Section 4: Chemical Calculations
Topic
Page Numbers
Relative Atomic Mass
2
Relative Molecular Mass
4
Calculating the Formula of Simple Compounds
7
Calculating the Mass of Products/Reactants from an Equation
11 - 12
Percentage Yield
18
Atom Economy
23-24
Comparing Atom Economy for Methods of Titanium Extraction
27-28
BY THE END OF THIS TOPIC YOU SHOULD BE ABLE TO:
Describe relative atomic mass (Ar) and know that
different elements have different masses
Calculate the relative molecular (formula) mass (Mr) of
a compound by using its formula
Calculate the formula of a simple compound from given
data
Be able to calculate reacting masses of reactants or
products using a balanced symbol equation
Describe and calculate the percentage yield of a
reaction
Calculate the atom economy of a reaction from a given
equation
Assess and compare the level of waste in chemical
reactions using atom economy
1
Relative Atomic Mass
Different elements have different numbers of protons,
neutrons and electrons.
The mass of an element comes from the neutrons and protons
in its nucleus (an electron’s mass is negligible).
 The relative atomic mass is the largest number in the
element’s box on the periodic table.
 It is usually at the top of the box.
 It is often written as
Ar.
 It tells us how heavy an element is compared to other
elements.
 Because it is a way of comparing elements it doesn’t have
any units.
2
Relative Atomic Mass Questions
Question 1:
Write down the Ar for the following elements:
Osmium
Sodium
Platinum
Titanium
Chlorine
Argon
Lithium
Nitrogen
Helium
Carbon
Hydrogen
Sulphur
Question 2:
Circle the element that is heavier:
a)
b)
c)
d)
e)
Carbon / Oxygen
Sodium / Lithium
Gold / Copper
Hydrogen / Helium
Lead / Tin
Question 3:
Write down the elements with the following Ar values:
Ar
Element
1
12
16
23
222
3
Relative Molecular Mass
 The relative molecular mass can be worked out by
adding up the relative atomic masses of the
elements in the molecule.
 It can also be written as
mass
Mr or relative formula
Example 1: Calculating the relative molecular mass of CH4
CH4 contains 1 x C and 4 x H atoms (remember the small
number applies to the H only)
Ar for C = 12
Ar for H = 1
So Mr = (1 x 12) + (4 x1)
= 16
Calculate the Mr of the following substances:
(the first 2 have been done for you)
(a) H2O = 1 + 1 + 16
= 18
(d) O2 =
(b) CO2 = 12 + 16 + 16
= 44
(e) H2 =
(c) NaCl =
(g) NaOH =
(h) KOH =
(i) MgO =
(j) CO =
(k) C2H4 =
(l) C2H6 =
(m)CH4 =
(n) C4H10 =
(o) C5H12 =
(p) C6H6 =
(q) S8 =
(r) Al2O3 =
(f) HCl =
4
Example 2: Calculating the relative molecular mass of Al(OH)3
Al(OH)3 contains 1 x Al, 3 x O and 3 x H (remember the small
number applies to everything inside the bracket)
Ar for Al = 27
Ar for O = 16
Ar for H = 1
So Mr = (1 x 27) + (3 x 16) + (3 x 1)
= 78
.
Now try these examples – the first one has been done for you.
(a) Ca(OH)2 = 40 + (16 + 1)2 = 74
(b) (NH4)2SO4 =
(b)
(NH4)2Cr2O7 =
(d) Fe(OH)2 =
(e) Fe(OH)3 =
(f) (NH4)2S203 =
(g) (NH4)3PO4 =
(h) C6H2CH3(NO2)3 =
5
Fill in this worksheet…
6
Calculating the Formula of Simple Compounds
You can use relative atomic masses (Ar) to calculate the
formula of compounds when you know the mass of reactants
you used.
Example:
