J mol−1 = 118.71 kJ mol−1
Essential Maths Skills for AS/A-level Chemistry
37
5 Geometry and trigonometry
Representing shapes of
molecules in 2D and 3D form
Guided question (p.87)
1 There are 4 fluorine atoms bonded, so there are 4 electrons coming from the fluorine
atoms giving a total of 12 electrons.
■■ 12 electrons = 6 electron pairs
■■ 6 electron pairs – 4 bonded atoms = 2 lone pairs
In BrF4− there are 4 bonding pairs and 2 lone pairs so the shape is square planar.
Practice questions (p.87)
2 a Tetrahedral
b Pyramidal
c Bent
d Tetrahedral
e Pyramidal
f Pyramidal
3 a +
H
N
H
b
H
Figure A.13
F
–
B
H
F
F
F
Figure A.14
Essential Maths Skills for AS/A-level Chemistry
38
c
+
O
H
H
H
Figure A.15
Guided question (p.89)
1 Step 1: there are three H atoms and CHClCHOHCHO, so it is not a chiral centre as
there are not four different atoms or groups attached.
Step 2: CH3, H, Cl, CHOHCHO. There are four different groups, so it is a chiral
centre.
Step 3: CH3CHCl, H, OH, CHO. There are four different groups, so it is a chiral
centre.
Step 4: there are two chiral centres.
Practice questions (p.90)
2 a
H
H
C
C
H2N
COOH
CH3
b no chiral centre
c no chiral centre
d
NH2
Figure A.16
H
H
C
C
H2N
HOOC
H 3C
Cl
COOH
HOOC
Cl
NH2
Figure A.17
Essential Maths Skills for AS/A-level Chemistry
39
Exam-style question answers
AS and A-level questions
1 3 marks awarded, as below:
To calculate percentage atom economy, use the equation:
% atom economy =
molecular mass of desired product × 100
sum of molecular masses of all reactants
The desired product is iron and according to the equation 2Fe are produced, so the
molecular mass of iron is (2 × 56.0) = 112.0. (1 mark)
The reactants are Fe2O3 and 3CO so the sum of the molecular masses of the
reactants is:
[(2 × 56.0) + (3 × 16.0)] + [3 × (12 + 16)] = (112.0 + 48) + (3 × 28)
= 160 + 84 = 244 (1 mark)
% atom economy =
molecular mass of desired product × 100
sum of molecular masses of all reactants
= 112.0 × 100 = 45.9% (1 mark)
244.0
2 3 marks awarded, as below:
The desired product is hydrazine and its molecular mass (Mr) is
(2 × 14.0) + (4 × 1.0) = 32.0 (1 mark)
The reactants are 2NH3 and NaOCl so the sum of the molecular masses of the
reactants is:
[(2 × 14.0) + (6 × 1.0)] + (23.0 + 16.0 + 35.5) = 34 + 74.5 = 108.5 (1 mark)
% atom economy =
molecular mass of desired product × 100
sum of molecular masses of all reactants
= 32.0 × 100 = 29.5% (1 mark)
108.5
3 3 marks awarded, as below:
moles of KMnO4 = 5.53 ÷ 158 = 0.035 (1 mark)
the ratio is 2 moles KMnO4 : 1 mole O2
moles of O2 = 0.035 ÷ 2 = 0.0175 (1 mark)
0.0175 = mass ÷ Mr
Mr = 32.0 (1 mark)
mass of O2 = 0.0175 × 32.0 = 0.56 g
Essential Maths Skills for AS/A-level Chemistry
40
4 a 3 marks awarded, as below:
