Worked solutions to textbook questions
Chapter 4 Relative atomic mass and the mole
E1.
A sample of argon (Ar) known to contain isotopes of mass numbers 36, 39 and 40 is
introduced in a mass spectrometer. The sample is bombarded with electrons to form
positively charged ions.
40
+
+
a Which ion, 36
18 Ar or 18 Ar , is likely to be deflected most in the magnetic field?
Give a reason for your answer.
+
b Some atoms lose two electrons in the ionisation chamber. Which ion, 36
18 Ar or
36
18
Ar 2+, is likely to be deflected most in the magnetic field? Give a reason for
your answer.
AE1.
a
36
18
Ar + has the lighter mass, so will be deflected the most.
b
36
18
Ar 2+ has the higher positive charge and so will be deflected the most.
Q1.
Use the data in Table 4.2 on page 55 to calculate the relative atomic mass of:
a oxygen
b silver
c hydrogen
A1.
(15.995  99.76)  (16.999  0.04)  (17.999  0.20)
100
= 15.999
a
Ar(O) =
b
Ar(Ag) =
c
Ar(H) =
(106.9  51.8)  (108.9  48.2)
100
= 108
(1.008  99.986)  (2.014  0.014)  (3.016  0.0001)
100
= 1.008
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2
Q2.
The element lithium has two isotopes:
• 6Li has a relative isotopic mass of 6.02.
• 7Li has a relative isotopic mass of 7.02.
The relative atomic mass of lithium is 6.94. Calculate the percentage abundance of the
lighter isotope.
A2.
Let the percentage abundance of the lighter isotope be x%.
 percentage abundance of the heavier isotope will be (100 – x)%.
 6.94 =
( x  6.02)  ((100  x)  7.02)
100
 x = 8.0
 proportion of the lighter isotope = 8.0%
Q3.
Calculate the relative molecular masses of:
a sulfuric acid (H2SO4)
b ammonia (NH3)
c ethane (C2H6)
A3.
a
b
c
Mr(H2SO4) = 2 × Ar(H) + Ar(S) + 4 × Ar(O)
= 2 × 1.008 + 32.06 + 4 × 16.00
= 98.1
Mr(NH3) = Ar(N) + 3 × Ar(H)
= 14.007 + 3 × 1.008
= 17.0
Mr(C2H6) = 2 × Ar(C) + 6 × Ar(H)
= 2 × 12.01 + 6 × 1.008
= 30.1
Q4.
Calculate the relative formula mass of:
a potassium chloride (KCl)
b sodium carbonate (Na2CO3)
c aluminium sulfate (Al2(SO4)3)
A4.
a
b
c
RFM(KCl) = Ar(K) + Ar(Cl)
= 39.01 + 35.45
= 74.5
RFM(Na2CO3) = 2 × Ar(Na) + Ar(C) + 3 × Ar(O)
= 2 × 22.99 + 12.01 + 3 × 16.00
= 106
RFM(Al2(SO4)3) = 2 × Ar(Al) + 3 × Ar(S) + 12 × Ar(O)
= 2 × 26.98 + 3 × 32.06 + 12 × 16.00
= 342
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Worked solutions to textbook questions
Q5.
Calculate the number of:
a atoms in 2.0 mol of sodium atoms (Na)
b molecules in 0.10 mol of nitrogen molecules (N2)
c atoms in 20.0 mol of carbon atoms (C)
d molecules in 4.2 mol of water molecules (H2O)
e atoms in 1.0 × 10–2 mol of iron atoms (Fe)
f molecules in 4.62 × 10–5 mol of CO2 molecules
A5.
Number of particles = amount (mol)  NA
a Number of sodium atoms (Na) = 2.0  6.02  1023
= 1.2  1024
b Number of nitrogen molecules (N2) = 0.10  6.02  1023
= 6.02  1022
c Number of carbon atoms (C) = 20.0  6.02  1023
= 1.20  1025
d Number of water molecules (H2O) = 4.2  6.02  1023
= 2.5  1024
e Number of iron atoms (Fe) = 1.0  10–2  6.02  1023
= 6.0  1021
f Number of CO2 molecules = 4.62  10–5  6.02  1023
= 2.78  1019
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4
Q6.
Calculate the amount of substance (in mol) represented by:
a 3.0 × 1023 molecules of water (H2O)
b 1.5 × 1023 atoms of neon (Ne)
c 4.2 × 1025 atoms of iron (Fe)
d 4.2 × 1025 molecules of ethanol (C2H5OH)
A6.
Remember: Avogadro’s number, NA, is 6.02  1023.
Amount (mol) =
a
number of particles
NA
Amount (H2O molecules) =
3.0  10 23
NA
= 0.5 mol
b
1.5  10 23
Amount (Ne atoms) =
NA
= 0.25 mol
c
Amount (Fe atoms) =
4.2  10 25
NA
= 70 mol
d
4.2  10 25
Amount (C2H5OH molecules) =
NA
= 70 mol
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Q7.
Calculate the amount (in mol) of:
a sodium atoms represented by 1.0 × 1020 sodium atoms
b aluminium represented by 1.0 × 1020 aluminium atoms
c chlorine molecules represented by 1.0 × 1020 chlorine molecules
A7.
Remember: Avogadro’s number, NA, is 6.02  1023.
Amount (mol) =
a
number of particles
NA
Amount (Na atoms) =
1.0  10 20
NA
= 1.7  10–4 mol
b
Amount (Al atoms)
=
1.0  10 20
NA
= 1.7  10–4 mol
c
1.0  10 20
Amount (Cl2 molecules) =
NA
= 1.7  10–4 mol
Q8.
Calculate the amount (in mol) of:
a chlorine atoms in 0.4 mol of chlorine (Cl2)
b hydrogen atoms in 1.2 mol of methane (CH4)
c hydrogen atoms in 0.12 mol of ethane (C2H6)
d oxygen atoms in 1.5 mol of sodium sulfate (Na2SO4)
A8.
a
b
c
d
Each Cl2 molecule has 2 Cl atoms.
 amount (Cl atoms) = 0.4  2
= 0.8 mol
Each methane molecule has 4 H atoms.
 amount (H atoms) = 4  1.2
= 4.8 mol
Each ethane molecule has 6 H atoms.
 amount (H atoms) = 6  0.12
= 0.72 mol
Each sulfate ion has 4 O atoms.
 amount (O atoms) = 4  1.5
= 6.0 mol
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Q9.
Calculate the molar mass of:
a nitrogen (N2)
b ammonia (NH3)
c sulfuric acid (H2SO4)
d iron(III) nitrate (Fe(NO3)3)
e acetic acid (CH3COOH)
f sulfur atoms (S)
g vitamin C (ascorbic acid C6H8O6)
h hydrated copper(II) sulfate (CuSO4.5H2O)
A9.
a
b
c
d
e
f
g
h
M(N2) = 2  14.007
= 28.0 g mol–1
M(NH3) = (1 × 14.007) + (3 × 1.008)
= 17.0 g mol–1
M(H2SO4) = (2  1.008) + (1 × 32.064) + (4  15.994)
= 98.1 g mol–1
M(Fe(NO3)3) = (1  55.847) + (3 × 14.007) + (9  15.994)
= 242 g mol–1
M(CH3COOH) = (4  1.008) + (2 × 12.012) + (2  15.994)
= 60.0 g mol–1
M(S) = (1  32.1)
= 32.1 g mol–1
M(C6H8O6) = (6  12.012) + (8  1.008) + (6  15.994)
= 176.1 g mol–1
M(CuSO4.5H2O) = (1  63.54) + (1 × 32.064) + (4  15.994) + (5 × 18.01)
= 250 g mol–1
Q10.
Calculate the mass of:
a 1.0 mol of sodium atoms (Na)
b 2.0 mol of oxygen molecules (O2)
c 0.10 mol of methane molecules (CH4)
d 0.25 mol of aluminium oxide (Al2O3)
A10.
It is useful to remember the formula m = nM, where m is the mass in grams, n the
amount of substance in mol, and M the molar mass. Use a periodic table to work out
the molar masses.
a m(Na atoms) = 1.0 mol × 23 g mol–1
= 23.0 g
b m(O2) = 2.0 mol × 32.0 g mol–1
= 64.0 g
c m(CH4) = 0.10 mol × 16.0 g mol–1
= 1.60 g
d m(Al2O3) = 0.25 mol × 102 g mol–1
= 25.5 g
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Worked solutions to textbook questions
Q11.
Calculate the amount, in mol, of:
a H atoms in 5 g of hydrogen
b H2 molecules in 5 g of hydrogen
c Al atoms in 2.7 g of aluminium
d CH4 molecules in 0.4 g of methane
e O atoms in 0.10 g of oxygen
f O2 molecules in 0.10 g of oxygen
g P atoms in 1.2 × 10–3 g of phosphorus
h P4 molecules in 1.2 × 10–3 g of phosphorus
A11.
m
, where m is the mass in grams, n the
M
amount of substance in mol, and M the molar mass. Use a periodic table to work out
the molar masses.
5
a  n(H atoms) = = 5 mol
1
It is useful to remember the formula n =
5
= 2.5 mol
2
b
 n(H2) =
c
 n(Al atoms) =
d
 n(CH4) =
e
 n(O atoms) =
f
 n(O2) =
g
 n(P atoms) =
h
1.2  10 3
 n(P4) =
= 9.7  10–6 mol
124
2 .7
= 0.10 mol
27
0 .4
= 0.025 mol
16
0.10
= 0.0063 mol
16
0.10
= 0.0031 mol
32
1.2  10 3
= 3.9  10–5 mol
31
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8
Q12.
