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AS Topic 1
Calculations Workbook
Student workbook
NAME: ___________________________
2
Sr
Rb
88
87
* Actinide series
89
Ac*
[227]
actinium
* Lanthanide series
radium
Ra
Fr
francium
56
[226]
55
[223]
57
La*
lanthanum
Ba
caesium
barium
39
138.9
38
137.3
37
132.9
Cs
yttrium
Y
21
88.9
scandium
Sc
45.0
(3)
strontium
rubidium
20
87.6
Ca
K
19
85.5
40.1
39.1
calcium
12
11
potassium
magnesium
sodium
Mg
24.3
23.0
Na
4
Li
3
9.0
Be
6.9
beryllium
(2)
(1)
lithium
2
1
(7)
(8)
(9)
(10)
[231]
Pa
91
232
Th
90
thorium protactinium
59
58
Pr
92
uranium
U
238
60
Nd
144
106
Sg
[266]
74
W
tungsten
43
44
(11)
45
93
94
Pu
[242]
62
Sm
150
108
95
Am
[243]
63
Eu
152
109
Mt
[268]
77
Ir
iridium
192.2
96
curium
Cm
[247]
64
Gd
157
110
Ds
[271]
78
Pt
platinum
195.1
46
palladium
Pd
28
106.4
nickel
Ni
58.7
(12)
81
Tl
thallium
204.4
49
indium
In
114.8
97
82
Pb
lead
207.2
50
tin
Sn
118.7
32
31
Ge
germanium
Ga
gallium
S
83
Bi
bismuth
209.0
51
antimony
Sb
121.8
84
Po
polonium
[209]
52
tellurium
Te
127.6
34
33
Se
selenium
As
arsenic
16
79.0
15
74.9
sulfur
P
phosphorus
8
32.1
oxygen
O
16.0
(16)
6
7
31.0
nitrogen
N
14.0
(15)
5
Ar
85
At
astatine
98
Cf
[251]
66
Dy
99
Es
[254]
67
Ho
165
dysprosium holmium
163
100
fermium
Fm
[253]
68
Er
erbium
167
101
102
No
[254]
70
Yb
ytterbium
173
103
Lr
[257]
71
Lu
lutetium
175
mendelevium nobelium lawrencium
Md
[256]
69
Tm
thulium
169
86
Rn
radon
[222]
54
[210]
53
Xe
xenon
I
iodine
131.3
126.9
36
35
Kr
krypton
Br
bromine
18
83.8
17
79.9
argon
Cl
chlorine
10
39.9
9
35.5
Ne
F
neon
20.2
19.0
fluorine
2
helium
He
(18)
4.0
0 (8)
(17)
7
Elements with atomic numbers 112-116 have been reported
but not fully authenticated
80
Hg
mercury
200.6
48
cadmium
Cd
30
112.4
Zn
zinc
65.4
14
72.6
13
69.7
berkelium californium einsteinium
Bk
[245]
65
Tb
terbium
159
111
Rg
[272]
79
gold
Au
197.0
47
silver
Ag
29
107.9
copper
Cu
63.5
hassium meitnerium darmstadtium roentgenium
Hs
[277]
76
Os
osmium
190.2
rhodium
Rh
27
102.9
neptunium plutonium americium
Np
[237]
61
Pm
[147]
107
Bh
[264]
75
Re
rhenium
186.2
Ru
26
101.1
cobalt
Co
58.9
praseodymium neodymium promethium samarium europium gadolinium
141
140
Ce
105
cerium
42
183.8
Tc
25
[98]
iron
Fe
55.8
molybdenum technetium ruthenium
Mo
24
95.9
Mn
54.9
dubnium seaborgium bohrium
Db
[262]
73
Ta
tantalum
180.9
41
niobium
Nb
23
92.9
Cr
52.0
(6)
vanadium chromium manganese
V
50.9
(5)
104
rutherfordium
Rf
[261]
72
Hf
hafnium
178.5
40
zirconium
Zr
22
91.2
titanium
Ti
47.9
(4)
Si
silicon
Al
aluminium
6
28.1
5
27.0
atomic (proton) number
carbon
C
12.0
(14)
4
B
boron
10.8
(13)
3
atomic symbol
name
relative atomic mass
Key
1
hydrogen
H
1.0
The Periodic Table of Elements
Contents
Section 1 : Atoms
4
Section 2 : The mole
7
Section 3 : Moles of gases
11
Section 4 : The Avogadro constant
16
Section 5 : Finding formulae
18
Section 6 : Solutions
24
Section 7 : Scaled quantities
27
Section 8 : Equations and amounts
29
Students should read the introduction and examples at the start
of each section then attempt to complete all the questions that
are NOT in a shaded box.
