Calculations in Chemistry
Molecular formula shows the actual number of atoms of various elements present in one
molecule of a substance.
For examples: Molecular formula of nitric acid is HNO3
Molecular formula of copper(II)sulphate is
CuSO4
Molecular mass is the comparative value which shows how many times a molecule of a
substance is heavier than 1/12th the mass of a carbon atom. Molecular mass is represented by
the symbol Mr. Molecular mass/atomic mass has no units
If molecular mass/atomic mass is expressed as grams, that amount contains a fixed number of
molecules/atoms. This number is same (constant) whatever the substance you take. This fixed
number is called Avogadro constant with a value of 6.02 x 1023.
So 32 is the Mr of oxygen. 32g of Oxygen contains 6.02 x 1023 oxygen molecules.
98 is the molecular mass of sulphuric acid. 98g of sulphuric acid contains 6.02 x 10 23 sulphuric
acid molecules.
The amount of a substance contains Avogadro number of particles is called one mole of
the substance.
It means 32g of Oxygen is 1 mole of oxygen molecules contains Avogadro number of
oxygen molecules.
98g is one mole of sulphuric acid containing Avogadro number of sulphuric acid
molecules.
If you know mass (g) you can calculate the number of moles.
Number of moles =
𝑴𝒂𝒔𝒔(𝒈)
𝑴𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
Mass = Moles x Molar mass
Number molecules = moles x 6.02 x 1023
Number of moles =
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒎𝒐𝒍𝒆𝒄𝒖𝒍𝒆𝒔
𝟔.𝟎𝟐 𝒙 𝟏𝟎^𝟐𝟑
Molar volume
One mole of any gas at room temperature and atmospheric pressure occupies 24 dm3 volume.
24dm3 = 24000 cm3
Volume of a gas = number of moles x 24000 cm3
Number of moles of a gas =
𝒗𝒐𝒍𝒖𝒎𝒆 𝒈𝒊𝒗𝒆𝒏 𝒄𝒎 ^𝟑
𝟐𝟒𝟎𝟎𝟎 𝒄𝒎^𝟑
It is important to convert the values to number of moles first before doing further
calculations.
Mole Ratio
When we write a balanced equation, moles of Reactants and products are in a definite mole
ratio. This mole ratio can be used to calculate the amount of various substances in the reaction.
Example:
Hydrochloric acid reacts with Calcium carbonate according to the following equation.
2HCl + CaCO3  CaCl2 + CO2 + H2O
Here hydrochloric acid moles, calcium carbonate moles, calcium chloride moles, carbon dioxide
moles and water moles are in the ration 2:1:1:1:1. This means if 2 moles of Hydrochloric acid are
completely reacted, one mole of calcium chloride, carbon dioxide and water are formed.
Example:
In an experiment, 10g of calcium carbonate are completely reacted with hydrochloric acid.
How many grams of calcium chloride are formed?
What number of moles of carbon dioxide is produced?
What volume of carbon dioxide at room temperature and pressure are produced?
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First convert mass of calcium carbonate in to moles.
Moles of calcium carbonate = mass(g)/molar mass 10g/100g = 0.10moles
Mole ratio of calcium carbonate and calcium chloride is 1:1 ratio
So moles of calcium chloride formed is 0.10 moles.
Mass of calcium chloride = moles x Molar mass 0.10 x 111g = 11.10g
Mole ratio of calcium carbonate and carbon dioxide is 1:1
So moles of carbon dioxide produced = 0.10 moles
Volume of carbon dioxide = moles x molar volume (24000cm3) 0.10 x 24000cm3 =
2400cm3
When you take mole ratio, the substance which is completely reacted is called the limiting
reagent. The reactant which is not completely used up is in excess.
Example:
In an experiment, 10g of Zinc powder ware mixed with 30g of copper(II) sulphate in
solution. The reaction takes place according to the following equation:
Zn9s) + CuSO4(aq)  ZnSO4(aq) + Cu(s)
Find out the number of moles of each chemical present.
