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BBC ­ GCSE Bitesize ­ Quantitative chemistry
Science
Quantitative chemistry
You should be able to calculate the masses of reactants
and products from balanced equations, and the
percentage composition by mass of an element in a
compound.
Higher tier students should also be able to calculate the
percentage yield of a reaction, and the empirical formula of
a compound from information about reacting masses.
Percentage composition
Percentage composition is just a way to describe what
proportions of the different elements there are in a compound.
If you have the formula of a compound, you should be able to
work out the percentage by mass of an element in it.
Example
The formula for sodium hydroxide is NaOH. It contains three
different elements: Na, O and H. But the percentage by mass
of each element is not simply 33.3 per cent, because each
element has a different relative atomic mass. You need to use
the Ar values to work out the percentages. Here is how to do it:
Question
What is the percentage by mass of oxygen (O) in sodium
hydroxide (NaOH)?
First, work out the relative formula mass of the
compound, using the Ar values for each element. In the
case of sodium hydroxide, these are Na = 23, O = 16, H =
1. (You will be given these numbers in the exam.)
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BBC ­ GCSE Bitesize ­ Quantitative chemistry
Next, divide the Ar of oxygen by the Mr of NaOH, and
multiply by 100 to get a percentage.
Answer
Mr of NaOH is 23 + 16 + 1 = 40
(16 ÷ 40 ) × 100 = 0.4 × 100 = 40%
So the percentage by mass of oxygen in sodium hydroxide is
40%.
Remember, if there is more than one atom of the element in
the compound, you need to multiply your answer by the
number of atoms. If your answer is more than 100 per cent,
you have gone wrong!
Conservation of mass
Massmass: The amount of matter an object contains. Mass is
measured in 'kg'. is never lost or gained in chemical reactions.
We say that mass is conserved. In other words, the total mass
of productsproduct: A product is a substance formed in a
chemical reaction. at the end of the reaction is equal to the total
mass of the reactants [reactant: Substances present at the
start of a chemical reaction ] at the beginning.
This fact allows you to work out the mass of one substance in a
reaction if the masses of the other substances are known. For
example:
Percentage yield
The principle of conservation of mass lets you calculate the
theoretical mass of product expected in a chemical reaction.
However, it is not always possible in practice to get the entire
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BBC ­ GCSE Bitesize ­ Quantitative chemistry
calculated amount of product. This is because:
Reversible reactions may not go to completion
Some product may be lost when it is removed from the
reaction mixture
Some of the reactants [reactant: Substances present at
the start of a chemical reaction ] may react in an
unexpected way
Yield
The yield of a reaction is the mass of product obtained:
The theoretical yield is the maximum theoretical mass
of product in a reaction (calculated using the idea of
conservation of mass)
The actual yield is the mass of product you get when
you actually do the reaction
The percentage yield is the ratio of actual mass of products
obtained compared with the maximum theoretical mass.
Percentage yield ­ Higher tier
The percentage yield of a reaction is calculated using this
equation:
percentage yield = (actual mass of product) ÷ (theoretical mass
of product) × 100
For example, the maximum theoretical mass of product in a
certain reaction is 20 g, but only 15 g is actually obtained.
Percentage yield = 15 ÷ 20 × 100 = 75%
Reversible reactions
Many reactions, such as burning fuel, are irreversible ­ they go
to completion and cannot be reversed easily. Reversible
reactions are different. In a reversible reaction, the
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BBC ­ GCSE Bitesize ­ Quantitative chemistry
productsproduct: A product is a substance formed in a
chemical reaction. can react to produce the original reactants
[reactant: Substances present at the start of a chemical
reaction ] again.
When writing chemical equations for reversible reactions, we
do not use the usual one­way arrow. Instead, we use two
arrows, each with just half an arrowhead ­ the top one pointing
right, and the bottom one pointing left. For example:
ammonium chloride
ammonia + hydrogen chloride
The equation shows that ammonium chloride (a white solid)
can break down to form ammonia and hydrogen chloride. It
also shows that ammonia and hydrogen chloride (colourless
gases) can react to form ammonium chloride again.
The animation below shows a reversible reaction involving
white anhydrous copper(II) sulfate and blue hydrated copper(II)
sulfate, the equation for which is:
anhydrous copper(II) sulfate + water
sulfate
hydrated copper(II)
The reaction between anhydrous copper(II) sulfate and water is
used as a test for water. The white solid turns blue in the
presence of water.
Now try a Test Bite.
Reacting masses calculations ­ Higher tier
If you have a balanced equation for a reaction, you can
calculate the masses of reactants and products.
Sample question
Look at this equation: CaCO3(s) → CaO(s) + CO2(g)
If we have 50g of CaCO3, how much CaO can we make?
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BBC ­ GCSE Bitesize ­ Quantitative chemistry
First, work out the Mr values for the two compounds:
Mr of CaCO3 is 40 + 12 + 16 + 16 + 16 = 100
Mr of CaO is 40 + 16 = 56
This means that 100 g of CaCO3 would yield 56 g of CaO in
this reaction. In the question we are told we have only half of
that amount of CaCO3, 50 g. So we will get half the amount of
CaO, 28 g.
So the mass of CaO we can make = 28 g
Notice in this that 22 g of CO2 would also be produced, as 50 ­
28 = 22
Empirical formula ­ Higher tier
You can use information about reacting masses to calculate
the formula of a compound. Here is an example:
Question
Suppose 3.2 g of sulfur reacts with oxygen to produce 6.4
g of sulfur oxide. What is the formula of the oxide?
Use the fact that the Ar of sulfur is 32 and the Ar of
oxygen is 16
Answer
Find the mass of each element. Conservation of mass tells
us that the mass of oxygen = the mass of sulfur oxide ­ the
mass of sulfur.
The mass of oxygen reacted = 6.4 ­ 3.2 = 3.2 g
So we have 3.2 g of sulfur and 3.2 g of oxygen.
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BBC ­ GCSE Bitesize ­ Quantitative chemistry
Now divide the mass of each element by its Ar value.
sulfur: 3.2 ÷ 32 = 0.1
oxygen: 3.2 ÷ 16 = 0.2
Finally, find the ratio of the elements.
You can do this by dividing the results by the smallest of the
numbers to give you the number of atoms of each element in
the compound.
In this case the smallest value is 0.1, so divide both results
by that.
S = 0.1 ÷ 0.1 = 1
O = 0.2 ÷ 0.1 = 2
(If one of the numbers ends in 0.5, multiply all the numbers
by 2 ­ this is because you cannot have half­atoms in a
compound.) So the ratio of sulfur to oxygen is 1:2
The number of atoms tells you that the formula for sulfur
oxide is SO2
Here is the calculation again in tabular form to help you
remember the steps:
Steps to calculation the formula of a compound
Step Action
S
O
1
Find masses
3.2 3.2
2
Look up given Ar values 32 16
3
Divide masses by Ar
0.1 0.2
4
Find the ratior
1
2
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Result: the formula for the oxide = SO2
Now try a Test Bite ­ Higher tier.
Back to Revision Bite
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