Version 6: June 2013
3
Section 1: Atoms
All matter is made of particles. At one time, it was thought that the tiniest
particle was the atom, which comes from the Greek word meaning
‘indivisible’.
We now know that atoms can be split and that there are particles smaller
than atoms, subatomic particles, electrons, protons and neutrons. You will
need to know about these particles, but remember that chemistry is all
about rearrangements of atoms that do not themselves change.
Atoms are very small. The hydrogen atom, the smallest and lightest of all
atoms, has a diameter of about 10-8 mm. 1 g of hydrogen atoms contains
about 6 x 1023 atoms.
An atom is the smallest, electrically neutral, particle of an element that
can take part in a chemical change.
A molecule is the smallest, electrically neutral, particle of an element or
compound that can exist on its own.
An ion is an atom, or group of atoms, which carries an electric charge.
You need to know these definitions and you need to be able to recognise the
formulae of atoms and molecules. Li, O, Cl, C are all formulae which
represent atoms. Some but not all of these can exist on their own. Oxygen,
for example, unless combined with something else normally exists as oxygen
molecules, O2, which contain two atoms.
The mass of an individual atom is very small and it is convenient to measure
atomic masses as relative masses.
Relative Atomic Mass (Ar) is the average mass of one atom in a naturally
occurring sample on a scale where the mass of one atom of carbon—12 is
exactly 12 atomic mass units.
Relative Molecular Mass (Mr) is the average mass of a single molecule on a
scale where the mass of one atom of carbon—12 is exactly 12 atomic mass
units.
Relative Formula Mass is a better phrase than Relative Molecular Mass
because many compounds such as salts exist as giant ionic structures rather
than separate molecules. Since such a compound does not exist as
molecules, it cannot have relative ‘molecular’ mass but the term is often
used nonetheless.
Any relative mass (Atomic, Molecular or Formula) does not have units.
4
Example 1
Calculate the Relative Formula Mass of sulfuric acid H2SO4
2 atoms of hydrogen each of mass 1
1 atom of sulfur of mass 32.1
4 atoms of oxygen of mass 16
=2x1
= 1 x 32.1
= 4 x 16
∴ Relative Formula Mass
=2
= 32.1
= 64
= 98.1
Example 2
Calculate the Relative Formula Mass of lead nitrate Pb(NO3)2
Care! This formula contains TWO nitrate groups.
1 atom of lead of mass 207.2
2 atoms of nitrogen of mass 14
6 atoms of oxygen of mass 16
= 1 x 207.2
= 2 x 14
= 6 x 16
= 207.2
= 28
= 96
∴ Relative Formula mass
= 331.2
Example 3
Calculate the Relative Formula Mass of CuSO4.5H2O
Care! This formula includes 5 molecules of water attached to each copper
sulfate unit. Many students make the mistake of thinking that there are 10
hydrogens and only 1 oxygen.
In CuSO4 1 atom of copper of mass 63.5
1 atom of sulfur of mass 32.1
4 atoms of oxygen each of mass 16
= 1 x 63.5
= 1 x 32.1
= 4 x 16
In 5H2O 10 atoms of hydrogen each of mass 1 = 10 x 1
5 x 1 atoms of oxygen each of mass 16
= 5 x 16
∴ Relative Formula Mass
= 63.5
= 32.1
= 64
= 10
= 80
= 249.6
Calculations of this type can be written as follows:
RFM [CuSO4.5H2O] = [ 63.5 + 32.1 + (4 x 16) + 5{(2 x 1) + 16} ] = 249.6
5
Exercise 1: Calculation of the Relative Mass of compounds
Calculate the Relative Formula Mass of the following compounds. You will
find relative atomic masses on the periodic table of elements.
1 H2O
2 CO2
3 C2H5OH
4 NH4VO3
5 Na2CO3
6 Pb(NO3)2
7 Fe2(SO4)3
8 (NH4)2SO4
9 CuSO4.5H2O
10 Na2S2O3.5H2O
11 NaClO
12 Pb3O4
13 CH3COCH3
14 NH4Cl
15 CaSO4
16 KMnO4
17 Cu(OH)2
18 Al2(SO4)3
19 CoCl2.6H2O
20 (NH4)2SO4.Fe2(SO4)3.24H2O
6
Section 2: The mole
Chemists often want to know how many particles are present in an amount
of substance. There are vast numbers of particles present in even very small
quantities of substance and chemists find it convenient to count particles
using a unit called a mole (symbol mol).
One mole is the amount of substance which contains the same number of
particles (atoms, ions, molecules, formulae or electrons) as there are
carbon atoms in exactly 12 g of pure carbon-12.
