Calderside Academy
Calculations
In
Higher
Chemistry
Calculations involving Excess
All reactants are needed for a chemical reaction to occur. As soon as one of the
reactants is used up the reaction will stop. Any of the other reactant which is
left is said to be “in excess”.
It is the reactant which is totally used up which determines the mass of
product formed.
Worked example
What mass of carbon dioxide is formed when 8g of methane is reacted with
128g of oxygen?
(1)Find out which reactant is in excess.
 Write a balanced equation
CH4 + 2O2
CO2 + 2H2O
 Mole ratio
1 mol 2 mol
 Calculate number of moles of each
reactant present
8/16 128/32
0.5
4
1 mol needs 2 mol
0.5mol needs 1mol
got 4 mol
 therefore oxygen is in excess
(2)Work out mass of product.
 Write a balanced equation
CH4 + 2O2
CO2 + 2H2O
 (ignore O2 as it is in excess)
 Mole ratio
1 mol
1 mol
 Change to mass (or other units
16g
44g
if required)
 Use proportion
8g
8 x 44
16
= 22g
Examples for practice
1.
2.
3.
What mass of magnesium oxide is produced when 2.4 g of magnesium is burned in 0.5
moles of oxygen?
2 Mg +
O2
2 MgO
What mass of hydrogen gas is produced when 1.2g of magnesium is added to 500cm3 of
hydrochloric acid, concentration 1 moll-1
Mg
+
2HCl
MgCl2 +
H2
What volume of carbon dioxide is produced when 3g of carbon are burned in 0.32g of
oxygen? (Take the molar volume of carbon dioxide to be 24 lmol-1)
1
Percentage Yield Calculations
Percentage yield =
actual yield
theoretical yield
x 100
The actual yield is given in the question.
The theoretical yield is calculated using a balanced equation.
Worked example
In a reaction 330 kg of carbon dioxide was produced when 120 kg of carbon was
burned. Calculate the percentage yield.
 Find theoretical yield using the balanced equation
C +
O2
12g
12kg
120kg

CO2
44g
44kg
440 kg
Calculate percentage yield
Actual yield
Theoretical yield
x 100
=
330 x 100
440
=
75%
Examples for Practice
1.
2.
3.
10.2g of 1,2-dibromopropane (C3H6Br2) was obtained from 2.6g of propene (C3H6).
C3H6 +
Br2
C3H6Br2
Calculate the percentage yield.
1g of hydrogen reacts with excess nitrogen to produce 5g of ammonia.
N2
+
3H2
2NH3
Calculate the percentage yield.
Under test conditions 1 tonne of sulphur dioxide reacts with excess oxygen to produce
0.8 tonnes of sulphur trioxide.
2SO2 +
O2
2SO3
Calculate the percentage yield.
2
Enthalpy of Combustion
Enthalpy of combustion of a substance is the energy released when one mole of
that substance burns in excess oxygen.
Enthalpy of combustion is exothermic.
Worked Example (Note: the method for all calculations is not always identical)
0.22g of propane, C3H8, is burned to heat 250 ml of water from 22oC to 32oC.
Calculate the enthalpy of combustion of propane.
 Use H = cmT
(Remember to make
sure mass of water
is in kg)
H
=
=
=
 Use proportion
(mass of 1 mole of propane)
0.22g
44g
cmT
4.18 x 0.25 x 10
10.45
10.45kJ
44 x 10.45
0.22
=