Calculate the number of atoms in:
a 23 g of sodium (Na)
b 4.0 g of argon (Ar)
c 0.243 g of magnesium (Mg)
d 10.0 g of gold (Au)
A12.
Use the formulas: Number of particles = n × NA, where NA = 6.02  1023 and n =
where m is the mass in grams, n the amount of substance in mol, and M the molar
mass. Use a periodic table to work out the molar masses.
23
a n(Na) =
= 1.0 mol
23
Number of Na atoms = 1.0 × 6.0  1023 = 6.0  1023 atoms
b
n(Ar) =
4 .0
= 0.10 mol
39.95
Number of Ar atoms = 0.10 × 6.0  1023 = 6.0  1022 atoms
c
n(Mg) =
0.243
= 0.01 mol
24.3
Number of Mg atoms = 0.01 × 6.0  1023 = 6.0  1021 atoms
d
n(Au) =
10.0
= 0.051 mol
196.97
Number of Mg atoms = 0.051 × 6.0  1023 = 3.0  1022 atoms
Q13.
Calculate:
a the number of molecules in:
i 16 g of oxygen (O2)
ii 2.8 g of nitrogen (N2)
b the number of oxygen atoms in 3.2 g of sulfur dioxide (SO2)
c the total number of atoms in 288 g of ammonia (NH3)
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m
M ,
Worked solutions to textbook questions
9
A13.
Use the formulas: Number of particles = n × NA, where NA = 6.02  1023, and n =
where m is the mass in grams, n the amount of substance in mol, and M the molar
mass. Use a periodic table to work out the molar masses.
a
i
n(O2) =
16
= 0.5 mol
32
Number of O2 molecules = 0.5 × 6.0  1023 = 3.0  1023 molecules
ii
n(N2) =
2 .8
= 0.1 mol
28
Number of N2 molecules = 0.1 × 6.0  1023 = 6.0  1022 molecules
b
n(SO2) =
3 .2
= 0.05 mol
64
Number of SO2 molecules = 0.05 × 6.0  1023 = 3.0  1022 molecules
Each molecule contains 2 oxygen atoms.
So, number of oxygen atoms = 6.0  1022 atoms.
c
n(NH3) =
288
= 16.9 mol
17
Number of NH3 molecules = 16.9 × 6.0  1023 = 1.0  1025 molecules
Each molecule contains 4 atoms (1 of N and 3 of H).
So, total number of atoms = 4.0  1025 atoms.
Q14.
Calculate the percentage by mass of:
a iron in iron(III) oxide (Fe2O3)
b uranium in uranium oxide (U3O8)
c nitrogen in ammonium chloride (NH4Cl)
d oxygen in copper nitrate (Cu(NO3)2)
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m
,
M
Worked solutions to textbook questions
10
A14.
a
Percentage by mass of an element
=
mass of element in 1 mol of compound
× 100%
mass of 1 mol of the compound
Use a periodic table to work out the molar masses.
For example, M(Fe2O3) = 159.6 g mol–1.
 %(Fe) =
2  55.8  100
159.6
= 69.9%
b
%(U) =
3  238.03  100
842.09
= 84.8%
c
%(N) =
14  100
53.5
= 26.2%
d
%(O) =
6  16  100
187.5
= 51.2%
Q15.
Determine the empirical formula of the compounds with the following compositions:
a 2.74% hydrogen, 97.26% chlorine
b 42.9% carbon, 57.1% oxygen
c 10.0 g of a compound of magnesium and oxygen that contains 6.03 g of
magnesium
d 3.2 g of a hydrocarbon that contains 2.4 g of carbon
A15.
a
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The number of moles of each atom is found by using n =
, where
M
m is the mass in grams, and M is the molar mass in g mol–1.
Mass
Molar mass
m
M
Divide all by the smallest
amount
Round off to whole numbers
Amount using n =
 empirical formula is HCl
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H
2.74 g
1 g mol–1
2.74
n=
= 2.74 mol
1
2.74
=1
2.74
1
Cl
97.26 g
35.5 g mol–1
97.26
n=
= 2.74 mol
35.5
2.74
=1
2.74
1
Worked solutions to textbook questions
11
b
Mass
Molar mass
m
M
Divide all by the smallest
amount
Round off to whole numbers
Amount, using n =
C
42.9 g
12 g mol–1
42.9
n=
= 3.575 mol
12
3.575
=1
3.57
1
O
57.1 g
16 g mol–1
57.1
n=
= 3.57 mol
16
3.57
=1
3.57
1
 empirical formula is CO
c
When 6.03 g is the mass of magnesium in 10.0 g of the compound, the mass of
oxygen is (10.0 – 6.03) = 3.97 g.
Mass
Molar mass
m
M
Divide all by the smallest
amount
Round off to whole numbers
Amount, using n =
Mg
6.03 g
24.3 g mol–1
6.03
n=
= 0.248 mol
24.3
0.248
=1
0.248
1
O
3.97 g
16 g mol–1
3.97
n=
= 0.248 mol
16
0.248
=1
0.248
1
 empirical formula is MgO
d
When 2.4 g is the mass of carbon in 3.2 g of the hydrocarbon, the mass of
hydrogen is (3.2 – 2.4) = 0.8 g.
Mass
Molar mass
m
M
Divide all by the smallest
amount
Round off to whole numbers
Amount, using n =
 empirical formula is CH4
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C
2.4 g
12 g mol–1
2 .4
n=
= 0.2 mol
12
0 .2
=1
0 .2
1
H
0.8 g
1 g mol–1
0 .8
n=
= 0.8 mol
1
0 .8
=4
0 .2
4
Worked solutions to textbook questions
12
Q16.
Determine the molecular formula of each compound in the table below.
Table 4.8
Empirical formula
CH
HO
CH2O
NO2
CH2
a
b
c
d
e
Relative molecular mass
78
34
90
46
154
A16.
a
The molecular formula is always a whole-number multiple of the empirical
formula. The empirical formula provides the simplest whole-number ratio of
atoms in a compound. The number of moles of each atom is found by using
m
n=
, where m is the mass in grams and M is the molar mass.
M
Molar mass of a CH unit (empirical formula) = 12 + 1 = 13 g mol–1
Molar mass of the compound (molecular formula) = 78 g mol–1
78
 number of CH units in one molecule =
=6
13
 molecular formula is C6H6 (which is benzene)
b
Molar mass of an HO unit (empirical formula) = 1 + 16 = 17 g mol–1
Molar mass of the compound (molecular formula) = 34 g mol–1
34
 number of OH units in one molecule =
=2
17
 molecular formula is H2O2
c
Molar mass of a CH2O unit (empirical formula) = 12 + (2 × 1) + 16 = 30 g mol–1
Molar mass of the compound (molecular formula) = 90 g mol–1
90
 number of CH2O units in one molecule =
=3
30
 molecular formula is C3H6O3
d
Molar mass of an NO2 unit (empirical formula) = 14 + (16 × 2) = 46 g mol–1
Molar mass of the compound (molecular formula) = 46 g mol–1
46
 number of NO2 units in one molecule =
=1
46
 molecular formula is NO2
e
Molar mass of a CH2 unit (empirical formula) = 12 + (1 × 2) = 14 g mol–1
Molar mass of the compound (molecular formula) = 154 g mol–1
154
 number of CH2 units in one molecule =
= 11
14
 molecular formula C11H22
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Worked solutions to textbook questions
13
Q17.
A hydrocarbon contains 85.7% carbon. Its relative molecular mass is 70.
a Determine the empirical formula of the hydrocarbon.
b Determine the molecular formula of the hydrocarbon.
A17.
a
The molecular formula is always a whole-number multiple of the empirical
formula. The empirical formula provides the simplest whole-number ratio of
atoms in a compound. The number of moles of each atom is found by using
m
n=
, where m is the mass in grams and M is the molar mass in g mol–1.
M
A hydrocarbon contains only carbon and hydrogen.
Hence, %H = 100 – 85.7 = 14.3%
Mass
Molar mass
m
M
Divide all by the smallest
amount
Round off to whole numbers
Amount, using n =
H
14.3 g
1 g mol–1
14.3
n=
= 14.3 mol
1
14.3
=2
7.14
2
C
85.7 g
12 g mol–1
85.7
n=
= 7.14 mol
12
7.14
=1
7.14
1
 empirical formula is CH2
b
Molar mass of a CH2 unit (empirical formula) = 12 + 2 = 14 g mol–1
Molar mass of the compound (molecular formula) = 70 g mol–1
70
 number of CH2 units in one molecule =
=5
14
 molecular formula is C5H10
Q18.
A sample of the carbohydrate glucose contains 1.8 g carbon, 0.3 g hydrogen and 2.4 g
oxygen. Calculate the empirical formula of the compound. Deduce its molecular
formula, given that its relative molecular mass is 180.
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14
A18.
The molecular formula is always a whole-number multiple of the empirical formula.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The number of moles of each atom is found by using n =
, where m is
M
the mass in grams and M is the molar mass in g mol–1.
Mass
Molar mass
Amount, using n =
Divide all by the
smallest amount
Round off to whole
numbers
m
M
C
1.8 g
12 g mol–1
1.8
n=
= 0.15 mol
12
0.15
=1
0.15
H
0.3 g
1 g mol–1
0 .3
n=
= 0.30 mol
1
0.30
=2
0.15
O
2.4 g
16 g mol–1
2 .4
n=
= 0.15 mol
16
0.15
=1
0.15
1
2
1
 empirical formula is CH2O
Molar mass of a CH2O unit (empirical formula) = 12 + (2 × 1) + 16 = 30 g mol–1
Molar mass of the compound (molecular formula) = 180 g mol–1
180
 number of CH2O units in one molecule =
=6
30
 molecular formula is C6H12O6
Chapter review
Q19.