This number is known as the Avogadro constant (symbol L or NA), and is
equal to 6.02 x 1023 mol−1.
The molar mass of a substance is the mass in grams of one mole of
substance. It is equivalent to the relative formula mass given units of
gmol-1.
number of moles =
mass in g
molar mass in gmol-1
Speaking or writing about moles, you should always state whether you are
dealing with atoms, molecules, ions, formulae etc.
Example 1
Calculate the number of moles of oxygen atoms in 64 g of oxygen atoms.
The Ar of O atoms is 16 so the molar mass of O atoms is 16 gmol−1.
number of moles of O atoms
=
64 g
16 gmol-1
=
4 mol
Example 2
Calculate the number of moles of chlorine molecules in 14.2 g of chlorine.
The Mr of Cl2 is [35.5 x 2] = 71 so the molar mass of Cl2 molecules is 71
gmol-1.
number of moles of Cl2 molecules
7
=
14.2 g
71 gmol-1
=
0.2 mol
Example 3
Calculate the number of moles of CuSO4.5H2O in 100 g of the solid.
The RFM of CuSO4.5H2O is [63.5 + 32.1 + (4 x 16) + 5{(2x1) + 16}] = 249.6 so
the molar mass of CuSO4.5H2O is 249.5 gmol-1.
number of moles of molecules
=
100 g
249.6 gmol-1
=
0.4006 mol
Example 4
Calculate the mass of 3 moles of sulfur dioxide SO2.
The Mr SO2 is [32.1 + (2 x 16)] = 64.1 so the molar mass of SO2 is 64.1 gmol-1.
∴ mass of 3 moles of SO2
= 3 mol x 64.1 gmol-1
= 192.3 g
Example 5
Calculate the mass of 0.05 moles of Na2S2O3.5H2O .
The RFM of Na2S2O3.5H2O is [(23 x 2) + (32.1 x 2) + (16 x 3)] + 5[(2 x 1) +16]
= 248.2 so the molar mass is 248.2 g mol–1
∴ 0.05 moles of Na2S2O3.5H2O
= 0.05 mol x 248.2 gmol-1
= 12.41 g
8
Exercise 2a: Calculation of the number of moles of
material in a given mass of that material
In this set of calculations all the substances are from the list of compounds
whose RFM you calculated in Exercise 1.
In each case calculate the number of moles of the material in the stated
mass.
1 9.00 g of H2O
2 88.0 g of CO2
3 23.4 g of NH4VO3
4 2.25 g of Na2CO3
5 1.43 g of Pb(NO3)2
6 13.77 g of (NH4)2SO4
7 230 g of C2H5OH
8 3.10 g of CH3COCH3
9 9.84 g of Fe2(SO4)3
10 3.16 g of KMnO4
11 3.40 g of CaSO4
12 50.9 g of CuSO4.5H2O
9
Exercise 2b: Calculation of the mass of material in a given
number of moles of at material
In this set of calculations all the substances are from the list of compounds
whose RFM you calculated in Exercise 1.
In each case calculate the mass in grams of the material in the number of
stated moles.
1 2 moles of H2O
2 0.50 moles of C2H5OH
3 0.11 moles of Na2CO3
4 0.06 moles of Pb(NO3)2
5 1.1 moles of (NH4)2SO4
6 0.20 moles of CuSO4.5H2O
7 3 moles of CO2
8 0.20 moles of CH3COCH3
9 0.050 moles of KMnO4
10 0.10 moles of CaSO4
11 0.025 moles of NH4VO3
12 0.10 moles of NaClO
10
Section 3: Moles of gases
For reactions involving gases it is often more convenient to measure a gases
volume rather than its mass. The volume occupied by any gas is easily
related to the number of particles it contains by the molar volume.
The molar volume of any gas is the volume occupied by one mole at room
temperature and atmospheric pressure (r.t.p). It is equal to 24 dm3mol-1 at
r.t.p.
volume of gas (in dm3) = amount of gas (in mol) x 24 dm3mol-1
Avogadro’s Law states that equal volumes of all gases, under the same
conditions of temperature and atmospheric pressure contain the same
number of molecules. If the volume is 24 dm3, at room temperature and
pressure, this number, is the Avogadro constant.
Remember that 1 dm3 (a.k.a. 1l) = 1000 cm3 (a.k.a. 1000 ml)
Example 1
Calculate the volume of 2 moles of oxygen molecules.