Remember the sign
2090 kJmol-1
Enthalpy of combustion is exothermic so we need a
negative sign
Enthalpy of combustion of propane is -2090kJmol-1
Examples for Practice
1. A gas burner containing butane, C4H10, is used to heat 0.15kg of water from
22.5oC to 31oC. 0.35g of butane is used during the experiment. Calculate
the enthalpy of combustion of butane
2. A can containing 100 cm3 of water at 21oC is heated to 29oC when a burner
containing an alcohol is lit underneath it. If 0.02 mol of the alcohol is
burned in the process, calculate the enthalpy of combustion.
3.
A Bunsen burner uses methane, CH4, which has an enthalpy of combustion
of -882kJmol-1. If 0.4g of methane was completely burned to heat a can
containing 500cm3 of water, what would be the maximum temperature rise
which would be produced?
3
Enthalpy of Neutralisation
Enthalpy of neutralisation is the energy released when one mole water is
formed in the neutralisation of an acid with an alkali.
Enthalpy of neutralisation is exothermic.
Worked Example (Note: the method is not always identical)
50cm3 of 1 moll-1 hydrochloric acid, HCl, is mixed with 50cm3 of 1 moll-1
potassium hydroxide, KOH, both at 20oC. The temperature of the resulting
solution rises to 26.9oC. Calculate the enthalpy of neutralisation.
 Equation
NaOH
+
HCl
NaCl
+
H2 O
 Use H = cmT
H =
cmT
(Remember to add
=
4.18 x 0.1 x 6.9
the volume of acid & alkali
=
2.88 kJ
and make sure it is in kg)
 Find number of moles of water formed
 Use proportion
 Remember the sign
n = CV(l)
=
=
1 x 0.05
0.05 mol
0.05 mol
1 mole
2.88kJ
1 x 2.88
0.05
=
57.6 kJmol-1
Enthalpy of neutralisation is exothermic so we need
a negative sign.
Enthalpy of neutralisation is
- 57.6 kJmol-1
Examples for Practice
1.
2.
3.
80cm3 of 0.5 moll-1 sodium hydroxide solution, NaOH, and 80cm3 of 0.5
moll-1 hydrochloric acid,HCl, were mixed and a temperature rise of 3.4oC
was recorded. Calculate the enthalpy of neutralisation.
25cm3 of 1 moll-1 H2SO4 is neutralised by 50cm3 of 1 moll-1 KOH. A
temperature rise of 9.1oC is noted. Find the enthalpy of neutralisation.
50cm3 of 0.2 moll-1 NaOH is neutralised by 50cm3 of 0.2 moll-1 HCl.
Calculate the resulting temperature rise. (Take the enthalpy of
neutralisation to be -57.3kJmol-1.)
4
Enthalpy of Solution
Enthalpy of solution of a substance is the energy change when one mole of that
substance dissolves in excess water.
Enthalpy of solution may be exothermic or endothermic.
Worked Example (Note: the method is not always identical)
4g of ammonium nitrate, NH4NO3, is dissolved completely in 100cm3 water in an
insulated container. The temperature of the water falls from 19oC to 16oC.
Calculate the enthalpy of solution of ammonium nitrate.
 Use H = cmT
(Remember to make
sure mass of water
is in kg)
H
 Use proportion
(find the mass of 1 mole of NH4NO3)
=
=
=
cmT
4.18 x 0.1 x 3
1.254 kJ
4g
80g
1.254kJ
80 x 1.254
4
=
25.08 kJmol-1
 Remember the sign
Enthalpy of solution may be exothermic or
endothermic.
The temperature decreased, so the reaction is
endothermic.
Enthalpy of solution of ammonium nitrate is
+ 25.08 kJmol-1
Examples for Practice
1.
2.
3.
14.9g of potassium chloride, KCl, is dissolved in 0.2kg of water. The
temperature falls from 22.5oC to 18.5oC. Calculate the enthalpy of solution
of potassium chloride.
0.05 mol of a compound is dissolved in 500cm3 of water causing the
temperature to rise from 19oC to 21oC. Find the enthalpy of solution.
The enthalpy of solution of ammonium chloride, NH4Cl, is +15.0kJmol-1.
What mass of ammonium chloride would require to be dissolved to
decrease the temperature of 200cm3 of water by 4oC?
5
Avogadro and the Mole
Avogadro’s constant is the number of elementary particles in one mole of the
substance. It has the value 6.02 x 1023.
Worked example
Calculate the number of atoms in 880g of carbon dioxide.

identify the elementary particle
carbon dioxide exists as molecules
1 mole
6.02 x 1023 molecules
44g
6.02 x 1023 molecules
880g
880 x 6.02 x 1023
44
1.204 x 1025 molecules
=