The standard on which all relative masses are based is the 12C isotope, which is given
a mass of 12 exactly. Explain why then, in the table of relative atomic masses in
Appendix 4 on page 489, the relative atomic mass of carbon is listed as 12.011.
A19.
The relative atomic mass of carbon is the weighted average of the isotopic masses of
all carbon isotopes (i.e. 12C, 13C and 14C). Small amounts of 13C and 14C make this
average slightly greater than 12, the relative isotopic mass of the 12C isotope.
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15
Q20.
When a sample of palladium is placed in a mass spectrometer, the following peaks are
recorded at the relative atomic masses and corresponding percentage abundances
given in the table.
Relative isotopic mass
Percentage abundance
101.9049
0.9600%
103.9036
10.97%
104.9046
22.23%
105.9032
27.33%
107.9039
26.71%
109.9044
11.80%
Calculate the relative atomic mass of palladium.
A20.
Ar(Pd) =
101.9049  0.96  103.9036  10.97  104.9046  22.23  105.9032  27.33  107.9039  26.71  109.9044  11.80
100
= 106.4
Q21.
The following table gives isotopic composition data for argon and potassium.
Element
Argon
Atomic
number
18
Potassium
19
a
b
Relative isotopic mass
35.978
37.974
39.974
38.975
39.976
40.974
Relative abundance (%)
0.307
0.060
99.633
93.3
0.011
6.69
Determine the relative atomic masses of argon and potassium.
Explain why the relative atomic mass of argon is greater than that of potassium,
although potassium has a larger atomic number.
A21.
a
Ar(Ar) =
Ar(K) =
b
(35.978  0.307)  (37.974  0.060)  (39.974  99.633)
= 39.96
100
(38.975  93.3)  (39.976  0.011)  (40.974  6.69)
= 39.11
100
Although potassium atoms have one more proton than argon atoms, the most
abundant isotope of argon has 22 neutrons, giving it a relative atomic mass close
to 40. The most abundant isotope of potassium has only 20 neutrons, giving it a
relative atomic mass close to 39.
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16
Q22.
The relative atomic mass of rubidium is 85.47. The relative isotopic masses of its two
isotopes are 84.94 and 86.94. Calculate the relative abundances of the isotopes in
naturally occurring rubidium.
A22.
Let the percentage abundance of the lighter isotope be x%.
 percentage abundance of the heavier isotope will be (100 – x)%.
 85.47 =
( x  84.94)  ((100  x)  86.94)
100
 x = 26.5
 proportion of the isotopes are 26.5% and 73.5%.
Q23.
Determine the percentage abundance of the lighter isotope of each of the following
elements.
a Gallium: relative isotopic masses 68.95 and 70.95, respectively; Ar = 69.72
b Boron: relative isotopic masses 10.02 and 11.01, respectively; Ar = 10.81
A23.
a
Let the percentage abundance of the lighter isotope be x%.
 percentage abundance of the heavier isotope will be (100 – x)%.
 69.72 =
( x  68.95)  ((100  x)  70.95)
100
 x = 61.5
 proportion of the lighter isotope = 61.5%
b
20.2% (calculated by using the same process as for part a)
Q24.
What is the relative molecular mass (Mr) of the following?
a water (H2O)
b white phosphorus (P4)
c carbon monoxide (CO)
A24.
a
b
c
The relative molecular mass, Mr, is the sum of the relative atomic masses, Ar, of
the elements in the compound.
 Mr(H2O) = 2 + 16
= 18
Mr(P4) = 4 + 31
= 124
Mr(CO) = 12 + 16
= 28
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17
Q25.
How would the molar mass (M) of a compound differ from its relative molecular mass
(Mr)?
A25.
The molar mass, M, has the same numerical value as the relative molecular mass, Mr,
which is the sum of the relative atomic masses, Ar, of the elements in the compound.
The molar mass, M, is the actual mass of one mole and so has the unit g mol–1.
Q26.
What is the molar mass (M) of each of the following?
a iron (Fe)
b sulfuric acid (H2SO4)
c sodium oxide (Na2O)
d zinc nitrate (Zn(NO3)2)
e glycine (H2NCH2COOH)
f aluminium sulfate (Al2(SO4)3)
g hydrated iron(III) chloride (FeCl3.6H2O)
A26.
a
b
c
d
e
f
g
55.8 g mol–1
Mr(H2SO4) = 2 + 32 + 64 = 98
 M(H2SO4) = 98 g mol–1
Mr(Na2O) = (2 × 23) + 16 = 62
 M(Na2O) = 62 g mol–1
Mr(Zn(NO3)2) = 65.4 + (2 × 14) + (6 × 16) = 189.4
 M(Zn(NO3)2) = 189.4 g mol–1
Mr(H2NCH2COOH) = (5 × 1) + 14 + (2 × 12) + (2 × 16) = 75
 M(H2NCH2COOH) = 75.0 g mol–1
Mr(Al2(SO4)3) = (2 × 27) + (3 × 32) + (12 × 16) = 342
 M(Al2(SO4)3) = 342 g mol–1
Mr(FeCl3.6H2O) = 55.9 + (3 × 35.5) + (12 × 1.01) + (6 × 16) = 271
 M(FeCl3.6H2O) = 271 g mol–1
Q27.
What is the mass of:
a 0.060 mol of ethane (C2H6)?
b 0.32 mol of glucose (C6H12O6)?
c 6.8 × 10–3 mol of urea ((NH2)2CO)?
d 6.12 mol of copper atoms (Cu)?
A27.
It is useful to remember the formula m = nM, where m is the mass in grams, n the
amount of substance in mol, and M the molar mass.
a m(C2H6) = 0.060  (24 + 6) = 1.8 g
b m(C6H12O6) = 0.32  ((6 × 12) + (12 × 1) + (6 × 16)) = 58 g
c m((NH2)2CO) = 6.8 × 10–3  ((2 × 14) + (4 × 1) + 12 + 16)) = 0.41 g
d m(Cu) = 0.6.12  63.5 = 389 g
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18
Q28.
What is the amount (in mol) of each of the following?
a carbon atoms in 1.201 g carbon
b sulfur molecules (S8) in 10.0 g sulfur
c methane molecules (CH4) in 20.0 g methane
d aspirin molecules (C6H4(OCOCH3)COOH) in 300 mg aspirin
e aluminium oxide (Al2O3) in 3.5 tonnes of aluminium oxide (1 tonne = 1000 kg)
A28.
a
b
c
d
e
m
, where m is the mass in grams, n the
M
amount of substance in mol, and M the molar mass in g mol–1. Use a periodic
table to work out the molar masses.
1.201
 n(C) =
= 0.10 mol
12
10.0
n(S8) =
= 0.0391 mol
256
20.0
n(CH4) =
= 1.25 mol
16
0.300
n(C6H4(OCOCH3)COOH) =
= 0.001 67 mol
180
3 500 000
n(Al2O3) =
= 3.4  104 mol
102
It is useful to remember the formula n =
Q29.
a
b
c
If 6.0  10 23 atoms of calcium have a mass of 40.1 g, what is the mass of one
calcium atom?
If 1 mol of water molecules has a mass of 18 g, what is the mass of one water
molecule?
What is the mass of one molecule of carbon dioxide?
A29.
a
Mass of one atom =
molar mass
mass of 1 mole
=
NA
number of particles in a mole
Mass of one calcium atom =
40.1
= 6.67  10–23 g
6.0  10 23
b
Mass of one water molecule =
c
Mass of one CO2 molecule =
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18
= 3.0  10–23 g
6.0  10 23
44
= 7.3  10–23 g
6.0  10 23
Worked solutions to textbook questions
19
Q30.
For each of the following molecular substances, calculate:
a the amount of substance, in moles
b the number of molecules
c the total number of atoms
i 4.2 g of phosphorus (P4)
ii 75.0 g of sulfur (S8)
iii 0.32 g of hydrogen chloride (HCl)
iv 2.2 × 10–2 g of glucose (C6H12O6)
A30.
i
a
b
c
ii a
b
c
iii a
b
c
iv a
b
c
n(P4) = m/M = 4.2/(4 × 31.0) = 0.034 mol
N(P4) = n × 6.02 × 1023 = 2.04 × 1022 molecules
Total number of atoms = 4 × 2.04 × 1022 = 8.2 × 1022 atoms
n(S8) = m/M = 75.0/(8 × 32.1) = 0.292 mol
N(S8) = n × 6.02 × 1023 = 1.75 × 1023 molecules
Total number of atoms = 0.292 × 8 × 6.02 × 1023 = 1.41 × 1024 atoms
n(HCl) = m/M = 0.32/(1.01 + 35.5) = 0.0088 mol
N(HCl) = n × 6.02 × 1023 = 5.3 × 1021 molecules
Total number of atoms = 0.0088 × 2 × 6.02 × 1023 = 1.1 × 1022 atoms
n(C6H12O6) = m/M = 2.2 × 10–2/(6 × 12.01) + (12 × 1.01) + (6 × 16.0)
= 1.22 × 10–4 mol
N(C6H12O6) = n × 6.02 × 1023 = 7.3 × 1019 molecules
Total number of atoms = 1.22 × 10–4 × 24 × 6.02 × 1023 = 1.8 × 1021 atoms
Q31.
What mass of iron (Fe) would contain as many iron atoms as there are molecules in
20.0 g of water (H2O)?
A31.
m
It is useful to remember the formula n = M , where m is the mass in grams, n the
amount of substance in mol, and M the molar mass in g mol–1. Use a periodic table to
work out the molar masses of iron and water.
