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| Quantitative Chemistry |
Relative Atomic Mass | Relative Formula Mass | Percentage Composition of Compounds | Calculating the Mass of Reactants or Products | Calculating Formulae | Electrolysis Calculations
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Relative Atomic Mass
Atoms of different elements have different masses.
Chemists often need to know how much of a chemical will be used up or produced in a reaction.
Atoms, ions and molecules are too small to count or weigh, so the average mass of a large number of atoms of an element is used. The number of atoms is called a mole and the mass of one mole is called the relative atomic mass, Ar. This value is usually given in the periodic table as the top number written next to the symbol for each element.
For carbon,  , the relative atomic mass is 12.
The relative atomic masses of some elements
Element |
Symbol |
Ar |
Element |
Symbol |
Ar |
Hydrogen |
H |
1 |
Magnesium |
Mg |
24 |
Carbon |
C |
12 |
Sulphur |
S |
32 |
Oxygen |
O |
16 |
Copper |
Cu |
64 |
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Relative Formula Mass
Using the Ar values, we can work out the relative formula mass, Mr, of a molecule.
The formula for water is H2O.
A water molecule contains two atoms of hydrogen and one of oxygen.
Mr water = 1+1+16
= 18
Here are some more examples:
Compound
|
Formula |
Number of atoms |
Ar of atoms |
Mr of compound |
Magnesium oxide |
MgO |
1 Mg
1O |
Mg = 24
O = 16 |
24 + 16
= 40 |
Sulphur dioxide |
SO2 |
1 S
2 O |
S = 32O
= 16 |
32 + 16
= 48 |
Copper carbonate |
CuCO3 |
1 Cu
1C
3O |
Cu = 64C
= 12O
= 16 |
64 + 12 + (3 x 16)
= 124 |
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Percentage Composition of Compounds
From the formula of a compound you can also calculate the percentage of each element present, if you know the Ar values for each element.
The formula for methane is CH4.
Its relative formula mass is 12 + 1 + 1 + 1 + 1 = 16.
| Percentage mass of carbon in methane = |
mass of carbon in methane |
x 100 |
|
total mass |
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| Percentage mass of hydrogen in methane = |
mass of hydrogen in methane |
x 100 |
|
total mass |
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Calculating the Mass of Reactants or Products
From a balanced equation it is possible to calculate the amounts of all the chemicals involved.
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CaCO3(s) |
+ |
2HCl(aq) |
—> |
CaCl2(aq) |
+ |
CO2(g) |
+ |
H2O(l) |
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Mr: |
40 + 12 + (3 x16) |
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(2 x 1) + (2 x 35,5) |
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40 + (2 x 35,5) |
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12 + (2 x 16) |
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(2 x 1) + 16 |
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= 100 |
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= 73 |
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= 111 |
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= 44 |
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= 18 |
The total mass of the reactants always equals the total mass of the products.
Example: Calculate the mass of carbon dioxide produced when 80 g of calcium carbonate reacts with excess hydrochloric acid.
100 g calcium carbonate produces 44 g of carbon dioxide.
Therefore 1 g of calcium carbonate will produce 44/100 g of carbon dioxide.
80 g of calcium carbonate will produce 80 x |
44 g of carbon dioxide. |
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100 |
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= 35,2 g |
Sometimes you need to calculate the volume of a gas produced in a reaction.
The relative formula mass of any gas has a volume of 24 litres at 25 ºC and 1 atmosphere pressure.
So to calculate the volume of carbon dioxide produced in this reaction:
| 44 g of CO2 has a volume of 24 litres. |
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Therefore 1 g of CO 2 has a volume of  litres. |
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| 35,2 g of CO2 will have a volume of |
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x 35,2 |
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= |
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19,2 litres |
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| Calculating Formulae
The formula of a compound can be found if the masses or percentages of the reacting chemicals are known.
For example, the formula of magnesium oxide can be found by burning a known mass of magnesium in a crucible and weighing the amount of magnesium oxide formed.
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Sample results: |
mass of crucible + lid |
= 20,2 g |
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mass of crucible + lid + magnesium |
= 22,6 g |
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mass of crucible + lid + magnesium oxide |
= 24,2 g |
mass of magnesium burnt |
= 22,6 – 20,2 = 2,4 g |
mass of magnesium oxide formed |
= 24,2 – 20,2 = 4,0 g |
therefore mass of oxygen |
= 4,0 – 2,4 = 1,6 g |
2,4 g of magnesium combined with 1,6 g oxygen.
The number of moles of each element is found from
number of moles |
= |
mass |
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mass of one mole (Ar) |
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Ar magnesium = 24 |
Ar oxygen = 16 |
2,4 |
magnesium reacted with 1,6 oxygen |
| 24 |
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16 |
0,1 moles of magnesium react with 0,1 moles of oxygen.
The formula of magnesium oxide must therefore be MgO, because there are equal moles of each element present.
1 g of hydrogen reacts with 16 g of sulphur to form hydrogen sulphide.
Ar hydrogen = 1 |
Ar sulphur = 32 |
1 moles of hydrogen react with 1,6 moles of sulphur. |
1 |
32 |
1 mole of hydrogen reacts with 0,5 moles of sulphur.
The formula of hydrogen sulphide is H2S.
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Electrolysis Calculations
If you are given the mass of one of the products of electrolysis, you can calculate the mass of the product at the other electrode if you know the half reactions.
For example, these are the half reactions for the electrolysis of sodium chloride solution:
2Cl- |
- 2e- |
—» |
Cl2 |
2H+ |
- 2e- |
—» |
H2 |
If you know that 10 g of hydrogen are formed you can calculate the mass of chlorine produced.
Number of moles = |
mass |
| |
mass of 1 mole |
Ar hydrogen = 1, so Mr H2 = 2
Ar chlorine = 35,5, so Mr Cl2 = 71
| Number of moles of hydrogen = |
10 |
= 5 |
| |
2 |
|
For every mole of hydrogen produced, the same moles of chlorine are produced.
| 5 |
= |
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| mass of chlorine |
= |
5 x 71 |
| |
= |
355 g |
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