number of moles of Pb3O4 = 2.74 = 0.04 (1 mark)
685.6
ratio of Pb3O4 : PbO = 2 : 6 = 1 : 3
moles of PbO = 0.004 × 3 = 0.012 (1 mark)
mass of PbO = 0.012 × 223.2 = 2.678 g (1 mark)
b 2 marks awarded, as below:
ratio Pb3O4 : O2 = 2 : 1
moles of O2 = 0.004 = 0.002 (1 mark)
2
volume of O2 = 0.002 × 24 = 0.048 dm3 (1 mark)
5 5 marks awarded, as below:
50.0 × 1
moles of HCl =
= 0.05 (1 mark)
1000
ratio HCl : M(OH)2 = 2 : 1
moles of M(OH)2 = 0.05 = 0.025 (1 mark)
2
M(OH)2
0.025 moles = mass of
Mr
3 − 0.55
0.025 =
Mr
2.45
0.025 =
Mr
Mr × 0.025 = 2.45 (1 mark)
2.45
= 98
(1 mark)
Mr =
0.025
Ar of M = 98 – [(2 × 16) + (2 × 1)] = 98 – 34 = 64 (1 mark)
Identity of M = Cu (1 mark)
6 a 2 marks awarded, as below:
0.2 = 0.005
moles of Ca =
40.0
ratio 1Ca : 1H2 ⇒ 0.005 : 0.005 (1 mark)
b 2 marks awarded, as below:
moles of Ca = 0.2 = 0.005
40.0
ratio 1Ca : 2HNO3 ⇒ 0.005 : (2 × 0.005) = 0.01 moles of HNO3 (1 mark)
n= c × v
1000
0.01= 2 × v
1000
0.01 × 1000 = 2 × v
10 = 2v
v = 5 cm3 (1 mark)
hydrogen is a gas, so use the equation
n= v
24
0.005 = v
24
v = 0.005 × 24 = 0.12 dm3 (1 mark)
Essential Maths Skills for AS/A-level Chemistry
41
c 2 marks awarded, as below:
Mr Ca(NO3)2 = 40.0 + [(14.0 + (3 × 16.0)) × 2] = 164.0 (1 mark)
%N = 2 × 14.0 × 100 = 17.1 (1 mark)
164.0
7 4 marks awarded, as below:
The result of 22.8 is an outlier, it is not concordant and should not be used in the
calculation of the average titre.
22.4 + 22.5
average titre =
= 22.45 cm 3 (1 mark)
2
moles of HCl = v × c = 22.45 × 0.2 = 0.00449 (1 mark)
1000
1000
ratio 2HCl : 1Ba(OH)2
There is twice as much HCl so divide the moles by 2.
0.0449 = 0.002245 (1 mark)
2
0.002245= v × c = 25.0 × c
1000
1000
0.002245 × 1000 = 25.0 × c
2.245 = 25.0 c
c = 2.245 = 0.0898 mol dm−3 (1 mark)
25.0
8 3 marks awarded, as below:
31.25 × 0.16 = 0.005
(1 mark)
moles of Na 2CO3 =
1000
ratio Na2CO3 : H2SO4 = 1 : 1
moles of H2SO4 = 0.005 (1 mark)
n= c × v
1000
n × 1000 = c
v
0.005 × 1000
= 0.2 moldm –3 (1 mark)
25.0
9 a 3 marks awarded, as below:
moles of NaHCO3 = 3.36 = 0.04 (1 mark)
84
moles of Na 2CO3 = 0.04 = 0.02 (1 mark)
2
mass of Na2CO3 = 0.02 × 106 = 2.12 g (1 mark)
b 2 marks awarded, as below:
moles of CO2 = 0.02 (1 mark)
Carbon dioxide is a gas so use the equation:
moles = volume
24
volume of CO2 = 0.02 × 24 = 0.48 dm3 (1 mark)
Essential Maths Skills for AS/A-level Chemistry
42
6.0 = 0.08108
74.0
ratio 1 C4H9OH : 1 CH3CH2COOC4H9
0.08108 =
Mr = 130
0.08108 × 130 = mass of ester = 10.54 g
% yield = 7.4 × 100 = 70.2%
10.54
The answer is B.
10 moles of C4 H 9OH =
mass of ester
Mr
(1 mark)
11 K c =
[SO3 ]2
[SO 2 ]2 [O 2 ]
The units of concentration as indicated by the square brackets are mol dm−3.
(moldm –3 )2
1
Kc =
=
= mol–1 dm 3
(moldm –3 )2 (moldm –3 ) mol dm -3
The answer is C.