Remember you do not need to work out the molar mass to do this
calculation as it does not matter what gas it is.
volume of 2 moles of O2
= 2 mol x 24 dm3mol-1
= 48 dm3
Example 2
Calculate the volume of 0.0056 moles of chlorine molecules.
volume of 0.0056 moles of Cl2 = 0.0056 mol x 24 dm3mol-1
= 0.1344 dm3
= 134.4 cm3
11
Example 3
Calculate the number of moles of hydrogen molecules in 240 cm3 of the gas.
number of moles of H2
=
240 cm3
24000 cm3mol-1
= 0.01 mol
Example 4
Calculate the number of moles of a gas in 1000 cm3 of the gas.
number of moles of gas =
1000 cm3
24000 cm3mol-1
= 0.0417 mol
Example 5
Calculate the volume of 88 g of carbon dioxide gas.
This is a two-stage calculation a) you need to calculate how many moles of
carbon dioxide gas are present and then b) you need to convert this to a
volume.
Molar mass of CO2 is [12 + (2 x 16) ] = 44 gmol-1
∴ number of moles of CO2 in 88 g =
88 g
44 gmol-1
= 2 mol
∴ volume of gas
= 2 mol x 24 dm3mol-1
= 48 dm3
12
Example 6
Calculate the mass of 1000 cm3 of sulfur dioxide.
Again this is a two-stage calculation a) you need to calculate the number of
moles of sulfur dioxide and then b) convert this to a mass using the molar
mass.
number of moles of SO2
=
1000 cm3
24000 cm3mol-1
= 0.0417 mol
molar mass of SO2 is [ 32 + (2 x 16) ] = 64 gmol-1
mass of 0.0417 mol
= 0.0417 mol x 64 gmol-1
= 2.669 g
Example 7
Calculate the Molar Mass of a gas for which 100 cm3 of the gas at room
temperature and pressure has a mass of 0.0667 g.
For calculations of this type you need to find the mass of 1 mole of the gas,
(ie 24 000 cm3 ) ; this is the molar mass of the gas.
100 cm3 of gas has a mass = 0.0667 g.
∴ 24000 cm3 of gas would have mass = 24000 cm3 x 0.0667 g
100 cm3
= 16 g
∴ The molar mass of the gas is 16 gmol–1
13
Exercise 3a: Calculation of the volume of a given number
of moles of a gas
In each case calculate the volume of the number of moles of molecules of
the stated gas.
1 1 mole of CO2
2 0.1 moles of NH3
3 0.5 moles of C2H4
4 2 moles of SO2
5 0.12 moles of NO2
6 0.0056 moles of C2H6
7 0.0090 moles of C3H8
8 0.040 moles of C2H2
9 0.123 moles of NO
10 3.4 moles of HBr
Exercise 3b: Calculation of the number of moles of gas in a
given volume of that gas
In each case calculate the number of moles of molecules of the stated gas.
1 200 cm3 of CO2
2 50 cm3 of NH3
3 1000 cm3 of C2H4
4 2000 cm3 of SO2
5 234 cm3 of NO2
6 2.7 dm3 of HBr
7 256 cm3 of Cl2
8 42 dm3 of F2
9 900 cm3 of N2
10 2 dm3 of H2
14
Exercise 3c: Calculation of the volume of a given mass of
gas
In each case calculate the volume in cm3 of the mass of gas given.
1 2 g of CO2
2 5 g of NH3
3 10 g of C2H4
4 20 g of SO2
5 2.3 g of NO2
6 2.26 g of HBr
7 10 g of Cl2
8 20 g of CH4
9 200 g of H2
10 240 g of O2
Exercise 3d: Calculation of the mass of a given volume of
gas
Calculate the mass of the volume of gases stated below.
1 200 cm3 of CO2
2 500 cm3 of NH3
3 1000 cm3 of C2H4
4 2000 cm3 of SO2
5 23 dm3 of NO2
6 226 cm3 of HBr
7 2.5 dm3 of Cl2
8 200 cm3 of CH4
9 2000 cm3 of H2
10 1 dm3 of O2
15
Section 4: The Avogadro Constant
It is usually sufficient for chemists to count particles in numbers of moles
but occasionally they want to know the absolute number of particles.