M(Fe) = 55.8 g mol–1 and 18.0 g mol–1
20
 n(H2O) =
= 1.11 mol
18
 n(Fe) needed = 1.11 mol
 m(Fe) needed = 1.11  55.8 = 62.0 g
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Q32.
For each of the following ionic substances, calculate:
a the amount of substance, in moles
b the amount of each ion, in moles
i 5.85 g of NaCl
ii 45.0 g of CaCl2
iii 1.68 g of Fe2(SO4)3
A32.
m
It is useful to remember the formula n = M , where m is the mass in grams, n the
amount of substance in mol, and M the molar mass. Use a periodic table to work out
the molar masses.
i
ii
5.85
= 0.100 mol
58.8
a
n(NaCl) =
b
n(Na+) = n(NaCl) = 0.100 mol Na+
n(Cl–1) = n(NaCl) = 0.100 mol Cl–
a
n(CaCl2) =
b
n(Ca2+) = n(CaCl2) = 0.405 mol
n(Cl–) = 2  n(CaCl2) = 2 × 0.405 mol = 0.81 mol
iii a
b
45.0
= 0.405 mol
111
n(Fe2(SO4)3) =
1.68
= 0.004 20 mol
399.6
n(Fe3+) = 2 × n(Fe2(SO4)3) = 0.008 40 mol Fe3+
n(SO42–) = 3 × n(Fe2(SO4)3) = 0.0126 mol SO42–
Q33.
For each of the following numbers of molecules, calculate:
a the amount of substance, in moles
b the mass, in grams, of substance
i 4.50  1023 molecules of water (H2O)
ii 9.00  1024 molecules of methane (CH4)
iii 2.3  1028 molecules of chlorine (Cl2)
iv 1 molecule of sucrose (C12H22O11)
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21
A33.
m
, where m is the mass in grams, n the
M
amount of substance in mol, and M the molar mass. Remember also that the
number of particles in 1 mol, Avogadro’s number, NA = 6.02  1023.
number of particles
Use the formula: n =
. Use a periodic table to work out the
NA
molar masses.
4.50  10 23
a n(H2O) =
= 0.75 mol
6.02  10 23
b m(H2O) = 0.75  18 = 13.5 g
i
It is useful to remember the formula n =
ii
a
b
iii a
b
iv a
b
15.0 mol
240 g
3.8  104 mol
2.7  106 g
1
= 1.7  10–24 mol
6.02  10 23
m(C12H22O11) = 1.7  10–24  342 = 5.7  10–22 g
n(C12H22O11) =
Q34.
a
b
If 0.50 mol of a substance has a mass of 72 g, what is the mass of 1.0 mol of the
substance?
If 6.0  1022 molecules of a substance have a mass of 10 g, what is the molar
mass of the substance?
A34.
m
, where m is the mass in grams, n the
n
amount of substance in mol, and M the molar mass in g mol–1. Remember also that the
number of particles in 1 mol is Avogadro’s number, NA = 6.02  1023.
It is useful to remember the formula M =
Use the formula: n =.
number of particles
NA
72
= 144 g mol–1
0 .5
a
M(substance) =
b
6.0  10 22
n(substance) =
= 0.1 mol
NA
M(substance) =
10
= 100 g mol–1
0 .1
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Q35.
Calculate the molar mass of a substance if:
a 2 mol of the substance has a mass of 80 g
b 0.1 mol of the substance has a mass of 9.8 g
c 1.7 mol of the substance has a mass of 74.8 g
d 3.5 mol of the substance has a mass of 371 g
A35.
m
, where m is the mass in grams, n the
n
amount of substance in mol, and M the molar mass in g mol–1.
80
a M(substance) =
= 40 g mol–1
2
All other parts follow the same process.
b 98 g mol–1
c 44 g mol–1
d 106 g mol–1
It is useful to remember the formula M =
Q36.
Which of the following metal samples has the greatest mass?
a 100 g copper
b 4.0 mol of iron atoms
c 1.2  1024 atoms of silver
A36.
It is useful to remember the formula m = nM, where m is the mass in grams, n the
amount of substance in mol, and M the molar mass. Use a periodic table to find the
molar masses of iron and silver. M = 55.8 g mol–1 and 108 g mol–1, respectively.
Remember also that the number of particles in 1 mol is NA.
m(Fe) = 4.0  55.8 = 223 g
1.2  1024
n(Ag) =
= 2.0 mol
NA
 m(Ag) = 2.0  108 = 216 g
 the mass of the iron is the greatest
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23
Q37.
A new antibiotic has been isolated and only 2.0 mg is available. The molar mass is
found to be 12.5 kg mol–1.
a Express the molar mass in g mol–1.
b Calculate the amount of antibiotic (in mol).
c How many molecules of antibiotic have been isolated?
A37.
It is useful to remember the formula m = nM, where m is the mass in grams, n the
amount of substance in mol, and M the molar mass in g mol–1. Remember also that the
number of particles in 1 mol is, Avogadro’s number, NA = 6.02  1023.
a M(antibiotic) = 12 500 g mol–1 = 1.25  104 g mol–1
2.0  10 3
b n(antibiotic) =
1.25  10 4
= 1.6  10–7 mol
c  number of molecules = nNA
= 1.6  10–7  6.02  1023
= 9.6  1016 molecules
Q38.
Calculate the percentage by mass of each element in the following compounds:
a Al2O3
b Cu(OH)2
c MgCl2.6H2O
d Fe2(SO4)3
e perchloric acid (HClO4)
A38.
mass of element in 1 mol of compound
× 100.
mass of 1 mol of compound
Use a periodic table to work out the molar masses. A useful check of these answers is
provided by seeing that they add up to 100%, or somewhere close to that value.
a
M(Al) = 27 g mol–1, M(O) = 16 g mol–1, M(Al2O3) = 102 g mol–1.
Percentage by mass of an element =
b
%(Al) =
2  27
× 100 = 52.9%
102
%(O) =
3  16
× 100 = 47.1%
102
Cu 65.1%; O 32.8%; H 2.1%
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c
24
M(Mg) = 24.3 g mol–1, M(Cl) = 35.5 g mol–1, M(MgCl2.6H2O) = 203.3 g mol–1
%(Mg) =
24.3
× 100 = 12.0%
203 .3
%(Cl)
=
2  35.5
× 100 = 34.9%
203.3
%(H)
=
12  1
× 100 = 5.9%
203 .3
%(O)
=
6  16
× 100 = 47.2%
203 .3
d
Fe 27.9%; S 24.1%; O 48.0%
e
H 1.0%; Cl 35.3%; O 63.7%
Q39.
Determine the percentage of carbon in the following compounds:
a naphthalene (C10H8)
b acetic acid (CH3COOH)
c urea (NH2CONH2)
d aspirin (C6H4(OCOCH3)COOH)
A39.
mass of element in 1 mol of compound
× 100.
mass of 1 mol of compound
Use a periodic table to work out the molar masses. A useful check of these answers is
provided by seeing that they add up to 100%, or somewhere close to that value.
a M(C10H8) = 128 g mol–1, M(H) = 1 g mol–1, M(C) = 12 g mol–1
10  12
%(C) =
× 100 = 93.8%
128
b %C = 40%
c %C = 19.9%
d %C = 60.0%
Percentage by mass of an element =
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Q40.
Determine the empirical formulas of the compounds with the following compositions:
a 42.9% carbon, 57.1% oxygen
b 27.2% carbon, 72.8% oxygen
c 54.5% carbon, 9.1% hydrogen, 36.4% oxygen
d 1.72 g iron, 1.48 g sulfur, 3.02 g oxygen
e 9.6 g carbon, 0.67 g hydrogen, 4.7 g chlorine
f 4.42 g carbon, 0.842 g hydrogen
A40.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The number of moles of each atom is found by using n =
, where m is
M
the mass in grams and M is the molar mass.
a
C
O
Mass
42.9 g
57.1 g
–1
Molar mass
12 g mol
16 g mol–1
m
42.9
57.1
Amount, using n =
n=
= 3.575 mol
n=
= 3.569 mol
M
12
16
3.575
3.569
Divide all by smallest
=1
=1
amount
3.569
3.569
Round off to whole
1
1
numbers
 empirical formula is CO
b
Mass
Molar mass
m
M
Divide all by smallest
amount
Round off to whole
numbers
Amount, using n =
C
27.2 g
12 g mol–1
27.2
n=
= 2.267 mol
12
2.267
=1
2.267
O
72.8 g
16 g mol–1
72.8
n=
= 4.55 mol
16
4.55
=2
2.267
1
2
 empirical formula is CO2
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c
Mass
Molar mass
Amount, using n =
m
M
Divide all by smallest
amount
Round off to whole
numbers
C
54.5 g
12 g mol–1
54.5
n =
12
= 4.54 mol
4.54
= 1.99
2.275
O
36.4 g
16 g mol–1
36.4
n =
16
= 2.275 mol
2.275
=1
2.275
H
9.1 g
1 g mol–1
9 .1
n =
1
= 9.1 mol
9.1
=4
2.275
2
1
4
Fe
1.72 g
55.8 g mol–1
1.72
n =
55.8
= 0.031 mol
0.031
=1
0.031
S
1.48 g
32 g mol–1
1.48
n =
32
= 0.046 mol
0.046
= 1.5
0.031
O
3.02 g
16 g mol–1
3.02
n =
16
= 0.189 mol
0.189
=6
0.031
2
3
12
 empirical formula is C2H4O
d
Mass
Molar mass
Amount, using n =
m
M
Divide all by smallest
amount
Round off to whole
numbers
 empirical formula is Fe2S3O12, which is Fe2(SO4)3
e
Mass
Molar mass
Amount, using n =
m
M
Divide all by smallest
amount
Round off to whole
numbers
C
9.6 g
12 g mol–1
9 .6
n =
12
= 0.800 mol
0.800
=6
0.132
Cl
4.7 g
35.5 g mol–1
4 .7
n =
35.5
= 0.132 mol
0.132
=1
0.132
H
0.67 g
1 g mol–1
0.67
n =
1
= 0.67 mol
0.67
=5
0.132
6
1
5
 empirical formula is C6H5Cl
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f
Mass
Molar mass
m
M
Divide all by smallest
amount
Round off to whole
numbers
Amount, using n =
C
4.42 g
12 g mol–1
4.42
n=
= 0.368 mol
12
0.368
=1
0.368
H
0.842 g
1 g mol–1
0.842
n=
= 0.842 mol
1
0.842
= 2.3
0.368
1×7=7
2.3 × 7 = 16
 empirical formula is C7H16
Q41.