1 carbon dioxide molecule (CO2) has 3 atoms
1 molecule
3 atoms
1.204 x 1023
3 x 1.204 x 1023
=
3.612 x 1023
atoms
Examples for Practice
1.
2.
3.
How many ions are present in 2.25 moles of sulphuric acid, H2SO4?
How many neutrons are present in 1.6g of oxygen gas?
A sample of ethane, C2H6, contains 2.408 x 1024 atoms of carbon. What
mass of ethane is present?
6
Faraday
The quantity of electrical charge flowing in a circuit is related to the current
and the time.
Q
=
I x t
Where
Q
is the electrical charge in coulombs (C)
I
is the current in amps (A)
t
is the time in seconds (s)
96,500C
=
1 Faraday
=
charge carried by 1 mole of electrons
n x F coulombs are required to deposit 1 mole of a substance
where
n
is the charge on the ion
F
is the Faraday (96,500C)
Worked example (Note: the method is not always identical)
What mass of nickel is deposited in the electrolysis of nickel(II)sulphate
solution if a current of 0.4A is passed for 2 hours?
 Q = It
=
0.4 x (2 x 60 x 60)
=
2880C
 Ion electron equation
Ni2+ +
 nF
2 x 96,500 C
193,000 C
1 mole
1 mole
193,000 C
2880 C
59 g
2880 x 59
193,000
0.88g
1 mole
2e
Ni
=
Examples for Practice
1.
In the electrolysis of copper(II)sulphate, a current of 0.1A flowed for 60
minutes. Calculate the mass of copper deposited.
2.
Electrolysis of silver(I)nitrate produced 1.08 g of silver per hour. What
current was flowing?
3.
A vanadium compound was electrolysed using a current of 0.15A for 45
minutes, during which 0.0534g of vanadium was deposited. Calculate the
charge on the vanadium ion.
7
Hess’s Law
Hess’s Law states that the enthalpy change for a chemical reaction depends
only on the enthalpies of the reactants and products and is independent of the
route taken for the reaction.
Worked example
Calculate H for the following reaction.
C(s)
Using
(1)
C(s)
(2)
H2(g)
(3)
CH4(g)
+
2H2(g)
CH4(g)
+
+
+
O2
½ O2(g)
2O2(g)
CO2(g)
H2O(l)
CO2(g) + 2H2O(l)
 Rearrange equations to try to get equation we want.
(1)
C(s) +
O2
CO2(g)
(2) x 2
2 H2(g) + O2(g)
2 H2O(l)
(3) rev
CO2(g) + 2H2O(l)
CH4(g) +

2O2(g)
Equations should cancel out to give equation we want
C(s)
+
2H2(g)
CH4(g)
H = -394 kJmol-1
H = -286 kJmol-1
H = -882 kJmol-1
H = -394 kJmol-1
H = -572 kJmol-1
H = +882 kJmol-1
H = -84 kJmol-1
Examples for Practice
1.
Calculate H for the following reaction.
2C(s) +
Using
(1)
C(s)
(2)
H2(g)
(3)
C2H6(g)
2.
3.
+
+
+
3H2(g)
C2H6(g)
O2
½ O2(g)
3½ O2(g)
CO2(g)
H2O(l)
2CO2(g) + 3H2O(l)
H = -394 kJmol-1
H = -286 kJmol-1
H = -1542 kJmol-1
Calculate H for the following reaction.
2C(s) +
3H2(g)
+
½ O2
C2H5OH(g)
Use enthalpies of combustion of carbon, hydrogen and ethanol in the
databook.
Calculate H for the following reaction.
2CH4(g)
C2H2(g)
+
3H2(g)
Use enthalpies of combustion of methane, hydrogen and ethyne, C2H2, in
the databook.
8
Molar Volume
Molar volume is the volume occupied by 1 mole of a gas under certain conditions.
Worked example (Note: the method is not always identical)
Under certain conditions carbon monoxide has a density of 0.8 gl-1. Find the
molar volume under these conditions.
(mass of 1 mole of CO)
0.8 g
occupies
1 litre
28 g
occupies
28 x 1
0.8
= 35 litres
Examples for Practice
1.
Under certain conditions, neon has a density of 0.9 gl-1. Calculate the molar
volume under these conditions.
2.
A gas has a molar volume of 29.1 l mol-1 and a density of 2.2 gl-1. Calculate
the molecular mass of the gas.
3.
From the data below, calculate the formula mass of gas X.
Mass of empty plastic bottle
Mass of plastic bottle + X
Capacity of plastic bottle
Molar volume of gas X
=
=
=
=
9
112.80 g
113.52 g
1 litre
23.6 litres
Reacting Volumes
Since 1 mole of any gas occupies the same volume under the same conditions, we
can use a balanced equation to calculate the volume of gases.
Worked example
What volume of carbon dioxide would be produced if 20cm3 of ethane is burned
in excess oxygen?

