Concentration calculations

Concentration of solutions is measured in moles per decimetre cubed (mol dm^-3), often abbreviated to molar or M. This is the number of moles of solute contained in one decimetre cubed of solution. Remembering that per means divided by, gives us the equation for concentration (mol dm^-3 = mol ÷ dm³):

When calculating a concentration check that your answer looks reasonable. Dilute acids and alkalis are around 1 - 2 M. A concentrated acid may reach 20 M, but most solutions are around 1 M or less. Note that the volume must be in dm³, but it is often given in cm³. You need to remember that 1000 cm³ = 1 dm³.

Example

30.0 g of potassium bromide is dissolved in water and made up to a volume of 250 cm³. What is the concentration of the solution?

KBr = 119

Number of moles of KBr = 30.0/119 = 0.252 mol

Volume of solution = 250/1000 = 0.25 dm³

\Concentration of solution = 0.252/0.25 = 1.01 mol dm^-3

Exercise

1. 1.041 g of anhydrous sodium carbonate was dissolved in water and made up to a volume of 100 cm³. What is the concentration of the solution?

2. 48.6 cm³ of a saturated solution of potassium chloride was evaporated and found to contain 15.1 g of solid. What is the concentration of the potassium chloride?

The following rearranged equation is used extensively in chemistry calculations and needs remembering:

Number of moles (in mol) = concentration (in mol dm^-3) × volume (in dm³)

This is fairly obvious if you think of an easy example. A 2M solution contains 2 moles in each dm³. If you had 4 dm³ of this solution, you'd have 4 lots of 2 moles. That is, 4 × 2 = 8 moles.

Often you will be given the volume in cm³, and it is important to remember to convert this to dm³. The conversion factor from cm³ to dm³ is to divide by 1000 as 1000 cm³ = 1 dm³. Try to get a feel for the size of these volumes. 1 dm³ is also known as 1 litre, which is the typical volume of a large carton of fruit juice (look in your fridge!). 1 cm³ is what it says - a cube with sides of 1 cm. This isn't very big - sketch it out on a piece of paper.

Example

25.0 cm³ of 1.00 M sulfuric acid contains 25/1000 × 1 = 0.025 mol of sulfuric acid

Knowing the number of moles of substance, we can now convert to the mass of substance by multiplying by the appropriate molar mass.

H₂SO₄ = 98

\mass of sulfuric acid = 0.025 × 98 = 2.45 g

Exercise

3. What mass of nitric acid is needed to make 5 dm³ of 2 M solution?

4. What mass of copper(II) sulfate-5-water is contained in 38.6 cm³ of 0.963 mol dm^-3 solution? Give your answer to three significant figures.