3.2g of sulphur reacts with oxygen to produce 6.4g of sulphur
oxide. What is the formula of the oxide?
Step 1 – Write down the masses of reactants:
S = 3.2g
O = 6.4 - 3.2 = 3.2g
Step 2 – Divide Masses by Ar
3.2 = 0.1
32
3.2 = 0.2
16
Step 3 – Divide everything by the smallest number to get the
ratio
0.1 = 1
0.1
Ratio S : O = 1:2
0.2 = 2
0.1
Step 4 - Write out the formula
SO2
Step
1
2
3
4
Process
Write down masses of reactants*
Divide by Ar
Divide by smallest number to get ratio
Write down formula
Use this table as a
prompt until you are
confident with the
method.
* If you are
given % just
use the % as
the mass in
grams.
7
Answer these questions to calculate the formula of simple
compounds:
1. 10.2g of an oxide of aluminium contains 5.4g of aluminium.
Calculate the formula of the oxide (i.e. empirical formula)
2. A compound contains 29.11% sodium, 40.51% sulphur and
30.38% oxygen by mass. Calculate its simplest formula
(empirical formula)
3. 14.2g of a compound that contains 3 elements (sodium,
sulphur and oxygen). The mass of sodium is 4.6g and
sulphur is 3.2g of sulphur. The rest is oxygen. Calculate
the mass of oxygen, and then calculate the simplest
formula of the compound.
8
4. A compound contains 40.21% of potassium, 26.80%
chromium and 32.99% oxygen by mass. Calculate its
empirical formula
5. A sample of a compound containing 3 elements (potassium,
chromium, oxygen) weighs 22.62g. This amount contains
6g of potassium and 8g of chromium and the rest is
oxygen. Calculate the mass of oxygen and then calculate
the simplest formula (empirical formula)
9
Calculating the Mass of Products from a
Reaction Equation
You can use the reaction equation and information about
relative atomic mass to calculate the mass of products you
would expect to get.
Example
N2 + 3H2 → 2NH3
If we have 7g of N2, how much NH3 can we make?
Step 1
Write down the Mr or Ar underneath each chemical and
multiply by any balancing number
N2
28
+
3H2
3x2 = 6
→
2NH3
2x17 = 34
This means that for every 28g of N2 we would expect to get
34g of NH3.
Step 2
Decide what fraction or multiple of the Mr the reacting amount
is:
By examination 7 is 28/4
Step 3
Work out how much product you would get
34 / 4 =
8.5g
Excess means you have more than enough of that reactant so it
won’t affect your calculation
10
Calculating the Mass of Reactants from a
Reaction Equation
The previous method can also be used to work out the mass of
reactants used in a chemical reaction.
Example
The reaction between iron (Fe) and sulphur (S) to make iron
(II) sulphide (FeS).
Fe + S  FeS
How much Fe is needed to make 176g of FeS?
Step 1
Write down the Mr or Ar underneath each chemical and
multiply by any balancing number
Fe + S  FeS
56
32
88
This means that for every 88g of FeS made, we need 56g of Fe
and 32g of S
Step 2
Decide what fraction or multiple of the Mr the reacting amount
is:
By examination: 176 is 2 x 88
Step 3
Work out how much reactant you would need
56 x 2 =
112g
Q. How much S would be needed to make 176g of FeS?
11
Now try these questions yourself. Remember to start by writing the Mr
or Ar under each formula in the equation, and multiplying by the
balancing number.
5.6g of iron react with excess sulphur to make WHAT MASS of
iron(II) sulphide?
Fe
+
S