To calculate the absolute number of particles simply multiply the number of
moles by the the Avogadro constant (6.02 x 1023 mol−1).
number of particles = number of moles x 6.02 x 1023 mol−1
Example 1
Calculate the number of atoms in 2 moles of sodium atoms (Na)
number of atoms = 2 mol x 6.02 x 1023 mol−1
= 1.204 x 1024
Example 2
Calculate the number of atoms in 1.15 g of sodium (RAM(Na)=23)
number of moles of Na = 1.15 = 0.05 mol
23
number of atoms = 0.05 mol x 6.02 x 1023 mol−1
= 3.01 x 1023
Example 3
Calculate the number of H atoms in 54 g of water molecules ( RFM[H2O]=18 )
number of moles of H2O = 54 = 3 mol
18
number of moles of H atoms = 3 x 2 = 6 mol
number of moles of H atoms = 6 mol x 6.02 x 1023 mol−1
= 3.612 x 1024
16
Exercise 4: Calculations using the Avogadro Constant
1 How many Be atoms are present in 6 g of beryllium Be?
2 How many N atoms are there in 3 moles of dinitrogen tetraoxide N2O4?
3 How many H atoms are there in 1 g of water H2O?
4 A mineral water has a concentration of potassium of 6 mg/litre. How many
potassium atoms are present in a 500ml bottle of water?
5 What is the mass of 1 million atoms of gold?
6 How many O atoms are there in 8.2 g of Ca(NO3)2
7 A tablet contains 500mg of paracetamol (C8H9O2N). How many molecules
of paracetamol are in one tablet?
8 It requires 243 kJ of energy to break the bonds in 1 mole of Cl2 molecules.
How much energy is required to break the bond in a single Cl2 molecule?
17
Section 5: Finding formulae
Part 1 : Empirical Formulae
A chemical formula tells us the ratio of atoms or ions within a compound.
The formula which expresses this as the simplest whole number ratio is
called the empirical formula (empirical means from observation).
An empirical formula can be derived from experimental data if one knows
the mass of each element within the compound. This data might be given in
the form of percentage composition or as actual masses measured in an
experiment. In either case one must divide the data given for each element
by the respective Ar in order to establish the ratio of moles present. This can
then be simplified to a whole number ratio to provide the empirical
formula.
When trying to determine the simplest whole number ratio it can be useful
to divide each number of moles by the smallest number of moles.
In calculations of this type at A Level you may meet compounds that are
different but have very similar percentage composition of their elements.
When you carry out this sort of calculation you should never round up the
figures until you get right to the end of the calculation. For example NH4OH
and NH2OH have a very similar composition and if you round up the data
from one you may well get the other.
Example 1
An oxide of sodium contains 58.97% sodium. What is its empirical formula?
As an oxide of sodium it must contain Na and O only. Since the percentage
of Na is 58.97 the percentage of O must be 100 − 58.97 = 41.03%.
Na
O
% by mass
58.97
41.03
÷ by Ar
58.97
23.0
41.03
16.0
2.564
2.564
1
1
simplest ratio
∴ the empirical formula is NaO
18
Example 2
Calculate the empirical formula of a compound with the percentage
composition; C 39.13%; O 52.17%; H 8.70%.
C
O
H
% by mass
39.13
52.17
8.70
÷ by Ar
39.13
12.0
52.17
16.0
8.70
1.0
3.26
3.25
8.70
3.26
3.25
3.25
3.25
8.70
3.25
1.00
1.00
2.66
÷ by smallest
It is clear at this stage that dividing by the smallest has not resulted in a
simple ratio. You must not round up or down at this stage. You must look
at the numbers and see if there is some factor that you could multiply each
by to get each one to a whole number. In this case, if you multiply each by 3
you will get:
C
O
H
3
3
8
∴ the empirical formula is C3H8O3
You need to be careful about this; the factors will generally be clear and
will be 2 or 3. You must not round 1.33 to 1 or 1.5 to 2.
19
Example 3
10.00 g of chromium is reacted with excess oxygen to create 14.61 g of
chromium oxide. Determine the empirical formula of the oxide.
Mass of oxygen added = 14.61 - 10.00 = 4.61 g
Cr
O
mass
10.00
4.61
÷ by Ar
10.00
52.0
4.61
16.0
0.192
0.288
0.192
0.192
0.288
0.192
1
1.5
2
3
÷ by smallest
simplest ratio
∴ the empirical formula is Cr2O3
Example 4
24.64 g of a hydrated salt of MgSO4.xH2O, gives 12.04 g of anhydrous MgSO4
on heating. What is the value of x?
Mass of water given off = 24.64 − 12.04 = 12.60 g
MgSO4
H 2O
mass
12.04
12.60
÷ by Mr
12.04
120.4
12.60
18.0
0.10
0.70
1
7
simplest ratio
∴ the empirical formula is MgSO4.7H2O
20
Exercise 5a: Calculation of empirical formulae from
experimental data
Determine the empirical formula of the following compounds from the data
provided. All percentage compositions are by mass.
1 Ca 40%; C 12%; O 48%
2 Na 32.4%; S 22.5%; O 45.1%
3 H 3.66%; P 37.8%; O 58.5%
4 When 1.17 g of potassium is heated in oxygen 2.13 g of an oxide is
produced
5 22.3 g of an oxide of lead produced 20.7 g of metallic lead on reduction
with hydrogen.
