Parts per million (ppm) is a rather awkward unit of concentration and for this reason often appears in questions!
To work with this unit, you need to be confident with converting between various SI units of mass e.g., Mg, kg, g, mg and μg
| Unit | Multiplier |
| Mg = 1 metric tonne | x 106 |
| kg | x 103 |
| g | x 100 |
| mg | x 10-3 |
| μg | x 10-6 |
A million corresponds to 106 so we need to get the values into any pair of units with a factor of 106 between them. This means that 1 ppm is equivalent to:
1 g of something in 1 Mg of something else.
1 mg of something in 1 kg of something else.
1 μg of something in 1 g of something else.
(etc)…
Once you have the masses in (any) suitable pair of units, you simply divide the mass by the total mass to find ppm:
8 mg of Pt in 2 kg of rock:
i.e., 8 mg / 2 kg = 4 ppm.
100 mg of As in 50 kg of alloy:
i.e., 100 mg / 50 kg = 2 ppm
30 mg of Pb in 100 g (=0.1 kg) of water:
i.e., 30 mg / 0.1 kg = 300 ppm.
30 mg (= 30,000 μg) of Pb in 100 g of water:
i.e., 30,000 μg / 100 g = 300 ppm.
80 kg (=80,000 g) of Fe in 5 Mg of rock:
i.e., 80,000 g / 5 Mg = 16,000 ppm.
0.7 μg of Mo in 1,000 mg (= 1 g) of a diamond:
i.e., 0.7 μg / 1 g = 0.7 ppm.
The ppm unit can also be used in the other direction. To find the mass of something, you simply multiply the total mass by the ppm, then adjust the unit accordingly.
4 ppm of As in 80 Mg of alloy
i.e., 4 ppm x 80 Mg = 320 g
2 ppm of Pb in 8 kg of rock
i.e., 2 ppm x 8 kg = 16 mg
9 ppm of Na in 100 g of water
i.e., 9 ppm x 100 g = 900 μg
| Example 1: 5 tonnes of ore contains 100 ppm Pb in the form of PbO. Determine the ppm of PbO in the ore. Answer: 5 tonnes of ore = 5 Mg of ore. 100 ppm x 5 Mg = 500 g Now use the Ar of Pb (207.2) to work out the moles of Pb: n Pb = 500 / 207.2 = 2.41 mol This is equivalent to the moles of PbO: n PbO = 2.41 mol Now multiply the moles of PbO by its Mr (223.2) to work out the mass of PbO: m PbO = 2.41 x 223.2 = 538 g 538 g of Pb in 5 Mg of ore, i.e., 538 g / 5 mg = 108 ppm. |
If you are feeling confident with ppm, there is another route to the same answer. This approach ignores the 5 tonnes (which isn’t actually relevant), skips the mole calculation and instead simply uses the idea of % by mass:
| Example 1: 5 tonnes of ore contains 100 ppm Pb in the form of PbO. Determine the ppm of PbO in the ore. Answer Using the Ar of Pb (207.2) and Mr of PbO (223.2), PbO is 92.8% Pb by mass. This means that the ppm of PbO is proportionally higher, i.e., 100 / 0.928 = 108 ppm |
Another example of this question might involve aqueous solutions:
| Example 2: An aqueous CaCl2 solution has a concentration of 0.1 mol dm-3. Determine the concentration of Cl– ions in ppm. The density of the solution is 1 kg dm-3. Answer We need to calculate the mass of Cl– in a given mass of solution. These easiest way to do this is just to imagine that we have exactly 1 dm3 of the solution… V solution = 1 dm3 m solution = 1 kg n CaCl2 = 0.1 mol n Cl– = 0.1 x 2 = 0.2 mol m Cl– = 0.2 x 35.5 = 7.1 g = 7100 mg i.e., 7100 mg / 1 kg = 7100 ppm |