FeS
3.65g of HCl react with excess sodium hydroxide. What mass of water
is formed?
HCl
+
NaOH

NaCl
+
H 2O
8g of hydrogen are exploded in excess oxygen. What mass of water is
formed?
2H2
+
O2

2H2O
If 3.2 g of oxygen is reacted with excess hydrogen, what mass of water
is formed?
2H2
+
O2

2H2O
12
What mass of hydrogen and oxygen are needed to make exactly 3.6g of
water (2 answers needed)
2H2
+
O2

2H2O
When sulphur (found in coal) burns it forms acid rain. What mass of
sulphur dioxide is formed from 32 tonnes of sulphur and excess
oxygen?
S
+
O2

SO2
When a hydrocarbon burns in excess oxygen, 2 products are formed
(water and carbon dioxide). What mass of the greenhouse gas (CO2)
forms when 10g of heptane (C7H16), found in petrol, are burned in
excess O2?
C7H16
+
11O2

7CO2
+
8H2O
13
Extension Questions
Try these questions…
123g of sulphur are burned in excess oxygen. What mass of sulphur
dioxide forms?
S
+
O2

SO2
29 tonnes of iron (III) oxide are reacted in the blast furnace with
excess carbon monoxide. What mass of iron is produced?
Fe2O3
+
3CO

2Fe
+
3CO2
14
2.567g of pure sodium hydroxide are reacted with excess dilute
sulphuric acid. What mass of water is formed?
2NaOH
+
H2SO4

Na2SO4
+
2H2O
What mass of lead nitrate has to be thermally decomposed to give 50g
of lead oxide?
2Pb(NO3)2

2PbO
+
4NO2
+
O2
15
What mass of water is formed when 1kg (1000g) of octane (equivalent
to just over a litre of petrol) is burned in excess oxygen?
2 C8H18
+
25 O2

16 CO2
+
18 H2O
What mass of carbon dioxide is formed on a typical 25 mile car
journey? (Assume the petrol consists of 2kg {2000g} of pure heptane
and that half a gallon of it [2.25 litres] reacts with excess oxygen in
the air).
C7H16
+
11 O2

7 CO2
+
8 H2 O
16
Percentage Yield
In most chemical reactions you do not get the maximum
expected amount of product. This can be to do with the
chemistry of the reaction or the way the experiment took
place.
Example
Fe + S → FeS
56 g of iron reacts to give 80 g of iron sulphide
Step 1:
Calculate the theoretical yield (the maximum amount of
product we could expect) in the same way you did before.
Fe
56
+
S →
32
FeS
88
So 56g of iron should give us 88 g of iron sulphide.
Step 2:
Calculate the percentage yield:
= 80 x 100 = 90%
88
17
Look at your answers to the theoretical yield questions a few pages
back and answer these related questions…
3.65g of HCl react with excess sodium hydroxide. What is the % yield
if 1.5g of water is formed?
HCl
+
NaOH

NaCl
+
H 2O
8g of hydrogen are exploded in excess oxygen. What is the % yield if
only 65g of water is formed?
2H2
+
O2

2H2O
18
If 3.2 g of oxygen is reacted with excess hydrogen to give 3.2g of
water. What is the theoretical yield?
2H2
+
O2

2H2O
When sulphur (found in coal) burns it forms acid rain. What is the
theoretical yield when 32 tonnes of sulphur are burned to give 63
tonnes of sulphur dioxide?
S
+
O2

SO2
19
When a hydrocarbon burns in excess oxygen, 2 products are formed
(water and carbon dioxide). What is the percentage yield when 10g of
heptane, found in petrol, is burned in excess O2 to give 25g of the
greenhouse gas (CO2)?
C7H16
+
11O2

7CO2
+
8H2O
1 tonne of nitrogen is reacted with excess hydrogen to give 0.25 tonnes
of ammonia (NH3). What is the percentage conversion?
N2
+
3H2

2NH3
20
Try this typical GCSE question…
21
Atom Economy
 This is a measure of how successful a reaction has been
in converting raw materials into a useful product.
 It looks similar to percentage yield, but it’s not the
same thing.
 If most of what you produce in a reaction is the desired
product then there will, by definition, be very little
waste.
 That is, the “Atom Economy (Percentage)” will be high
(just below 100).
 In an ideal world, all industrial chemical processes should
have an atom economy of 100% since this would mean
there would be no waste (unwanted material).
 In practice, much “unwanted” material that used to be
dumped/buried/burned is now used.
 For example, the waste slag from a Blast Furnace is
nowadays turned into insulation blocks for new homes.
Atom Economy = Mass of useful product
x100
Total mass of reactants used
22
Calculating Atom Economy
Most chemical reactions produce useful products and waste
products. Atom economy allows us to calculate the percentage
of reactants that are converted into useful products.
Example 1:
What is the atom economy for making hydrogen by reacting
coal with steam?
C(s) + 2H2O(g)
→
CO2(g) + 2H2(g)
Step 1: Work out the Mr and Ar values for the reactants and
the useful products.
C(s) + 2H2O(g)
Mr / Ar
12
→
CO2(g) + 2H2(g)
2(2+16)
36
2x2
4
Step 2: Work out the percentage of useful product using the
following equation:
Atom Economy = Mass of useful product
x100
Total mass of reactants used
=
4
36 + 12
x 100 = 8.3%
So this reaction is not very economical and probably would not
be used to make hydrogen industrially.
23
Example 2:
Lithium hydroxide will react with nitric acid to produce lithium nitrate
plus water. Suppose that we are trying to make lithium nitrate, and that
the water we also make is the waste material. Firstly we write an
equation…
LiOH
+
HNO3