A compound used as a solvent for nitrocellulose, resins and dyes has the following
composition by mass: 32% carbon, 6.7% hydrogen, 18.7% nitrogen and 42.6%
oxygen. Find the empirical formula of the compound.
A41.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n = M , where m is the mass
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
masses of C, H, N and O.
Mass
Molar mass
Amount, using
m
n=
M
Divide all by
smallest amount
Round off to
whole numbers
C
32 g
12 g mol–1
32
n=
12
= 2.67 mol
2.67
= 1.99
1.34
H
6.7 g
1 g mol–1
6 .7
n=
1
= 6.7 mol
6.7
=5
1.34
N
18.7 g
14 g mol–1
18.7
n =
14
= 1.34 mol
1.34
=1
1.34
O
42.6 g
16 g mol–1
42.6
n=
16
= 2.66 mol
2.66
=1.99
1.34
2
5
1
2
 empirical formula is C2H5NO2
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Q42.
A compound of tungsten and sulfur is a useful solid lubricant. Deduce the empirical
formula of this compound if a particular sample is formed when 1.84 g of tungsten
reacts exactly with 0.64 g of sulfur.
A42.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the mass
M
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
masses of W and S.
W
S
Mass
1.84 g
0.64 g
–1
Molar mass
184 g mol
32 g mol–1
m
1.84
0.64
Amount, using n =
n=
= 0.01 mol
n=
= 0.020 mol
M
32
184
0.01
0.20
Divide all by smallest
=1
=2
amount
0.01
0.10
Round off to whole
1
2
numbers
 empirical formula is WS2
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Q43.
A clear liquid extracted from fermented lemons was found to consist of carbon,
hydrogen and oxygen. Analysis showed it to be 52.2% carbon and 34.8% oxygen.
a Find the empirical formula of the substance.
b If 2.17 mol of the compound has a mass of 100 g, find the molecular formula of
the compound.
A43.
a
Step 1: Calculate mass of each element present in 100 g.
m(C) = 52.2 g
m(O) = 34.8 g
m(H) = 100 – 52.2 – 34.8
= 13 g
Step 2: Calculate amount, in mol, of each element present.
m
M
52.2
=
12
= 4.35 mol
n(C) =
34.8
16
= 2.175 mol
n(O) =
13
1
= 13 mol
Step 3: Convert to whole-number ratios.
4.35
n(C) =
2.175
=2
2.2175
n(O) =
2.175
=1
13
n(H) =
2.175
=6
Step 4: Write as empirical formula.
C2H6O
n(H) =
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b
30
Step 1: Calculate molar mass of one empirical formula unit.
M = 2  12 + 6  1 + 16
= 46 g mol–1
Step 2: Calculate molar mass of compound.
m
M=
n
100
=
2.17
= 46 g mol–1
Step 3: Calculate number of empirical formula units in compound.
46
no. of units =
46
=1
Step 4: Write molecular formula.
C2H6O
Q44.
When 0.200 g of white phosphorus is burnt in oxygen, 0.456 g of an oxide of
phosphorus is formed. Deduce the empirical formula of this oxide.
A44.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the mass
M
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
masses of P and O.
Mass
Molar mass
m
M
Divide all by smallest
amount
Round off to whole
numbers
Amount, using n =
O
0.456 – 0.200 = 0.256 g
16 g mol–1
0.256
n=
= 0.016 mol
16
0.016
= 2.5
0.006 45
P
0.200 g
31 g mol–1
0.200
n=
= 0.006 45 mol
31
0.006 43
=1
0.006 45
5
2
 empirical formula is P2O5
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31
Q45.
A hydrocarbon is a compound that contains carbon and hydrogen only. Determine the
empirical formula of a hydrocarbon that is used as a specialty fuel and contains 90.0%
carbon.
A45.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the mass
M
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
masses of C and H.
Mass
Molar mass
m
M
Divide all by smallest
amount
Round off to whole
numbers
Amount, using n =
C
90.0 g
12 g mol–1
90
n=
= 7.5 mol
12
7.5
=1
7.5
H
100 – 90.0 = 10.0 g
1 g mol–1
10.0
n=
= 10 mol
1
10
= 1.3
7.5
1×3=3
1.3 × 3 = 4
 empirical formula is C3H4
Q46.
Find the relative atomic mass of nickel if 3.370 g nickel was obtained by reduction of
4.286 g of the oxide (NiO).
A46.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the mass
M
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
mass of O.
Mass
Molar mass
Amount, using n =
m
M
Ni
3.370 g
? g mol–1
3.370
n=
M (Ni)
O
4.286 – 3.370 = 0.916 g
16 g mol–1
0.916
n=
= 0.0573 mol
16
n( Ni)
3.370
1
=

0.0573
n(O)
M (Ni)
As the empirical formula is NiO,
 M(Ni) =
3.370
1
1

=
0.0573 1
M (Ni)
3.370
= 58.9 g mol–1
0.0573
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32
Q47.
4.150 g tungsten was burned in chlorine and 8.950 g tungsten chloride (WCl6) was
formed. Find the relative atomic mass of tungsten.
A47.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n = M , where m is the mass
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
mass of Cl.
Mass
Molar mass
m
M
Amount, using n =
W
4.150 g
? g mol–1
4.150
n=
M (W)
Cl
8.950 – 4.150 = 4.800 g
35.5 g mol–1
4.800
n=
= 0.135 mol
35.5
4.150
n( W )
=
M (W)  0.135
n(Cl)
As the empirical formula is WCl6,
 M(W) = 4.150 
4.150
1
=
M (W)  0.135 6
6
= 184.2 g mol–1
0.135
Q48.
If 3.72 g of element X reacts with exactly 4.80 g of oxygen to form a compound
whose molecular formula is shown, from other experiments, to be X4O10, what is the
relative atomic mass of X?
A48.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the mass
M
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
mass of O.
Mass
Molar mass
Amount, using n =
m
M
X
3.72 g
? g mol–1
3.72
n=
M (X)
n(X )
3.72
=
n(O)
M (X)  0.3
As the empirical formula is X4O10,
 M(X) =
3.72  10
= 31.0 g mol–1
0 .3  4
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3.72
4
=
M (X)  0.3 10
O
4.80 g
16 g mol–1
4.80
n=
= 0.3 mol
16
Worked solutions to textbook questions
33
Q49.
Determine the molecular formulas of compounds with the following compositions and
relative molecular masses:
a 82.75% carbon, 17.25% hydrogen; Mr = 58
b 43.66% phosphorus, 56.34% oxygen; Mr = 284
c 40.0% carbon, 6.7% hydrogen, 53.3% oxygen; Mr = 180
d 0.164 g hydrogen, 5.25 g sulfur, 9.18 g oxygen; Mr = 178
A49.
a
The molecular formula is always a whole-number multiple of the empirical
formula. The empirical formula provides the simplest whole-number ratio of
m
atoms in a compound. The amount of each atom is found by using n =
,
M
where m is the mass in grams and M is the molar mass in g mol–1. Use a periodic
table to find the molar masses of C and H.
Mass
Molar mass
m
M
Divide all by smallest
amount
Round off to whole
numbers
Amount, using n =
C
82.75 g
12 g mol–1
82.75
n=
= 6.89 mol
12
6.89
=1
6.89
H
17.25 g
1 g mol–1
17.25
n=
= 17.25 mol
1
17.25
= 2.5
6.89
1×2=2
2.5 × 2 = 5
 empirical formula is C2H5
Molar mass of a C2H5 unit (empirical formula) = 24 + 5 = 29 g mol–1
Molar mass of the compound (molecular formula) = 58 g mol–1
58
 number of C2H5 units in one molecule =
=2
29
 molecular formula is C4H10
b
P4O10
c
C6H12O6
d
The molecular formula is always a whole-number multiple of the empirical
formula. The empirical formula provides the simplest whole-number ratio of
m
atoms in a compound. The amount of each atom is found by using n =
,
M
where m is the mass in grams and M is the molar mass in g mol–1. Use a periodic
table to find the molar masses of O, S and H.
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Mass
Molar mass
Amount, using
m
n=
M
Divide all by
smallest amount
Round off to
whole numbers
34
H
0.164 g
1 g mol–1
0.164
n=
1
= 0.164 mol
0.164
=1
0.164
S
5.25 g
32 g mol–1
5.25
n=
32
= 0.164 mol
0.164
=1
0.164
O
9.18 g
16 g mol–1
9.18
n=
16
= 0.574 mol
0.574
= 3.5
0.164
1×2=2
1×2=2
3.5 × 2 = 7
 empirical formula is H2S2O7
Molar mass of a H3S3O7 unit (empirical formula) = (2 × 1) + (2 × 32) + (7 × 16)
= 178 g mol–1
Molar mass of the compound (molecular formula) = 178 g mol–1
178
 number of C2H5 units in one molecule =
=1
178
 molecular formula is H2S2O7
Q50.
