Higher tier only
Glucose has the molecular formula C6H12O6, the numbers in this formula tell you the molar ratio of each element present in the compound. So
for example water has the molecular formula H2O, this indicates that there are 2 moles of hydrogen for
every one mole of oxygen present in water molecules. Or for glucose there are 6 moles of carbon,
12 moles of hydrogen and 6 moles
of oxygen atoms present in one mole of glucose molecules. You can also work backwards, so if you are given masses you can easily calculate the
formula for compounds.
So for our example using the masses given in the example above we have:
| moles of zinc: 6.5g/65 = 0.1 moles | moles of oxygen: 1.6g/16 = 0.1 moles |
| moles of zinc: 0.1 moles/0.1 =1 | moles of oxygen: 0.1 moles/0.1 = =1 |
The empirical formula for a compound is the simplest whole number ratio of each element present
in the compound. The empirical formula is usually found from experimental data. As an
example consider the sugar glucose, its molecular formula as shown above 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 relative molecular mass (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 molecular formula you need to multiply the empirical formula by 6.
A sample of magnesium ribbon, mass 1.2g was placed inside a crucible and heated strongly using a Bunsen burner 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 formed? Simply follow the steps listed in the table 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 | |
| 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, the smallest number of moles present is 6. | 6/6=1 = 1 | 12/6 = 2 |
| empirical formula | empirical formula is CH2 | |
You may also come across questions where instead of being given masses for the reactants and products to work out an empirical formula you may instead be given percentages e.g.
Example 3b- A compound has a composition of 49.3% carbon, 6.9% hydrogen, and 43.8% oxygen by mass. Calculate its empirical formula and determine its molecular formula if the molar mass is 146g.
Simply follow the method in the table below, which is very similar to the method shown above to work out the empirical formula.
| write out the symbols for each element preset | C | H | O |
|---|---|---|---|
| % of each element present | 49.3% | 6.9% | 43.8% |
| assume you have 100g of the substance so % will become grams- mass of each element present | 49.3g | 6.9g | 43.8g |
| divide masses by Ar to get number of moles present | 49.3/12= 4.1 | 6.9/1 = 6.9 | 43.8/16=2.74 |
| divide by the smallest number of moles to get ratio, the smallest number of moles present is 6. | 4.1/2.74=1 = 1.5 | 6.9/2.74 = 2.5 | 2.74/2.74=1 |
| empirical formula | empirical formula is C1.5H2.50
To remove decimals double all the numbers to give: C3H502 |
||
The empirical formula is C3H502 which gives a Mr = 73, however in the original question the molecular mass of the compound was given as 146; now this is simply double the mass of the empirical formula so the actual molecular formula for the unknown compound is simply C6H1004.
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 using the reacting masses given in the question.
Now you may be able to simply write the balanced symbolic equation if you have seen this combustion reaction before, but that is NOT the point of this example. You need to use the reacting masses in the example given to calculate the mole ratios, that is the purple a,b,c and d numbers in the equation above, for each of the reactants and products, to do this use the method shown below:
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 of the reactants and products present.
| substance | methane (CH4 | oxygen (O)2 | carbon dioxide (CO2) | hydrogen oxide (H2O) |
|---|---|---|---|---|
| Mr | 12 + 4 =16 | 16 x2 =32 | 12 + 32=44 | 2 + 16=18 |
| 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 |
| 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:
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 reactants and products 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 |
| 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 |
| 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 |