6 C 75%; H 25%
7 H 2.44%; S 39.0%; O 58.5%
8 Fe 20.14%; S 11.51%; O 63.31%; H 5.04%
9 When 14.97 g of hydrated copper(II) sulfate, CuSO4.xH2O, is heated it
produces 9.60 g of anhydrous copper(II) sulfate. What is the formula of the
hydrated salt?
10 When 1.335 g of a chloride of aluminium is added to excess silver nitrate
solution 4.305 g of silver chloride is produced. Calculate the empirical
formula of the chloride of aluminium.
21
Part 2 : Molecular Formulae
If you know the molar mass (or Mr or RFM) of the compound then you can
determine not just the simplest ratio (empirical) formula, but also the
“real” or molecular formula. This tells us how many atoms are actually in a
molecule and not just the ratio of its elements.
Calculate the relative mass corresponding to the empirical formula, then
divide the known molar mass or RFM by this figure. This tells you how many
times the empirical unit is repeated in the molecular formula.
The factor by which the molar mass is greater than the empirical mass must
be a whole number (often 1, 2 or 3). If it is not a whole number, go back and
check your working.
Example 1
A hydrocarbon has the empirical formula CH2 and is known to have a molar
mass of 84 gmol-1. What is the molecular formula?
The empirical formula has a relative mass of [12 + (2 x 1)] = 14
The molar mass is 84 = 6 times greater than the empirical mass
14
So the molecular formula is 6 times CH2 , which is C6H12
22
Exercise 5b: Calculation of molecular formulae from
experimental data
In each case determine the molecular formula of the compound from the
data provided.
1 A compound with the empirical formula NO2 has the molar mass 92 gmol-1.
2 A hydrocarbon with Mr of 56 has the following composition: carbon 85.7%;
hydrogen 14.3%.
3 A hydrocarbon composed of 85.7% C has Mr of 42.
4 3.348 g of iron joined with 1.44 g of oxygen in an oxide of iron that has
molar mass of 159.6 gmol–1.
5 A compound with the composition N 12.28%; H 3.51%; S 28.07%; O 56.14%
has a molar mass of 228 gmol−1.
6 Analysis of a compound with a Mr = 58 shows that 4.8 g of carbon are
joined with 1.0 g of hydrogen.
7 A compound with the composition P 10.88%; I 89.12% has a molar mass of
570 g mol−1.
8 Analysis of a hydrocarbon showed that 7.8 g of the hydrocarbon contained
0.6 g of hydrogen and that the Mr = 78.
23
Section 6: The Concentration of Solutions
Many reactions take place in solutions of known concentration.
Concentration in solution is generally measured as moles per 1000 cm3 of
solution. For example, if a solution of sodium chloride has a concentration
of 1 moldm-3 this means that each 1000 cm3 of the solution contains 1 mole
of NaCl (which happens to be 58.5 g of NaCl).
This solution might have been made up by measuring out 58.5 g of the solid,
dissolving it in about 500 cm3 of water and then adding more water to make
the total volume of the mixture up to 1000 cm3 (1 dm3).
Concentration in moldm-3 is sometimes called molarity (symbol M) and a
solution of concentration of for example 2 M is described as “two molar”.
conc. in moldm-3 = amount of substance in mol
volume of solution in dm-3
amount in mol = conc. in moldm-3 x volume in dm3
Problems in the lab often involve volumes in cm3. Remember to divide the
volume in cm3 by 1000 to convert to dm3.
amount in mol = conc. in moldm-3 x volume in cm3
1000
Occasionally you may be required to use concentrations measured in gdm-3
rather than moldm-3. Converting between the two is just the same as
converting between moles and mass.
conc. in gdm-3 =
conc. in moldm-3 x molar mass in gmol-1
24
Example 1
Calculate the concentration of a solution produced by dissolving 0.2 moles
of substance in 100 cm3
100 cm3 = 100 dm3
1000
= 0.1 dm3
conc in moldm-3 = 0.2 mol
0.1 dm3
= 2 moldm-3
Example 2
Calculate the number of moles of sulfuric acid in 50 cm3 of solution of
concentration 2 moldm-3
number of moles = 2 x
50
1000
= 0.1 mol
Example 3
Calculate the volume containing 0.02 moles of sodium chloride if the
solution has a concentration of 0.5 moldm-3
volume in dm3 = number of mol
conc. in moldm-3
=
0.02 mol
0.5 moldm-3
= 0.04 dm3
= 40 cm3
25
Exercise 6: Calculations about concentrations
A: What is the concentration in moldm−3 of the following?