Using suitable examples, clearly distinguish between the following terms:
a relative isotopic mass
b relative atomic mass
c relative molecular mass
d relative formula mass
e molar mass
A50.
The relative isotopic mass (Ir) of an isotope is the mass of an atom of that isotope relative
to the mass of an atom of 12C, taken as 12 units exactly. For example, the relative isotopic
mass of the lighter of the two chlorine isotopes (see Table 4.3, page 56) is 34.969.
The relative atomic mass of an element is the weighted average of the relative masses
of the isotopes of the element on the 12C scale. For example, the relative atomic mass
of boron (see Table 4.4, on page 57) is 10.81.
The relative molecular mass (Mr) of a compound is the mass of one molecule of that
substance relative to the mass of a 12C atom, which is 12 exactly. For example, the
relative molecular mass of carbon dioxide is 44.0.
Relative formula mass is calculated by taking the sum of the relative atomic masses of
the elements in the formula. Relative formula mass (rather than relative molecular
mass) is the appropriate term to use for ionic compounds as these do not contain
molecules. For example, the relative formula mass of sodium chloride is 58.5.
The molar mass of an element is the mass of one mole of the element. It is equal to
the relative atomic mass of the element expressed in grams. For example, the molar
mass of magnesium (see Table 4.5, on page 62) is 24.3 g mol–1. Note that relative
atomic mass and molar mass of an element are numerically equal. However, relative
atomic mass has no units because it is the mass of one atom of the element compared
with the mass of one atom of the carbon-12 isotope.
The molar mass of a compound is the mass of one mole of the compound. It is equal to
the relative molecular or relative formula mass of the compound expressed in grams.
For example, the molar mass of sodium chloride (Table 4.5, page 62) is 58.5 g mol–1.
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Q51.
Caffeine, which is a stimulant found in coffee, tea and cola drinks, contains 49.48%
carbon, 5.15% hydrogen, 28.87% nitrogen and the rest oxygen.
a Determine the empirical formula of caffeine.
b If 0.20 mol of caffeine has a mass of 38.8 g, what is the molar mass of a caffeine
molecule?
c Determine the molecular formula of caffeine.
d How many moles of caffeine molecules are in 1.00 g caffeine?
e How many molecules of caffeine are in 1.00 g caffeine?
f How many atoms altogether are in 1.00 g caffeine?
A51.
a
The molecular formula is always a whole-number multiple of the empirical
formula. The empirical formula provides the simplest whole-number ratio of
m
atoms in a compound. The amount of each atom is found by using n =
,
M
where m is the mass in grams and M is the molar mass in g mol–1. Use a periodic
table to find the molar masses of C, H, N and O.
Take one mole of caffeine.
Mass of oxygen can be found by subtraction = 100 – (49.48 + 5.15 + 28.87)
= 16.5 g
Mass (g)
Molar mass
Amount,
m
using n =
M
Divide all by
smallest
amount
Round off to
whole
numbers
C
49.48
12.0 g mol–1
49.98
n =
12.0
= 4.17 mol
H
5.15
1.01 g mol–1
5.15
n =
1.01
= 5.10 mol
4.17
= 4.08
1.02
5.10
= 5.0
1.02
4
5
N
28.87
14.0 g mol–1
28.87
n =
14.0
= 2.06 mol
2.06
= 2.02
1.02
2
O
16.5
16.0
16.5
16.0
= 1.02 mol
n =
1.02
= 1.0
1.02
1
 empirical formula is C4H5N2O
b
c
Molar mass of caffeine = 38.8 × 1/0.2 = 194 g mol–1
The molecule must contain a whole number of (C4H5N2O) units.
Molar mass of a C4H5N2O unit is (4 × 12.0) + ( 3 × 1.01) + (2 × 14.0) + (16.0)
= 97 g mol–1.
If the compound has a molar mass of 194 g mol–1, then the number of (C4H5N2O)
units in a molecule = molar mass of compound/molar mass of one unit
= 194 g mol–1/97 g mol–1
=2
The molecular formula of caffeine is therefore 2 × (C4H5N2O), that is,
C8H10N4O2.
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d
e
f
36
n(caffeine) = m/Mr
= 1.00/194
= 5.15 × 10–3 mol
N(caffeine) molecules = n(caffeine) × NA
= 5.15 × 10–3 × 6.02 × 1023
= 3.10 × 1020 molecules of caffeine
N(caffeine) atoms = n(caffeine) × 24 × 6.02 × 1023
= 7.44 × 1025 atoms altogether
Q52.
The empirical formula of a metal oxide can be found by experimentation (see figure
below). The mass of the metal and the mass of the oxygen that reacts with it must be
determined. The six boxes below each contain one step in the experimental method.
A Ignite a burner and heat the metal. B Allow the crucible to cool, then
weigh it.
C Continue the reaction until no
D Clean a piece of metal with the
further change occurs.
emery paper to remove any oxide
layer.
E Place the metal in a clean,
F Weigh the metal and record its
weighed crucible and cover with
mass.
a lid.
a Place the steps in the correct order by letter.
b Wan and Eric collected the following data:
 Mass of the metal = 0.542 g
 Mass of the empty crucible = 20.310 g
 Mass of the crucible and metal oxide = 21.068 g
They found from the data that the metal oxide had a 1 : 1 formula (i.e. MO, where
M = metal). Complete the table, using the data given.
Metal
Mass (g)
Relative atomic mass
Moles
Ratio
c
Oxygen
16.0
What metal was used in the experiment?
Equipment that can be used to find the empirical formula of a metal oxide.
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A52.
a
b
D, F, E, A, C, B
Step 1: Calculate mass of compound after reaction.
m = 21.068 – 20.310
= 0.758 g
Step 2: Calculate mass of oxygen.
m = 0.758 – 0.542
= 0.216 g
Step 3: Calculate amount, in mol, of oxygen.
m
n =
M
0.126
=
16
= 0.0135 mol
Step 4: Use mole ratios to determine amount of metal.
Ratio is 1 : 1, so n = 0.0135 mol
Step 5: Calculate molar mass of metal.
m
M=
n
0.542
=
0.0135
= 40.1
Step 6: Fill in table.
Mass (g)
Relative atomic mass
Moles
Ratio
c
Metal
0.542
40.1
0.0135 mol
1
Oxygen
0.216
16.0
0.0135 mol
1
Refer to periodic table to identify metal as calcium.
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38
Unit 1 Area of Study 1 Review
Multiple-choice questions
Q1.
Which of these proposals corresponds to the ideas put forward by John Dalton in
1804?
I Matter consists of indivisible particles.
II Atoms of a particular element can vary in their mass.
III The proportion and type of atom is always the same in a particular compound.
A I only
B I and II
C I and III
D I, II and III
A1.
C. Dalton did not say that atoms of a particular element could vary in their mass. He
had no concept of the existence of isotopes.
Q2.
When he first constructed his periodic table, Mendeleev arranged the known
elements:
A in order of their atomic number
B according to their electronic configuration
C into vertical groups according to their mass number
D into horizontal periods according to their atomic mass
A2.
D. Atomic number; electronic configuration and mass number were not known at that
time.
Q3.
Ernest Rutherford contributed to knowledge about the structure of the atom by:
A discovering the composition of alpha particles
B discovering that protons are found in the nucleus of an atom
C proving the existence of neutrons in the nucleus of an atom
D proposing that electrons move in circular orbits around the nucleus
A3.
B. This was part of the research that Rutherford and his team of co-workers did that
showed that most of the mass, and the positive charge, of an atom is concentrated in
the nucleus.
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Q4.
Which of the following elements would have atoms with the largest atomic radius?
A Li
B Na
C F
D Cl
A4.
B. Sodium is on the left-hand side of period 3; hence, its outer-shell electron is subject
to the lowest core charge for that period.
Q5.
Which of the following elements would be expected to show greatest similarity in
chemical properties to the element that has 14 electrons in its neutral atoms?
A Al
B P
C Ga
D Ge
A5.
D. Ge is in the same group of the periodic table, with four outer-shell electrons.
Q6.
The Pauli exclusion principle states that:
A all atomic orbitals must hold two electrons
B an atomic orbital must hold a minimum of two electrons
C a new subshell is started whenever an atomic orbital holds two electrons
D an atomic orbital may hold a maximum of two electrons
A6.
D
Q7.
A double negatively charged ion has eight protons. The number of electrons in the ion
is:
A 10
B 8
C 6
D 2
A7.
A. The double negative charge means there are two more electrons than protons,
giving ten electrons in the ion.
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Q8.
35 Cl and 37 Cl. If the
A sample of chlorine was thought to consist of the isotopes 17
17
relative atomic mass of this sample of chlorine was found to be 35.5, it can be said
that:
35 Cl and 37 Cl
A there are equal amounts of 17
17
35
17 Cl
37 Cl
17
than
37 Cl
17
35
17 Cl
B
there is a greater abundance of
C
D
there is a greater abundance of
than
the sample consists of a different isotope with a relative isotopic mass of 35.5
A8.
B. There is a greater abundance of
closer to 35 than to 37.
35
17 Cl
than
37
17 Cl ,
as the relative atomic mass is
Q9.
The number of neutrons in
A 18
B 19
C 20
D 39
39 K+
19
is:
A9.
C. The number of neutrons is the mass number (39) minus the atomic number (19).
Q10.
Which one of the following has a different electronic configuration from the others?
A Na+
B K+
C Ne
D F–
A10.
