Calculations in Chemistry
IGCSE Chemistry
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Mole concept
Relative atomic mass and Relative
molecular mass
Why atomic mass or molecular mass has
no unit?
 Atoms and molecules are too small to
weigh.
 Relative atomic mass and Relative
molecular mass are comparative values.
 This shows how many times an atom or a
1
molecule heavier than 12 th of a Carbon
atom.
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The mole….
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Atomic mass or molecular mass with g after
the number is called a mole.
Example:
Atomic mass of Oxygen is 16.
16g of oxygen(O) is 1 mole of oxygen
atoms.
Molecular mass of Oxygen O2 is 32.
32g of Oxygen (O2) is 1 mole of oxygen
molecules.
Molecular mass of water is 18.
18g of water is called 1 mole of water.
mass( g )
Molarmass
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Number of moles =
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Mass (g) =Number of moles X Molar mass
Questions
1. How many moles are there in 4g of
Sodium hydroxide?
4g
 Number of moles =
40 g
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= 0.10moles
 2. Find the number of moles present in
1.28g of oxygen molecules.
 Number of moles = 1.28 g
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32 g
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= 0.04 moles
3. What is the mass of 0.01 moles of
sodium hydroxide?
 Mass = 0.01 X 40g
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= 0.40g
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4. How many grams 1.5 moles of
sulphuric acid weighs?
 Mass = 1.5 X 98g
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= 147g
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Try the following questions:
1. What is the mass of 0.02 moles of carbon dioxide gas?
 0.88g
 2. How many moles are present in 1g of calcium
carbonate,CaCO3?
 0.01moles
 3. Find the number of moles present in 3g of potassium
hydroxide, KOH.
 0.054moles
 4. Calculate the mass of 100 moles of copper oxide, CuO
 8000g
 5. 5. 2g of sodium hydroxide,NaOH or 3g of potassium
hydroxide, KOH contains more number of moles? Show
your working clearly.
 NaOH = 0.13 moles KOH = 0.054 moles. So NaOH
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Avogadro constant
One mole of any substance contains
23
6.02 X 10 atoms or molecules.
23
 6.02 X 10 is a constant and is called
Avogadro constant or Avogadro Number.
 In other words, Avogadro Number is the
number of atoms or number of molecules
present in ONE mole of a substance.
 Avogadro constant is used to find number
of particles or number of moles.
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Number of moles = Number of molecules
Avogadro constant
Number of molecules =
Number of moles X Avogadro constant
Calculations
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Remember always change mass or
number of molecules into moles.
Examples:
How many molecules of Carbon dioxide
are there in 0.44g of the gas?
 Answer:
 Number of moles = 0.44g/44g
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= 0.01moles
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Number of molecules = 0.01 X 6.02 X 10
21
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= 6.02 X 10
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23
Calculate how many oxygen atoms are
present in 1.96g of sulphuric acid?
Answer:
 Moles of sulphuric acid = 1.96g/98g
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= 0.02 moles
 Number of molecules of sulphuric acid
23
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= 0.02 X 6.02 X 10
22
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=1.20 X 10
 Each sulphuric acid contains 4 oxygen atoms.
 So number of oxygen atoms 0.02 moles of
H2SO4 is
22
22
 = 4 X 1.20 X 10
= 4.80 X 10
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Mole ratio
When chemical reactions take place, the reactants
and products are in a simple ratio (mole ratio)
 For example, Zinc and hydrochloric acid react
according to the following reaction:
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Zinc + Hydrochloric acid  Zinc chloride + Hydrogen gas
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Zn + 2HCl  ZnCl2 + H2
It means, zinc, Hydrochloric acid, Zinc chloride and
Hydrogen moles are in the ratio 1:2:1:1
One mole of Zinc reacts with 2 moles of Hydrochloric
acid to produce, One mole of Zinc chloride and 1 mole
of Hydrogen gas.
The reactants and products are always in a ratio by
moles and not by mass.
Find how many grams of copper oxide
can be made from 3.2g of copper.
Answer:
2Cu + O2  2CuO
The balanced equation shows that two moles of
copper reacts with 1 mole of oxygen to form 2
moles of copper(II) oxide.
 So change 3.2g to moles
 Moles of copper = 3.2g/64g
=0.05moles
 Copper :Copper oxide is 1:1 ratio.
 So moles of copper(II) oxide = 0.05moles
 Mass (g) = 0.05 X 80g
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= 4g
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When Posphorus burns in oxygen, Phosporus(V)oxide is
produced. In an experiment, 8g of phsphorus(V)oxide is
produced. Find the mass of Phosphorus reacted with
oxygen.
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Answer:
Equation: 4P + 5O2  2P2O5
Mole ratio is 4:5:2
Moles of phosphorus(V)oxide= 8g/142g
= 0.056moles
Mole ratio Phosphorus oxide:phosphorus
2:4
Moles of phosphorus burnt = 0.056 X 2
= 0.112
Mass of phosphorus = 0.112 X 31g
= 3.472g
Limiting reagent and excess
When there is a chemical reaction, the
reactant which is used up completely is the
limiting reagent.
 The chemical which is left some unreacted
at the end is called excess.