1 1 mole of HCl in 500 cm3 of solution
2 0.05 mole of HBr in 100 cm3 of solution
3 0.1 mole of Pb(NO3)2 in 250 cm3 of solution
4 2.5 mole of CuSO4 in 400 cm3 solution
5 1.96 g of H2SO4 in 250 cm3 of solution
6 1.58 g of KMnO4 in 250 cm3 of solution
7 10.2 g of Ca(NO3) in 500 cm3 of solution
8 1.00 g of NaOH in 250 cm3 of solution
B: Calculate the number of moles of compound in the given volume of
solution.
1 25 cm3 of 1.0 moldm−3 HCl
2 50 cm3 of 0.5 moldm−3 H2SO4
3 250 cm3 of 0.25 moldm−3 Na2S2O3
4 500 cm3 of 0.01 moldm−3 KI
5 125 cm3 of 2.0 moldm−3 NaOH
6 150 cm3 of 0.5 moldm−3 KOH
7 25 cm3 of 0.05 moldm-3 KMnO4
8 10 cm3 of 2 moldm-3 NaCl
C: Calculate the volume of solution that contains the given number of
moles.
1 1.25 moles from a solution of 1.0 moldm−3 HNO3
2 0.5 moles from a solution of 0.5 moldm−3 NaCl
3 2 moles from a solution of 0.25 moldm−3 H2SO4
4 0.05 moles from a solution of 0.01 moldm−3 HCl
5 0.25 mols from a solution of 2.0 moldm−3 LiOH
6 0.1 mole from a solution of 0.2 moldm-3 KI
7 0.01 mole from a solution of 0.05 moldm-3 HNO3
8 0.5 mole from a solution of 0.4 moldm-3 KCl
26
Section 7: Scaled quantities
Some properties like temperature or pressure do not depend on the quantity
of substance involved. Other properties do depend on the amount of
substance and must be recorded per amount of substance, often per gram,
per mole or per cubic decimetre.
Very often experiments are done using less (or occasionally more) than 1
mole of substance and the results of the experiment have to be scaled up
(or down) in order to quote the result per mole.
In other cases, experiments (such as titrations) determine the amount of
solute in a small portion of solution and one has to calculate how much
substance would be present in 1 dm3 in order to quote the concentration in
the normal unit of moldm-3.
Example 1
To thermally decompose 0.5 g of copper carbonate, ZnCO3, requires the
input of 930 J of energy. How much energy is required per mole of zinc
carbonate?
The molar mass of ZnCO3 is 65.4 + 12 + (3 x 16) = 125.4 gmol-1
To scale up 0.5 g to 125.4 g (1 mole) requires muliplication by x 125.4
0.5
Therefore to decompose 1 mole (125.4 g) would require 125.4 x 930 J
0.5
= 233244 Jmol-1
= 233 kJmol-1
27
Exercise 7: Scaled quantities
1. The production of ammonia by the Haber process;
N2 + 3H2 ⇌ 2NH3
releases 54 kJ of energy per mole of nitrogen used up. What is the energy
released per mole of ammonia produced?
2. The specific heat capacity of copper is 0.387 Jg-1K-1. What is the molar
heat capacity in Jmol-1K-1?
3. When 5 g of magnesium carbonate, MgCO3, is reacted with excess acid,
7260 J of energy is released. How much energy is released per mole of
magnesium carbonate?
4. The solubility of calcium sulphate, CaSO4, is 0.634 g per 100 cm3. What is
the solubility in moldm-3?
5. The energy released when 1 litre (1000 cm3) of methane gas is burnt is 37
kJ. What is the energy released per mole of methane gas?
6. It is determined by experiment that 20 cm3 of a solution contains 0.014
moles of substance. What is the concentration of the substance in moldm-3?
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Section 8: Equations and amounts
Equations can tell us how much of a chemical is reacting or is produced.
The equation...
2Mg(s) + O2(g) → 2MgO(s)
... tells us that 2 moles of (solid) magnesium atoms react with 1 mole of
(gaseous) oxygen molecules to produce 2 moles of (solid) magnesium oxide.
Since the molar mass of magnesium is 24 gmol-1, and that of oxygen is 16
gmol-1 then the equation also tells us that 48 g of magnesium react with 32 g
of oxygen (because an oxygen molecule contains two atoms) to give 80 g of
magnesium oxide. Notice that the total mass of the reactants (48 g + 32 g) is
equal to the total mass of the products (80 g); this must always be the case.
Since we know the ratio of reacting masses (and volumes in the case of
gases) we can calculate any reacting quantities based on the chemical
equation.