B. All have ten electrons except K+, which has eighteen electrons.
Q11.
In which groups of the periodic table would you not expect to find a metal?
A groups 14 and 15
B groups 14 and 18
C groups 13 and 17
D groups 17 and 18
A11.
D. These groups have a large number of electrons in the outer shell and cannot lose
electrons easily.
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Q12.
Which one of the following pairs of elements is most likely to combine to form a
compound with properties similar to those of sodium chloride?
A calcium and bromine
B carbon and oxygen
C copper and nickel
D phosphorus and chlorine
A12.
A. Calcium is a metal with a small number of electrons in the outer shell. Bromine is a
non-metal with a large number of electrons in the outer shell. Sodium chloride is also
a compound of a metal and a non-metal, so it will have similar properties to a
compound of calcium and bromine.
Q13.
Isotopes of an element contain:
A the same atomic number and the same mass number
B the same atomic number and a different mass number
C a different atomic number and the same mass number
D a different atomic number and a different mass number
A13.
B. Isotopes have the same number of protons (atomic number) and a different number
of neutrons (mass number is protons plus neutrons).
Q14.
Going down group 7 of the periodic table, the electronegativity:
A decreases because the atomic radius increases
B increases because the atomic number increases
C decreases because the atomic number increases
D increases because the number of subshells increases
A14.
A. The atomic radius increases; hence, the attraction for an additional electron
decreases.
Q15.
The relative atomic mass of magnesium, Ar(Mg), is 24.31. The most important reason
why it is not a whole number is that:
A magnesium atoms lose electrons when they react
B the relative atomic mass given is only an approximation
C not all atoms of magnesium have the same number of neutrons
D the mass of the magnesium atom is compared to the mass of the 12C isotope
A15.
C. Relative atomic mass is an average of the relative isotopic masses for an element.
Isotopes are atoms of magnesium with different numbers of neutrons.
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Q16.
Which of the following lists contain empirical formulas only?
A C2H2, CBr4, Ca(OH)2, KMnO4
B NH3, N2H4, C6H5CH3, HCOOH
C H2SO4, Al2(SO4)3, C6H5CH3, CH3Cl
D HCOOH, C2H6O, Fe2O3.xH2O, C6H12O6
A16.
C. The following are not empirical formulas: H2SO4, Al2(SO4)3, C6H5CH3, CH3Cl.
Q17.
A compound has an empirical formula of CH. A 0.25 mol sample of the compound
weighs 13 g. The molecular formula is:
A C8H8
B C6H6
C C4H4
D C2H2
A17.
C.
Molar mass of hydrocarbon is
13 g
= 52 g mol–1
0.25 mol
Molar mass of CH unit is 13 g mol–1. Number of units =
52
=4
13
 molecular formula is C4H4
Short-answer questions
Q18.
a
b
c
d
In what part of the periodic table are the metals found? Why are they found there?
The helium atom contains two electrons in the outer shell. Why is helium not
placed in group 2 with the other elements also containing two electrons in the
outer shell?
The heavier elements, atomic numbers 87–112, are all metals. Explain this fact.
Some metallic elements are very reactive. Which group of the periodic table
contains the most reactive metals?
A18.
a
b
c
d
Metals are found at the left and bottom of the periodic table. These elements have
a small number of electrons in their outer shell.
Helium has a full outer shell and cannot easily donate these electrons as metals
do.
Elements 87–112 all have just one or two outer-shell electrons, which can be
easily lost.
group 1
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Q19.
a Explain, using suitable examples, the relationship between the electronic
configuration of an element and its position in the periodic table.
b Consider the following: Na, Cl, Mn, Ca2+, O2–, Al3+. For each:
i give its electronic configuration
ii indicate its position in the periodic table (e.g. group 17, period 2, or transition
series, period 4)
A19.
a For any element, the shell of the highest order of an element containing electrons
determines the period of an element. For example, in calcium the fourth shell
contains electrons and, being the highest order in which electrons are found,
makes calcium a period 4 element.
The number of electrons in the outermost shell of an element determines the
group number of the element. For example, calcium has two electrons in its
outermost shell and so belongs to group 2.
Transition metals are those that have a ‘d’ subshell being filled.
Lanthanides and actinides have ‘f’ subshells being filled.
b Na
1s22s22p63s1
group 1 period 3
2 2 6 2 5
Cl
1s 2s 2p 3s 3p
group 17 period 3
Mn
1s22s22p63s23p63d54s2 transition series period 4
Ca2+
1s22s22p63s23p6
group 2 period 4
2–
O
1s22s22p6
group 16 period 2
3+
2 2 6
Al
1s 2s 2p
group 13 period 3
Q20.
Select your answers to the questions below from the following list of elements:
Cl, C, Na, Mg, K, O, F, Al, N, Ca.
Which elements:
a are in period 2 of the periodic table?
b are in period 3 of the periodic table?
c are in group 1 of the periodic table?
d are in group 2 of the periodic table?
e are in group 13 of the periodic table?
f are in group 16 of the periodic table?
g are classified as metals?
h have one valence electron?
i have three valence electrons?
j has the highest molar mass?
A20.
a
b
c
d
e
f
g
h
i
j
C, N, O and F
Na, Mg, Al and Cl
Na and K
Mg and Ca
Al
O
Na, Mg, Al, K and Ca
Na and K
Al
Ca
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Q21.
a
Supply the missing information in the table.
Atomic
Neutron
Mass
number
number
number
4
9
(i)
8
b
(ii)
Symbol
9
4 Be
(iii)
17
8
31
15
(iv)
(v)
(vi)
34
(vii)
78
O
P
(viii)
The neutron was not discovered until more than 30 years after the discovery of
the proton and the electron. Why was the neutron more difficult to detect?
A21.
a
b
i 5
ii 9
iii 17
iv 15
v 16
vi 31
vii 44
78
Se
viii 34
Most of the instruments used for investigating the structure of the atom are based
on the use or measurement of electric charge. As the neutron is an uncharged
particle, it was not detected by these instruments.
Q22.
a
b
c
d
Explain the meanings of the following terms:
i relative atomic mass
ii relative molecular mass
iii mole
iv Avogadro’s number
v molar mass
Write the electronic configuration for the element phosphorus.
When 0.100 g of white phosphorus is burned in oxygen, 0.228 g of an oxide of
phosphorus is produced. The molar mass of the oxide is 284 g mol–1.
i Determine the empirical formula of the phosphorus oxide.
ii Determine the molecular formula of the phosphorus oxide.
Would you expect the properties of the oxide of phosphorus to be more similar to
those of sodium chloride or those of water? Explain your answer.
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A22.
a
i
Relative atomic mass—the weighted mean of the relative masses of the
isotopes of an element on the 12C scale.
ii Relative molecular mass—the relative mass of a molecule on the 12C scale.
iii Mole—the amount of substance that contains the same number of specified
particles as there are atoms in exactly 12 g of 12C.
iv Avogadro’s number—the number of carbon atoms in exactly 12 g of 12C
(approximately 6.02  1023).
v Molar mass—mass in grams of a mole of a substance.
b
1s22s22p63s23p3
c
i
Step 1: Write the ratio by mass.
P
: O
0.100 g : (0.228 – 0.100) g
0.100 g : 0.128 g
Step 2: Calculate the ratio by amount (in moles).
0.100 g
0.128 g
:
1
16.0 g mol 1
30.974 g mol
0.003 229 mol
: 0.008 mol
Step 3: Divide by the smaller amount.
0.008
0.003 229
:
0.008
0.008
0.4036
: 1
Step 4: Express as integers by multiplying by 5.
2:5
 empirical formula of the compound is P2O5
ii
d
As the empirical formula is P2O5, the molecule must contain a whole number
of P2O5 units.
The molar mass of one of these units is
((2  30.974) + (5  16)) = 141.948 g mol–1.
The number of units in a molecule = molar mass of the compound/molar
mass of one unit
284 g mol 1
=
141.948 g mol 1
=2
 molecular formula of the compound is P4O10
Properties are more similar to those of water. Water is a compound of two nonmetals (hydrogen and oxygen), as is the oxide of phosphorus, since both
phosphorus and oxygen are non-metals.
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46
Q23.
The original version of the periodic table was devised by the Russian chemist Dmitri
Mendeleev in 1869.
a What were the two pieces of information that he used to devise the table?
In the modern form of the periodic table, the elements are arranged in order of atomic
number.
b What information about the structure of the atom is given by its atomic number?
c What is the link between the atomic number of an element and the block in the
periodic table in which it would be found?
d Consider the element with atomic number 14.
i Write the full electronic configuration.
ii In which group, period and block of the periodic table would it be found?
A23.
a
b
c
d
He arranged the known elements:
 in order of increasing atomic mass;
 with elements having similar chemical properties in vertical groups.
the number of protons
For a neutral atom, the number of protons equals the number of electrons, and the
electrons are arranged into shells and subshells of increasing energy. The block of
the periodic table corresponds to the highest energy subshell of electrons in the
particular atom.
i 1s22s22p63s23p2
ii group 4, period 3, p-block
Q24.
The electronic configurations of seven elements (A–G) are given below.
A 1s22s22p5
B 1s22s22p63s1
C 1s22s22p63s13p5
D 1s22s22p63s23p64s2
E 1s22s22p63s23p63d64s1
F 1s22s22p63s23p63d84s2
G 1s22s22p63s23p63d104s24p5
Indicate which one or more are likely to be:
a metals
b d-block elements
c group 17 elements
d period 3 elements
e elements not in the ground state
A24.
a
b
c
d
e
B, D, E, F
E, F
A, G
B, C
C, E
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47
Q25.
a
b
c
d
e
Give the ground-state electronic configuration of calcium (Ca).