(1 mark)
12 a Table A.15
moles of P = 0.775 = 0.025
31.0
Ratio (÷ 0.025) 1
2
moles of oxygen = 1 = 0.0625
16.0
2.5
5
The ratio works out at 1 : 2.5 but both are multiplied by 2 to give whole numbers.
Therefore the molecular formula is P2O5.
3 marks awarded: 1 for each correct row in the table
b Mr of P2O5 = 142 and the Mr of the oxide is 284. So 2 × P2O5 must be present in the
compound, therefore the molecular formula is P4O10.
1 mark awarded for the correct answer
13 First calculate the number of electrons around the central atom in CCl2F2. C has 4 outer
shell electrons, add on 4 for the 4 bonded halogen atoms, gives 8 electrons, which is
4 pairs. There are 4 bonded pairs due to the 4 halogens and no lone pairs. The 4 bonding
pairs of electrons, which will repel each other equally, result in a tetrahedral shape.
Chlorine has 7 outer shell electrons and adding on the 3 bonded electrons from
fluorine is 10, which gives 5 electron pairs; take away the 3 bonded electrons pairs,
which give 2 lone pairs. ClF3 has 3 bonding pairs of electrons and 2 lone pairs of
electrons. The lone pairs will repel more than the bonding pairs and the shape is T
shaped. Trigonal planar is also accepted.
2 marks awarded; 1 for each correct shape
Essential Maths Skills for AS/A-level Chemistry
43
14 3 marks awarded, as below:
actual yield × 100
theoretical yield
4.0 × 100
50 =
theoretical yield
theoretical yield = 4.0 × 100
50
400
theoretical yield =
= 8.0 g (1 mark)
50
moles of NO = mass of NO
Mr
percentage yield =
Mr = 30.0
moles of NO = 8.0 = 0.27
30.0
ratio 4NH3 : 4NO
Hence there are 0.27 moles of NH3 (1 mark)
mass of NH 3
moles of NH 3 =
Mr
Mr = (1 × 14.0) + (3 × 1.0) = 17.0
mass of NH 3
0.27 =
17.0
0.27 × 17.0 = 4.6 g (1 mark)
15 2 marks awarded, as below:
[CO][H 2 ]3
(1 mark)
Kc =
[CH 4 ][H 2O]
3
K c = 0.17 × 0.51 = 0.02255 = 0.29 (1 mark)
0.14 × 0.55
0.077
The units are
(mol dm –3 )4
= mol2 dm –6 .
(mol dm –3 )2
16 3 marks awarded, as below:
[E] (1 mark)
[A]3[B]2
Kc =
The equilibrium concentrations are calculated by dividing the amount, in moles, at
equilibrium by the volume of the container in dm3.
[A] = 28.0 = 0.2 moldm –3
140
[B] = 113.4 = 0.81 moldm –3 (1 mark)
140
Rearranging the equilibrium expression for Kc to make [E] the subject:
K c [A]3 [B]2 = [E] = 96.2 × 0.23 × 0.812
[E] = 0.50 mol dm−3 (1 mark)
Essential Maths Skills for AS/A-level Chemistry
44
17 2 marks awarded, as below:
Initial burette reading = 0.05 cm3.
Final burette reading = 26.35 cm3
The overall error in any volume measured in a burette always comes from the two
measurements, so the overall error = 2 × 0.05 cm3 = 0.1 cm3. (1 mark)
Titre value = 26.35 − 0.05 = 26.30 cm3
Percentage uncertainty = 2 × 0.05 × 100 = 0.38% (1 mark)
26.30
A-level only questions
1 6 marks awarded, as below:
A general rate equation would be: rate = k[NO]x[O2]y
From experiments 1 and 2: [NO] ×2; [O2] ×1; rate ×4
(×2)x = (×4)
factor by which
rate has increased
factor by which
concentration of
NO has increased
order (x) with respect to NO
which we are trying to
determine
Figure A.18
So the order with respect to NO is 2 as (×2)2 = (×4) (1 mark)
From experiments 2 and 4: [NO] ×1; [O2] ×2; rate ×2
(×2)y = (×2)
factor by which
rate has increased
order ( y) with respect to O2
which we are trying to
determine
factor by which
concentration of
O2 has increased
Figure A.19
So the order with respect to O2 is 1 as (×2)1 = (×2) (1 mark)
Therefore the rate equation is: rate = k[NO]2[O2] (1 mark)
To calculate a value of the rate constant at this temperature, substitute in any values
from the table. From experiment 1: [NO] = 4 × 10−3 mol dm−3; [O2] = 1 × 10−3 mol dm−3;
rate = 6 × 10−4 mol dm−3 s−1.