Example 1
What mass of oxygen would react with 16 g of magnesium and what mass of
magnesium oxide would be produced?
Use the molar mass of Mg to convert 16 g to a number of moles:
16 g contains
16 g
= 0.667 mol
-1
24 gmol
Recognise that 2 mol Mg react with 1 mol of O2, therefore 0.667 mol Mg
reacts with 0.333 mol O2
mass of 0.333 mol O2 = 0.333 mol x 32 gmol-1
= 10.67 g
Now recognise that 2 mol Mg react to produce 2 mol MgO, therefore 0.667
mol Mg produces 0.667 mol MgO (molar mass 40 gmol-1).
mass of 0.667 mol O2 = 0.667 mol x 40 gmol-1
= 26.7 g
29
Example 2
What is the total volume of gas produced by the action of heat on 1 g of
silver nitrate if it decomposes according to the equation:
2AgNO3(s) → 2Ag(s) + 2NO2(g) + O2(g)
Use the molar mass of AgNO3 to convert 1 g to a number of moles:
1 g contains
1g
169.9 gmol-1
= 0.00589 mol
Recognise that 2 mol AgNO3 reacts to produce 3 mol of gas (2 mol of NO2 and
1 mol of O2) therefore
mol of gas
= 0.00589 mol x 3/2
= 0.008829 mol
Then calculate the volume using the fact that 1 mol occupies 24 dm3
volume of gas = 0.008829 mol x 24 dm3mol-1
= 0.2119 dm3
Example 3
What is the maximum mass of solid lead iodide that can be precipitated by
adding an excess of lead nitrate to 50 cm3 of 2 M potassium iodide solution:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
First calculate how many moles of KI are present in 50 cm3
moles of KI = 50 cm3
1000
x 2 mol
= 0.1 mol
Using the equation, recognise that this will produce 0.05 mol PbI2
mass of PbI2 = 0.05 mol x 461 gmol-1
= 23.05 g
30
Exercise 8: Calculations of products and reactants based
on equations
1 What mass of barium sulfate would be produced from 10 g of barium
chloride in the following reaction?
BaCl2 + H2SO4 → BaSO4 + 2HCl
2 What mass of potassium chloride would be produced from 20 g of
potassium carbonate?
K2CO3 + 2HCl → 2KCl + H2O + CO2
3 What is the maximum mass of iron (III) hydroxide that can be formed by
adding an excess of iron nitrate to 50 cm3 of 0.5 M sodium hydroxide
solution?
Fe(NO3)3 + 3NaOH → Fe(OH)3 + 3NaNO3
4 Sulfur dioxide can be removed from the waste gases of a power station by
passing it through a slurry of calcium hydroxide. The equation for this
reaction is:
SO2 + Ca(OH)2 → CaSO3 + H2O
What mass of calcium hydroxide would be needed to deal with 1000 dm3
of sulfur dioxide?
5 In a fermentation reaction, glucose is converted to alcohol and carbon
dioxide according to the following equation:
C6H12O6 → 2C2H5OH + 2CO2
What mass of alcohol and what volume of carbon dioxide would be
produced from 10 g of glucose?
6 What volume of hydrogen would be produced by 1 g of calcium in its
reaction with water?
Ca + 2H2O → Ca(OH)2 + H2
31
7 What mass of sodium hydroxide would be needed to neutralize a spillage
of 100 cm3 of sulphuric acid acid of concentration 2 moldm-3?
2NaOH + H2SO4 → Na2SO4 + 2H2O
8 Copper(II) oxide reacts with sulphuric acid to produce copper(II) sulphate.
If this is allowed to crystallise in plenty of water the formula of the
crystals is CuSO4.5H2O. What mass of copper oxide would be needed to
make 100 g of crystals?
9 Nitric acid is produced by the following series of reactions:
4NH3 + 5O2 → 4NO + 6H2O
4NO + 2O2 → 4NO2
4NO2 + O2 + 2H2O → 4HNO3
What mass of nitric acid would be produced from 34 tonnes of ammonia
and what volume of oxygen would be needed in the reaction?
10 Hardness in water is caused by dissolved calcium compounds. When
heated some of these break down and deposit calcium carbonate as
follows:
Ca(HCO3)2 → CaCO3 + H2O + CO2
This builds up as ‘fur’ on the inside of boilers. It can be removed by
reaction with hydrochloric acid.
What mass of calcium carbonate would be produced from 10000 dm3 of
water containing 0.356 g of calcium hydrogen carbonate per dm3 of water
and what volume of 10 moldm-3 hydrochloric acid solution would be
needed to remove the solid calcium carbonate from the inside of the
boiler?
32