Explain, giving an example of the new electronic configuration, what happens to
the energy levels of the electrons when the atom in the ground state is provided
with sufficient energy to become:
i an excited atom
ii a charged ion
The stable ion of calcium is Ca2+. What would be the difference in atomic radius
of a Ca atom and a Ca2+ ion?
Why would a Ca atom and a Ca2+ ion have almost the same mass?
Give the symbol of two elements that would have chemical properties similar to
those of calcium.
A25.
a
b
c
d
e
1s22s22p63s23p64s2
i Electrons are promoted to a higher energy level. An example of an electronic
configuration of an excited calcium atom could be (several answers
possible): 1s22s22p63s23p64p2.
ii With sufficient energy, a calcium atom can lose its valence electrons to form
a cation e.g. Ca2+ with electronic configuration 1s22s22p63s23p6.
The ion is much smaller as electrons occupy most of the atom’s volume and the
ion has one less electron shell than the atom.
The mass of electrons is negligible in comparison to that of the protons and
neutrons in an atom. The number of protons and neutrons is the same in the atom
and ion.
e.g. Mg, Sr
Q26.
Give explanations for the following:
a A sodium atom has a larger atomic radius than a chlorine atom, even though both
belong to period 3.
b A sodium ion (Na+) is much smaller than the sodium atom.
c Fluorine is more electronegative than iodine.
A26.
a
b
c
Members of period 3 all have their outer-shell electrons in the third shell but the
core charge increases across the period from +1 for sodium to +7 for chlorine.
Hence, the outer-shell electrons in the chlorine are attracted more strongly to the
nucleus, reducing the radius of the species.
From the electronic configuration for sodium, 1s22s22p63s1, it can be seen that the
Na atom has three shells of electrons but the Na+ ion has lost the single outershell electron and the remaining eight electrons of the second shell are then
attracted to a greater core charge.
The fluorine atom is smaller than that of iodine. Electrons are attracted to fluorine
atoms more strongly than to those of iodine because they are closer to the
positively charged nucleus in fluorine. So, fluorine is more electronegative.
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Q27.
14
8
15
19
40
32
20
4
The following is a list of atoms: 27
13 A, 20 B, 16 C, 7 D, 3 E, 7 F, 9 G, 9 H, 2 I.
a Which pairs of atoms are isotopes?
b Which atoms have equal numbers of protons and neutrons in the nucleus?
c Which is an isotope of sulfur?
d Which has one more electron than a magnesium atom?
e Which is a group 2 element?
f How many different elements are shown?
A27.
a
b
c
d
e
f
D and F, G and H
B, C, D and I
C
A
B
7
Q28.
A sample of magnesium carbonate weighs 21.8 g.
a Calculate the amount (mol) of magnesium carbonate present.
b Calculate the amount (mol) of oxygen atoms present.
c Calculate the number of carbon atoms present.
d Calculate the total number of atoms present.
e Calculate the percentage, by mass, of magnesium in magnesium carbonate.
A28.
a
b
Step 1: Calculate the molar mass of MgCO3.
M = 24.3 + 12 + 3  16
= 84.3 g mol–1
Step 2: Calculate the amount (mol) of MgCO3.
m
n =
M
21.8
=
84.3
= 0.2586 mol
Step 3: Give answers with correct number of significant figures.
n = 0.259 mol
Step 1: Calculate number of oxygen atoms per molecule.
No. atoms per molecule = 3
Step 2: Calculate amount (mol) of oxygen atoms.
n = 3  0.259
= 0.776 mol
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48
Worked solutions to textbook questions
c
d
e
49
Step 1: Calculate number of carbon atoms per molecule.
No. atoms per molecule = 1
Step 2: Calculate amount (mol) of carbon atoms.
n = 1  0.259
= 0.259 mol
Step 3: Calculate number of carbon atoms.
No. atoms = 0.259  6.02  1023
= 1.56  1023 atoms
Step 1: Calculate number of atoms per molecule.
No. atoms per molecule = 5
Step 2: Calculate amount (mol) of atoms.
n = 5  0.259
= 1.30 mol
Step 3: Calculate total number of atoms.
No. atoms = 1.30  6.02  1023
= 7.80  1023 atoms
M (Mg)
% of Mg in MgCO3 
100%
M (MgCO 3)
=
24.3
100%
84.3
= 28.8%
Q29.
Ethylene glycol is a compound often used as an antifreeze in cars in cold weather. Its
molar mass is 62 g mol–1. It has a percentage composition of 38.7% carbon, 9.7%
hydrogen and the rest oxygen. Determine:
a the empirical formula of ethylene glycol
b the molecular formula of ethylene glycol
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50
A29.
The molecular formula is always a whole-number multiple of the empirical formula.
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the mass
M
in grams and M is the molar mass in g mol–1. Use a periodic table to find the molar
masses of C, H and O.
a
C
O
H
Mass
38.7 g
100 – (38.7 + 9.7) = 51.6 g 9.7 g
Molar mass
12 g mol–1
16 g mol–1
1 g mol–1
Amount, using
38.7
9 .7
51.6
n =
n =
m
n =
= 3.225 mol
12
1
n=
16
= 3.225 mol
= 9.7 mol
M
3.225
3.225
9.7
Divide all by
=1
=1
=3
smallest amount 3.225
3.225
3.225
Round off to
1
1
3
whole numbers
b
 empirical formula is CH3O
Molar mass of a CH3O unit (empirical formula) = 12 + 3 + 16 = 31 g mol–1
Molar mass of the compound (molecular formula) = 62 g mol–1
62
 number of CH3O units in one molecule =
=2
31
 molecular formula is C2H6O2
Q30.
Methane (CH4) is the major component of natural gas.
a What is the mass of 0.50 mol methane?
b How many molecules are there in 0.100 g methane?
c How many atoms are there altogether in 0.10 g methane?
d How many protons are there in 0.10 g methane?
e What is the percentage by mass of carbon in methane?
f What mass of carbon would be present in 34 g methane?
A30.
It is useful to remember the formula m = nM, where m is the mass in grams, n the
amount of substance in mol, and M the molar mass in g mol–1. Use a periodic table to
find the molar masses. M(C) = 12 g mol–1 and M(CH4) = 16 g mol–1. Remember also
that the number of particles in 1 mol is Avogadro’s number, NA = 6.02  1023.
a m(CH4) = 0.50  16 = 8.0 g
0.100
b n(CH4) =
16
= 0.006 25 mol
 number of molecules = nNA
= 0.006 25  6.02  1023
= 3.76  1021 molecules
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c
d
e
f
51
There are 5 atoms in each molecule of methane (1 of C and 4 of O).
So, total number of atoms in 3.76  1021 molecules = 5  3.76  1021 atoms
= 1.88  1022 atoms
There are 10 protons in each molecule of methane (6 in each C and 1 in each of
the 4 H atoms).
So, total number of protons in 3.76  1021 molecules = 10  3.76  1021 protons
= 3.76  1022 protons
12
%(C) =
× 100
16
= 75%
75
m(C in 34 g) = 75% of 34 = 34 
100
= 26 g
Q31.
Cobalt (Co) has an atomic number of 27.
a Give the ground state electronic configuration of cobalt.
b To what section of the periodic table does cobalt belong?
Cobalt reacts with oxygen to form a compound, cobalt oxide, formula CoO. Cobalt
oxide contains 78.6% by mass of cobalt.
c Use the above information to calculate the relative atomic mass of cobalt.
A31.
a
b
c
1s22s22p63s23p63d74s2
first transition metal series or d-block
The empirical formula provides the simplest whole-number ratio of atoms in a
m
compound. The amount of each atom is found by using n =
, where m is the
M
mass in grams and M is the molar mass in g mol–1. Use a periodic table to find the
molar mass of O.
O
21.4 g
16 g mol–1
m
21.4
Amount, using n =
n=
= 1.34 mol
M
16
78.6
As the empirical formula is CoO, the ratio is 1 : 1 =
: 1.34
M (Co)
78.6
 M(Co) = 1.34 = 58.7 g mol–1
Mass
Molar mass
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Co
78.6 g
? g mol–1
78.6
n=
M (Co)
Worked solutions to textbook questions
52
Q32.
a
Calculate the relative atomic mass of silicon from the data given below.
Relative isotopic mass
27.98
28.98
29.98
b
Relative abundance (%)
92.2
4.7
3.1
To what:
i group
ii period
of the periodic table does silicon belong?
A32.
a
Substitute into the formula for relative atomic mass.
(isotopic mass  relative abundance)
Ar(Si) =
 total relative abundance
=
(27.98  92.2)  (28.98  4.7)  (29.97  3.1)
100
=
2579.756  136.206  92.907
100
=
2808.869
100
= 28.1
b
i
ii
14
3
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53
Q33.
The successive ionisation energies for an element have been plotted in the diagram.
a
b
c
How many shells of electrons exist for this atom?
Use this information to write the electronic configuration for this element in its
ground state.
What is the likely charge on the stable cation of the element?
A33.
a
b
c
3
1s22s22p63s1
1+
Q34.
The graphs show trends in some of the properties of elements as you go from top to
bottom down group 17 of the periodic table.
Which graph best represents the trend for:
a electronegativities?
b first ionisation energies?
c number of outer-shell electrons?
Figure 3.13
d atomic radii?
A
B
C
A34.
a
b
c
d
C. Electronegativity decreases as atomic radius increases down a group.
C. First ionisation energy decreases as atomic radius increases down a group.
A. Number of outer-shell electrons is constant within a group.
B: Atomic radius increases down a group as nth number of electron shells
increases.
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