Higher tier only
Finding the formula for compounds using mole ratios
When you work out the formula for a compound e.g. water, H2O or carbon dioxide, CO2 or glucose, C6H12O6
the numbers in the formula give you the molar ratio of each element present in the compound. So
for water the formula tells you that there are 2 moles of hydrogen for
every mole of oxygen. Or for glucose there are 6 moles of carbon,
to 12 moles of hydrogen to 6 moles
of oxygen. You can also work backwards, so if you are given masses you can easily calculate the
formula for compounds.
Example 1. When a compound containing zinc and oxygen was broken down into its elements, 6.5g of zinc
and 1.6g of oxygen were obtained. Calculate the formula for this compound.
step 1 - calculate the number of moles of each element produced - use the formula shown below:
So for our example:
| moles of zinc: 6.5g/65 = 0.1 moles |
moles of oxygen: 1.6g/16 = 0.1 moles |
step 2- The final step is simply to cancel the number of moles down to the simplest ratio possible. To
do this divide by the smallest number of moles present. Since number of
moles is the same for both zinc and oxygen, simply
divide both by 0.1
| moles of zinc: 0.1 moles/0.1
=1 |
moles of oxygen: 0.1 moles/0.1
= 1 |
So the ratio of the number of moles of zinc to oxygen is 1:1, so the formula is simply ZnO - simple!
Empirical formula
The empirical formula for a compound is the simplest whole number ratio of each element present
in the compound The empirical formula is usually work out from experimental data. As an
example consider glucose, its molecular formula is C6H12O6, however we can simplify this formula
down by dividing by 6 to give : C1H2O1 or simply CH2O. This simplest formula is called the
empirical formula.
This empirical formula could represent any number of substances e.g. by multiplying it by 2 we get
C2H4O or by multiplying by 3 we get C3H6O3. To get the
actual molecular formula from the empirical formula you need the Mr of the compound. Glucose has a
Mr of 180. The Mr of the empirical formula CH2O is 30. So if we divide 180 by this mass
(180/30 =6) then this tells you that in order to get the actual formula you need to multiply by 6.
Finding the empirical formula from experimental work
A sample of magnesium ribbon, mass 1.2g was placed inside a crucible and heated strongly until all the
magnesium had completely oxidised and formed magnesium oxide. The crucible was allowed to cool and the mass of
the magnesium oxide formed was found to be 2g. How can we calculate the
empirical formula of the magnesium oxide.
Simply follow the steps below:
| write out the symbols for each element preset |
Mg |
O |
| mass of each element present |
1.2g |
0.8g |
| divide masses by Ar to get number of moles present |
1.2g/24= 0.05 |
0.8g/16 = 0.05 |
| divide by the smallest number of moles to get ratio (since in this case both numbers are the same we divide by 0.05) |
0.05/0.05 = 1 |
0.05/0.05 = 1 |
| empirical formula |
empirical formula is MgO |
example3 - A hydrocarbon molecule has an Mr of 84. It is known that 72g of its mass is due to carbon.
Find the empirical and molecular formula for this hydrocarbon molecule.
If the molecule has a molar mass of 84g and 72g are due to carbon, then 12g must be due to the hydrogen present. So simply
follow the method above:
| write out the symbols for each element preset |
C |
H |
| mass of each element present |
72g |
12g |
| divide masses by Ar to get number of moles present |
72/12= 6 |
12/1 = 12 |
| divide by the smallest number of moles to get ratio (since in this case both numbers are the same we divide by 0.05) |
6/6=1 = 1 |
12/6 = 2 |
| empirical formula |
empirical formula is CH2 |
The empirical formula CH2 has an Mr of 14. The Mr of the compound is
known to be 84. So 84/16 = 6. So to get the molecular formula the empirical
formula needs to be multiplied by 6. This gives
a molecular formula of C6H12.
Example 4 - 32g of methane (CH4) was combusted in 128g of oxygen to produce 88g of carbon dioxide and
72g of water vapour. Write a balanced symbolic equation for this reaction.
Ar of C=12 Ar of H=1 Ar of O=16
Step 1- As above the first step is to calculate the number of moles of each substance present. Use the formula
in the blue box above.
| substance |
methane (CH4 |
oxygen ()2) |
carbon dioxide (CO2) |
hydrogen oxide (H2O) |
| Mr |
12 + 4 =16 |
16 x2 =32 |
12 + 32=44 |
2 + 16=18 |
Step 2- Divide the mass of each substance by its Mr
to calculate the number of moles present:
| substance |
methane (CH4 |
oxygen ()2) |
carbon dioxide (CO2) |
hydrogen oxide (H2O) |
| mass/Mr |
32/16 |
128/32 |
88/44 |
72/18 |
| number of moles |
2 |
4 |
2 |
4 |
Step 3- Divide by the smallest number of moles present.
In this case the smallest number of moles present is 2 moles, so divide
by 2 to get a mole ratio.
| substance |
methane (CH4) |
oxygen (O2) |
carbon dioxide (CO2) |
hydrogen oxide (H2O) |
| mass/Mr |
32/16 |
128/32 |
88/44 |
72/18 |
| number of moles |
2 |
4 |
2 |
4 |
| divide by 2 to get mole ratio |
1 |
2 |
1 |
2 |
The bottom row in the table gives the mole ratio for each of the reactants and products in the balanced symbolic equation,
so simply put these into the equation:
methane(g) + oxygen(g) → carbon dioxide(g) + hydrogen oxide(g)
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Example 5 - 8g of iron(III) oxide (Fe2O3) was reduced by 4.2g of carbon monoxide to produce 5.6g of iron
and 6.6g of carbon dioxide. Work out a balanced equation for this reaction.
Ar of C=12 Ar of H=1 Ar of O=16 Ar of Fe=56
Method is identical to the one above:
Step 1 - Work out the Mr for each of the substances in the reaction.
| substance |
iron (III) oxide(Fe2O3) |
carbon monoxide (CO) |
iron (Fe) |
carbon dioxide (CO2) |
| Mr |
(56 x2) + (16 x3)=160 |
16 + 12 =28 |
56 |
12 + 32=44 |
Step 2- divide the mass of each substance by its Mr/Ar to calculate
the number of moles.
| substance |
iron (III) oxide(Fe2O3) |
carbon monoxide (CO) |
iron (Fe) |
carbon dioxide (CO2) |
| mass/Mr |
8/160 |
4.2/28 |
5.656 |
6.6/44 |
| number of moles present |
0.05 |
0.15 |
0.1 |
0.15 |
step 3- divide by the smallest number of moles present to get the moles ratio:
| substance |
iron (III) oxide(Fe2O3) |
carbon monoxide (CO) |
iron (Fe) |
carbon dioxide (CO2) |
| number of moles present |
0.05 |
0.15 |
0.1 |
0.15 |
| mole ratio |
0.05/0.05 = 1 |
0.15/0.05 =3 |
0.1/0.05=2 |
0.15/0.05= 3 |
So we have:
iron oxide(s) + carbon monoxide (g) → iron(s) + carbon dioxide(g)
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
Practice questions
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