Write a balanced equation
C2H6 + 3½O2
2CO2 + 3H2O

Mole ratio
1 mol
2 mols

Volume statements
1 vol
2 vols

Use proportion
20 cm3
2 x 20
1
= 40cm3
Remember – water is a gas if temperatures are above 100oC
Examples for Practice
1.
20 litres of propane is burned in 140 litres of oxygen. Calculate the volume
and composition of the resulting gas mixture. (All measurements made at
room temperature.)
2. Hexane reacts with steam (H2O) to form carbon dioxide and hydrogen.
Under certain conditions 4 x 104 litres of hexane is reacted with excess
steam. Assuming the reaction goes to completion, calculate the volume of
(a) carbon dioxide and (b) hydrogen produced.
3.
100 litres of hydrazine, N2H4, is burned in 400 litres of oxygen to form
nitrogen gas and water. What will be the volume and composition of the
resulting gas mixture if all measurements are made at (a) 20oC and (b)
300oC?
10
Calculations from Equations
A balanced equation shows the number of moles of each reactant and product
involved in the reaction. This allows us to find the number of moles of a
substance reacting or being produced.
Alternatively, number of moles can be converted into
mass
number of molecules/atoms
volume (for gases, given the molar volume)
etc
Worked example
What mass of carbon dioxide would be produced if 3 g of ethane is burned in
excess oxygen?

Write a balanced equation
C2H6 + 3½O2
2CO2 + 3H2O

Mole ratio
1 mol
2 mols

Change moles to units required 30 g
88 g

Use proportion
3 x 88
30
= 8.8 g
3g
Examples for Practice (Note: the method is not always identical)
All examples refer to the following equation:
Na2CO3
+
H2SO4
Na2SO4
+
CO2 +
1.
2.
H2 O
When 5.3g of sodium carbonate react with excess acid:
(a) what mass of water is formed?
(b) how many water molecules will be formed?
(c) what volume of carbon dioxide will be formed (Take the molar volume
of carbon dioxide as 22 l mol-1)
When excess sodium carbonate is added to 250 cm3 of 2 moll-1 sulphuric
acid:
(a) how many moles of sodium sulphate are formed?
(b) what mass of sodium sulphate is formed
(c) how many sodium ions are formed?
11
pH and Concentration
pH is related to the concentration of hydrogen ions.
e.g
[H+] = 1 x 10-1 moll-1, pH = 1
[H+] = 1 x 10-5 mol-1, pH = 5
Also, in aqueous solutions
[H+] [OH-] = 10-14 mol2l-2
Worked example
What is the concentration of OH- ions in a solution with pH 9

pH
=9

[H+]
= 1 x 10-9

[H+] [OH-] = 10-14,
therefore [OH-] = 1 x 10-5
Examples for Practice (Note: the method is not always identical)
1.
The concentration of H+ ions is 1 x10-3 moll-1. What is the pH of the
solution?
2.
The concentration of OH- is 1 x10-5 moll-1. What is the pH of the solution?
3.
The concentration of H+ is 1 x 10-6 moll-1. What is the concentration of
OH-?
4.
The concentration of H+ is 2 x 10-6. What is the concentration of
5.
The concentration of H+ is 2 x 10-6.
The pH of the solution will be between which two numbers?
12
OH-
ions?
Radioactivity and Halflife
Half-life is the time taken for half of a radioisotope to decay. This value is
constant for that particular isotope.
Usually questions about radioactivity involve
 half-life
 the time over which the radioactivity has been measured
 the quantity or intensity of the radiation
Worked example
A radioisotope has a half-life of 3 days. What fraction of the original isotope
will be present after 12 days?