Which substance is the limiting reagent?
What mass of Copper will be produced?
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Moles of Zinc = mass(g) / molar mass of Zn 10g/65g = 0.15moles
Moles of copper(II) sulphate = mass(g)/molar mass of copper sulphate 30g/160g = 0.19
moles
According to the balanced equation, Zinc and Copper(II)sulphate react at 1:1 ratio. So All
Zinc will be reacted. Zinc is the limiting reagent. Copper(II)sulphate is in excess.
Moles of copper produced = 0.15 moles (limiting reagent ratio)
Mass of copper produced = moles x molar mass of copper 0.15 x 64g = 9.6g
Gases react and if gaseous products formed, volumes of the gases are in a simple ratio
same like moles. This is because volume of a gas is proportional to the number of moles.
Example:
Hydrogen and oxygen react according to the following equation:
2H2(g) + O2(g)  2H2O(g)
In an experiment, 100 mL of hydrogen were exploded in 80 mL oxygen to produce water
vapour. Find out the limiting reagent. What volume of water vapour is produced?
What is the final volume?
All volumes are measured under same temperature and pressure.
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Hydrogen and Oxygen react according to the ratio 2:1 by moles/volume
Hydrogen is the limiting reagent because all hydrogen will be used up but only 40mL of
Oxygen are used. Oxygen left after the reaction is 40mL. All hydrogen is used up.
Hydrogen(limiting reagent) : Water vapour is 1:1
So volume of water vapour produced is 100mL.
Total volume of gases at the end of the reaction is volume of water vapour formed + volume
of Oxygen left.
100 + 40 = 140mL
See the conversion table given separately.
% Yield and % purity
% yiels and % purity can be calculated by the following equation:
𝑨𝒄𝒕𝒖𝒂𝒍 𝒚𝒆𝒊𝒍𝒆
% yield = 𝑪𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆𝒅 𝒚𝒊𝒆𝒍𝒅 𝒙 𝟏𝟎𝟎
Actual yield is the amount of substance actually made in the process
Calculated yield is the amount of substance you get through mole calculations.
Emperical formula and molecular formula.
Empirical formula is the smallest formula of a substance. Empirical formula shows which
elements are present in a substance and in which ratio atoms of different elements present in the
compound.
Molecular formula shows the actual number of atoms of each element present in a molecule.
Emperical formula can be calculated by finding the mole atoms of each element (divide with the
atomic masses) and making a mole ratio. You can make a mole ratio by dividing each number of
moles with the smallest moles. Once we have the empirical formula, we can calculate the
empirical formula mass. Divide the molecular mass with the empirical formula mass, we get a
number. Multiply the empirical formula with the number, we get the molecular formula.
For example:
In an organic compound made up of carbon, hydrogen and oxygen, C=40%, H=6.67%,
O=53.3%. Find the empirical formula of the compound.
Carbon
Hydrogen
Oxygen
% composition
40%
6.67%
53.3%
Moles
40/12 = 3.33
6.67/1 = 6.67
53.3/16 = 3.33
Mole ratio
3.33/3.33 = 1
6.67/3.33 = 2
3.33/3.33 = 1
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Empirical formula = CH2O
If the molecular mass is 90, find the Molecular formula.
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Empirical formula mass = 12 + 1 + 1 + 16 =30
Divide the Molecular mass with the Emperical formula mass to get a number.
90/30 = 3
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Multiply the Emperical formula with the number to get the molecular formula.
3 x CH2O
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Molecular formula is C3H6O3
Solutions
Solution is a mixture of a solvent and a solute. The concentration of the solution depends on how
much solute is present in the solution. Concentration is expressed as mole/dm -3 .
If 4g of sodium hydroxide is dissolved in 250mL solution what is the concentration of the solution?
First find the number of moles in 4g = 4g/40g = 0.1 moles
So 0.1 moles of sodium hydroxide is present in 250cm 3. Number of moles in 1000cm3 is 0.1 x 4 =
0.4 moles
Concentration is 0.4 moles/dm3