LiNO3
+
H2 O
Then we write the Mr (Relative Formula Masses) underneath
LiOH
+
HNO3

LiNO3
+
H2 O
24
63
69
18
Atom economy = (69/87) x 100 = 79.3%
(If you are wondering where 87 comes from, it’s 24 + 63)
Calculate the atom economy percentage for the Blast Furnace reaction.
The desired product is iron and the waste product is carbon dioxide
Fe2O3
+
3CO

2Fe
+
3CO2
24
Calculate the atom economy percentage for the Haber Process. The
desired product is ammonia (NH3).
N2
+
3H2

2NH3
Calculate the atom economy percentage for the Oswald Process (the
desired product is nitric acid).
NH3
+
2O2

HNO3
+
H 2O
25
Try this typical GCSE question…
26
Comparing the Atom Economy for 2
methods of Titanium (Ti) extraction
 Titanium is a very useful metal.
 It is light (low density), strong, hard and resists
corrosion.
 For these reasons it has many uses including in the
aerospace industry & for artificial hip joints.
Although there is plenty of titanium in the Earth’s crust it is
an expensive metal due to extraction costs and the low atom
economy.
There are 2 methods for extracting Titanium – the traditional
method and the modern electrolytic method.
Method 1 – The Traditional Method
The following reaction is used:
TiO2 + 2Cl2 + C + 4Na  CO2 + Ti + 4NaCl
Calculate the atom economy for this reaction:
(The desired product is Ti (CO2 & NaCl are waste products)
TiO2 + 2Cl2 + C + 4Na  CO2 + Ti + 4NaCl
27
 Do you agree that this is rather wasteful?
 Your answer should have been 14.7%.
 This means that 100 – 14.7 = 85.3% is potentially waste
material (or, at least, not very useful).
Method 2 – The Modern Electrolytic Method
 More recently a better process has been developed using
direct electrolysis.
 This will hopefully enable titanium to be produced at half
its current price, or even less.
 You do not need to know details of the process, but it can
be summarised as follows:
TiO2
Ti

+
O2
 Notice that the only by-product is oxygen, which is, of
course, very useful.
 Can YOU now work out the atom economy percentage for
this process?
 It ought to be much more efficient than 14.7%.
TiO2

Ti
+
O2
28
Relevant Internet and School Network Links
In school, try these web links and click on the white arrow in the green
square to view:
http://www.yenka.com/freecontent/item.action?quick=13i#
http://www.yenka.com/freecontent/item.action?quick=13j#
http://www.yenka.com/freecontent/item.action?quick=u0
School network Links:
 SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR
GCSE > INTRODUCING MOLES
 SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR
GCSE > MOLES IN CHEMICAL EQUATIONS
 SUBJECTS > SCIENCE > FOOTPRINTS > MATERIALS & THEIR
PROPERTIES > ELEMENTS, COMPOUNDS & MIXTURES
 SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR
GCSE > EXTRACTION OF TITANIUM
 SUBJECTS > SCIENCE > CHEMISTRY > ABSORB CHEMISTRY FOR
GCSE > MOLES & CHEMICAL FORMULAE

Useful websites:
http://web.visionlearning.com/MW_calculator.shtml
http://www.tufts.edu/as/wright_center/fellows/george/georgepage3.ht
m
Please note – some of these websites include extension materials and
useful concepts (such as “the mole”) which can be extremely useful but
you do not necessarily need to know for your exam!
29