Original quantity (whole amount)
1

After 1 half-life
(3 days)
0.5
(1/2)

After 2 half-lives
(6 days)
0.25
(1/4)

After 3 half-lives
(9 days)
0.125
(1/8)

After 4 half-lives
(12 days)
0.0625
(1/16)
Examples for Practice
1.
Po-210 has a half-life of 140 days and decays by alpha emission to give a
stable isotope.
(a)
(b)
(c)
2.
What mass of a 4.2g sample of Po-210 will remain unchanged after
280 days?
What mass will have decayed?
How many atoms will have decayed?
A sample of ancient wood is found to have a radioactive count rate due to
carbon-14 of 12 counts min-1. A sample of modern wood has a count rate of
48 counts min-1. Calculate the age of the wood, taking the half-life of
carbon-14 to be 5570 years.
13
Redox Titration
Redox titrations are used to find out information about one reactant, using
known information about another reactant.
Worked example
Iron(II)sulphate can be oxidised by potassium dichromate. Calculate the mass
of iron(II)sulphate which will react completely with 500cm3 of dichromate
solution, concentration 0.4 moll-1.


Write a balanced redox equation
6 Fe2+
+
Cr2O72+
14H+
6Fe3+ +
Calculate number of moles of ‘known’ reactant
N
=
C
x
V
=
0.4 x
0.5 =
2Cr3+ +
7H2O
0.2 moles Cr2O72-

Use mole ratio to work out number of moles of ‘unknown’ reactant
6 Fe2+
+
Cr2O726 moles
1 mole
0.2 x 6
0.2
1
= 1.2 mol

Change moles into information required
1 mole of FeSO4 contains 1 mole of Fe2+
therefore, we have 1.2 moles of FeSO4
1 mole of FeSO4
1.2 moles
weighs
weighs
6
152 g
182.4 g
Example for Practice
The chlorine level in a swimming pool was found by titrating samples of water
against iron(II)sulphate .
Cl2 +
2Fe2+
2Cl+
2Fe3+
A 50cm3 sample of water from a swimming pool required 12.5 cm3 of
iron(II)sulphate, concentration 2.95 moll-1, to reach the end-point. Calculate the
chlorine concentration in the swimming pool,
(a) in mol l-1 (b) in g l-1
14
Answers
Calculations involving Excess
1.
oxygen in excess
4g of magnesium oxide
2.
hydrochloric acid in excess 0.1g of hydrogen
3.
carbon in excess
0.24 litres of carbon dioxide
Percentage Yield Calculations
1.
81.6%
2.
88.2%
3.
64%
Enthalpy of Combustion
1.
-883.3 kJmol-1
2.
-167.2kJmol-1
3.
10.55oC
Enthalpy of Neutralisation
1.
-56.75kJmol-1
2.
-57kJmol-1
3.
1.37oC
Enthalpy of Solution
1.
+16.7kJmol-1
2.
-83.6kJmol-1
3.
11.9g
Avogadro and The Mole
1.
4.06 x 1024
2.
2.408 x 1023
3.
60g
Faraday
1.
0.118g
2.
0.268A
3.
4+
15
Hess’s Law
1.
-104kJmol-1
2.
-279 kJmol-1
3.
+376 kJmol-1
Molar Volume
1.
22.2 litres
2.
64g
3.
16.99g
Reacting Volumes
1.
60 litres of carbon dioxide, 40 litres of oxygen left over
2.
(a) 2.4 x 105 litres (b) 7.6 x 105 litres
3.
(a) 300 litres excess oxygen, 100 litres nitrogen
(b) 300 litres excess oxygen, 100 litres nitrogen, 200 litres water (steam)
Calculations from Equations
1.
(a) 0.9g
(b) 3.01 x 1022 (c)
2.
(a) 0.5 mol
(b) 71g
(c)
1.1 litres
6.02 x 1023
pH and Concentration
1.
3
2.
9
3.
1 x 10-8 moll-1
4.
5 x 10-9 moll-1
5.
between 5 and 6
Radioactivity and Halflife
1.
(a) 1.05g
(b) 3.15g
2.
11140 years
(c)
Redox Titration
(a) 0.369 moll-1
(b) 26.2 gl-1
16
9.03 x 1021 atoms