So substituting the values from experiment 1 into the rate equation gives:
6 × 10−4 = k(4 × 10−3)2(1 × 10−3)
6 × 10−4 = k(1.6 × 10−8)
−4
k = 6 × 10 −8 = 37 500 (1 mark)
1.6 × 10
Essential Maths Skills for AS/A-level Chemistry
45
The units for the rate constant are:
mol dm –3 s –1 = k (mol dm –3 )3 2
s –1 = k (mol dm –3 )2
k=
s –1
= 37 500 s –1 mol–2 dm 6 (1 mark)
–3 2
(mol dm )
rate = 37 500[NO]2[O2] (1 mark)
2 5 marks awarded, as below:
Generally, rate = k[A]x[B]y. From experiments 1 and 2: [A] ×3, [B] ×1, rate ×3
(×3)x = (×3)
factor by which
rate has increased
order (x) with respect to A
which we are trying to
determine
factor by which
concentration of
A has increased
Figure A.20
So the order with respect to A is 1 as (×3)1 = (×3) (1 mark)
There are no two experiments in which the concentration of A does not change so
you can choose any two experiments now that you know the order of reaction with
respect to A.
From experiments 1 and 3: [A] ×5, [B] ×0.5, rate ×1.25.
(×5)1 (×0.5)y = (×1.25)
factor by which
rate has increased
order (y) with respect
to B which we are
trying to determine
factor by which
concentration of
A has increased
order with
respect to
A=1
factor by which
concentration of
B has increased
Figure A.21
This simplifies to: ×5 (×0.5)y = ×1.25
×1.25
= × 0.25
×5
y=2
( × 0.5) y =
The order of reaction with respect to B is 2. (1 mark)
The rate equation is rate = k[A][B]2. (1 mark)
To calculate a value of the rate constant at this temperature, substitute in any values
from the table. From experiment 1: [A] = 1.7 × 10−2 mol dm−3; [B] = 2.4 × 10−2 mol dm−3;
rate = 7.40 × 10−5 mol dm−3 s−1.
So substituting these values from experiment 1 into the equation:
7.40 × 10 –5 = k (1.7 × 10 –2 )(2.4 × 10 –2 )2
7.40 × 10 –5 = k (9.792 × 10 –6 )
k=
7.4 × 10 –5
= 7.56 (1 mark)
9.792 × 10 –6
Essential Maths Skills for AS/A-level Chemistry
46
The units for the rate constant are:
mol dm –3 s –1 = k (mol dm –3 )(mol dm –3 )2
s –1 = k (mol dm –3 )2
k=
s –1
= 7.56 s –1 mol–2 dm 6 (1 mark)
(mol dm –3 )2
3 rate = k[CH3COCH3][H+]
mol dm –3 s –1 = k (mol dm –3 )(mol dm –3 )
s−1 = k mol dm−3
k = s−1 mol−1 dm3
The answer is B: mol−1 dm3 s−1.
1 mark awarded for the correct answer
4 5 marks awarded, as below:
∆G° has units of kJ mol−1, exactly like ∆H°, so the units of ∆S must be changed to
kJ K−1 mol−1. This is achieved by dividing the ∆S° value by 1000.
So ∆S° is + 0.1537 kJ K−1 mol−1 (1 mark)
∆G° = ∆H° − T∆S°
∆G° = 323 – (1250 × 0.1537) = +131 kJ mol−1 (1 mark)
Because ∆G° is positive at this temperature the reaction is not feasible. (1 mark)
∆G° = 0 when ∆H° − T∆S° = 0 (1 mark)
+323 – T(0.1537) = 0
T(0.1537) = 323
T =
T = 2102 K (1 mark)
T must be equal to or greater than 2102 K for the reaction to be feasible.