 When the limiting reagent is finished, the
reaction stops.
 All mole ratios should be done using the
limiting reagent.
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Example 1
3Mg + Al2O3  3MgO + 2Al
If 2 moles of Magnesium is used with 1 moles of
aluminium oxide, which is the limiting agent? Which
chemical is in excess? How many moles in excess?
Mole ratio of Mg and Al2O3 is 3:1
It means for 2 moles of Mg, 0.67 moles of Al2O3 is
enough. But we have 1 mole of Al2O3. So Al2O3
is excess and Mg is the limiting reagent.
Moles of Al2O3 (excess) left after reaction is
1.00-0.67 = 0.33moles.
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If mass is given, change all masses into moles first.
Example 2
2Cu + O2  2CuO
In a reaction, 1.92g of copper is oxidised with 0.64g of oxygen.
(a) Find which chemical is the limiting agent
(b) How many g of the other chemical (excess) remaining at the end?
(c) What mass of Copper (II)oxide produced?
Answer
Change mass in to moles
Number of moles of copper = 1.92g/64g
= 0.03 moles
Number of moles of oxygen = 0.64g/32g
= 0.02 moles
Cu : O2 mole ratio is 2:1
It means, for 0.03 moles of copper only 0.015 moles of oxygen is
enough. But we have 0.02 moles.
Copper is the limiting reagent and oxygen is excess.
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Excess oxygen = 0.020 – 0.015 = 0.005 moles
Mass of excess oxygen = 0.005 X 32g = 0.16g
Moles of copper oxide (make mole ratio with
limiting reagent) 2:2 That is 0.03moles
Mass of copper oxide = 0.03 X 80g
= 2.40g
Now try out the following:
2Na + S  Na2S
If 0.46g of sodium and 0.30g of sulphur reacted,
(a) find the limiting reagent and excess reagent.
Sulphur
(b) find the mass of excess reagent left.
0.0012 moles Na which is 0.028g
(c) find what mass of sodium sulfide formed.
Moles of Na2S is 0.0094 means 0.73g
Molar volume
Gases have volume. Volume of a gas
depends on two factors :
 Temperature and Pressure
 Volume decreases when pressure is
increased.
 Volume increases when the temperature
is increased.
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Molar Volume
One mole of any gas at Room Temperature
and Pressure has a volume of 24dm^3
(24000 cm^3)
 Room temperature is considered as
20 degree C and atmospheric pressure is
1 atmosphere (760 mm of mercury)
 It means 1 mole of oxygen gas is 24000 cm3
 3 moles of oxygen gas is 3X24000 cm3 at
RTP.
 24dm3(1mole) is called molar volume of
gases
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Moles X molar volume = volume of a gas
 Moles = volume of a gas/molar volume
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Remember to convert volume or mass to moles first.
Example 1
What is the volume of 1.1g of carbon dioxide gas at
RTP?
First convert mass in to moles
Moles of CO2 = 1.1g/44g = 0.025moles
Volume = moles X molar volume
= 0.025 X 24000 cm3
= 600cm3
Example 2
 Find out the mass of 8000 cm3 of sulphur
dioxide gas, SO2 at RTP.
 First change the volume into moles.
 Moles = Volume/molar volume
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8000 cm3/24000 cm3
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= 0.33 moles
 Now change the moles in to mass
 Mass = moles X molar mass
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0.33 X 64g
= 21.12g
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% Yield
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We can calculate and predict how much of a
product is formed during a reaction using the
mole ratio.
How're in a real experiment, the predicted
(calculated)amount may be different that what we
get in actual experiment.
This can be due to:
1. Some reactants may be left unreacted.
2. Change of substances in to other forms such as
gases
3.Some reactants may be lost to surrounding by
overflowing, splashing etc
4. Experimental errors and low quality measuring
devices.
%yield = Actual yield X 100
Theoretical yield
 Example 1:
 In an experiment, 20g of Calcium carbonate when
heated gave 10g of calcium oxide is formed. Find the
percentage yield of calcium oxide.
 CaCO3  CaO + CO2
 %yield = Actual yield X 100
Theoretical yield
 To work out the above problem, we need the actual
yield and theoretical yield(Calculated yield).
 Actual yield is given as 10g of Calcium oxide.
 Theoretical yield should be calculated by using mole
ratio method.
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CaCO3  CaO + CO2
We can find the mass of calcium oxide formed
from the mass of calcium carbonate given.
 But first convert mass in to moles.
 Moles of CaCO3 = 20g/100g
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= 0.20moles
 Mole ratio CaCO3:CaO is 1:1
 So, moles of CaO is also 0.20
 Mass of CaO = 0.20 X 56g
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= 11.20g
 Now insert the values
 % yield = Actual yield X 100%
Theoretical yield
That is
10g X 100%
= 89.28%
11.20g
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Example 2
 2Na + 2H2O  2NaOH + H2
 115g of sodium was reacted with water
which produced 4.2g of hydrogen gas.
What is the % yield of hydrogen gas?
Answer
 Moles of sodium = 5.00 moles
 Moles of H2
= 2.50 moles (2:1)
 Mass of H2
= 5.00g
 % yield
= 84%
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End of part 1