323 = 2101.4964 K
0.1537
5 4 marks awarded, as below:
For a change of state, ΔG° = 0 kJ mol−1
ΔG° = ΔH° − TΔS°
0 = −23.35 – 240(ΔS°) (1 mark)
∆S ° = –
ΔS° = ΣS°products – ΣS°reactants = S°(NH3(l)) – 193 = −97.29 J K−1 mol−1 (1 mark)
S°(NH3(l)) = −97.29 + 193 = 95.71 J K−1 mol−1 (1 mark)
23.35
= – 0.09729 kJ K –1 mol–1 = –97.29 J K –1 mol –1 (1 mark)
240
Essential Maths Skills for AS/A-level Chemistry
47
6 a 4 marks awarded, as below:
Table A.16
SO2Cl2
2
→
0.5
SO2
0
+
1.5
Cl2
0
1.5
partial pressure SO2 = (1.5 ÷ 3.5) × 150 = 64.286 (1 mark)
partial pressure Cl2 = 64.286 (1 mark)
partial pressure SO2Cl2 = (0.5 ÷ 3.5) × 150 = 21.429 (1 mark)
Kp =
[p(SO2) × p(Cl2)] (64.286)2
= 192.86 kPa (1 mark)
=
21.429
p(SO2Cl2)
b 3 marks awarded, as below:
moles of SO 2Cl2 = mass
Mr
Mr = (1 × 32.1) + (2 × 16.0) + (2 × 35.5) = 135.1
moles = 135.2 = 1 (1 mark)
135.2
1 mole SO2Cl2 : 4 moles HCl
4 moles of HCl in 1 dm3 = 4 mol dm−3 (1 mark)
pH = −log[H+ ] = −log 4 = −0.6 (1 mark)
7 3 marks awarded, as below:
pKa = 4.87 = −log Ka
Take the inverse log of 4.87 to find Ka
Ka = 1.35 × 10 –5 mol dm–3 (1 mark)
Ka =
[CH 3CH 2COO – ][H + ]
[H + ]2
=
[CH 3CH 2COOH]
[CH 3CH 2COOH]
1.35 × 10 –5 =
[H + ]2
0.05
[H+]2 = 1.35 × 10 –5 × 0.05 = 6.74 × 10 –7 take the square root of this to find [H+]
[H+] = 8.2 × 10 –4 (1 mark)
pH = –log(8.2 × 10 –4)
pH = 3.09 (1 mark)
8 2 marks awarded, as below:
[H + ]– =
K a × [weak acid] = 1.74 × 10 –5 × 0.1 = 1.74 × 10 –6
= 1.319 × 10 –3 mol dm –3 (1 mark)
pH = −log[H+] = −log (1.319 × 10−3) = 2.8798
pH = 2.88 (1 mark)
Essential Maths Skills for AS/A-level Chemistry
48
9 4 marks awarded, as below:
moles O3 at start = 10; at equilibrium = 7
moles O2 at start = 0; at equilibrium = 4.5
total moles at equilibrium = 11.5 (1 mark)
partial pressure of O3 = 10 × 7 = 6.09 (1 mark)
11. 5
partial pressure of O3 = 10 × 4.5 = 3.91 (1 mark)
11.5
10a
(3.913)3
= 1.62 atmospheres (1 mark)
(6.087)2
Concentration of Br2/mol dm–3
Kp =
0.50
0.40
0.30
0.20
0.10
0.00
0
10
20
30
40
50
60
Time/s
Figure A.22
3 marks awarded; 1 mark for x- and y-axes labelled with the correct units,
1 mark for plotting the points on the graph and 1 mark for drawing a smooth
curve through the points
b Gradient of tangent = 0.30 ÷ 50 = 0.006 (Note: this will depend on your graph,
the tangent you have drawn and the triangle you use to calculate the gradient.)
2 marks awarded; 1 mark for drawing the tangent and 1 mark for calculating
the gradient
Essential Maths Skills for AS/A-level Chemistry
49