Relative masses, percentage composition and solution concentration

For a substance made of molecules, the term relative molecular mass can also be used. Ionic compounds such as sodium chloride do not consist of separate molecules, so use relative formula mass for them.

Reading subscripts and brackets

A small number after an element symbol tells you how many atoms of that element to count. If there is no number, count one. Water, H₂O, therefore contains two hydrogen atoms and one oxygen atom per molecule. Using $A_r(\text{H})=1$ and $A_r(\text{O})=16$:

$$M_r(\text{H}_2\text{O}) = (2 \times 1) + 16 = 18$$

The same method works for more complicated formulae. Potassium carbonate, K₂CO₃, requires two potassium atoms, one carbon atom and three oxygen atoms. With $A_r(\text{K})=39$, $A_r(\text{C})=12$ and $A_r(\text{O})=16$:

$$M_r(\text{K}_2\text{CO}_3) = (2 \times 39) + 12 + (3 \times 16) = 138$$

A subscript outside brackets multiplies everything inside them. In calcium hydroxide, Ca(OH)₂, the two OH groups contribute two oxygen atoms and two hydrogen atoms. Using $A_r(\text{Ca})=40$:

$$M_r(\text{Ca(OH)}_2) = 40 + (2 \times 16) + (2 \times 1) = 74$$

In ammonium sulfate, (NH₄)₂SO₄, the outer 2 doubles both the nitrogen and hydrogen inside the brackets. The complete atom count is two N, eight H, one S and four O. Using $A_r(\text{N})=14$ and $A_r(\text{S})=32$, as well as the values above, gives $M_r=132$.

Finding an element’s percentage by mass

An element’s percentage by mass tells you how much of the compound’s total mass comes from that element. First calculate its contribution to the formula mass, then divide by the whole formula mass and multiply by 100:

$$\text{percentage by mass} = \frac{A_r\text{ of element} \times \text{number of its atoms in the formula}}{M_r\text{ of compound}} \times 100$$

For example, the two oxygen atoms in Ca(OH)₂ contribute $2 \times 16 = 32$ to its formula mass of 74. The oxygen percentage is therefore:

$$\frac{32}{74} \times 100 = 43.2\% \quad \text{(to one decimal place)}$$

This means that about 43.2% of the mass of any pure sample of calcium hydroxide is contributed by oxygen. It is a mass percentage, not a percentage of the atoms: different elements have different atomic masses.

Concentration compares solute mass with solution volume

When a substance dissolves, it is the solute; the liquid it dissolves in is the solvent. Together they form a solution. For example, when sodium hydroxide dissolves in water, sodium hydroxide is the solute and water is the solvent.

Concentration in grams per cubic decimetre tells you the mass of solute in each cubic decimetre of solution:

$$\text{concentration (g dm}^{-3}\text{)} = \frac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}$$

More solute in the same solution volume gives a higher concentration. Adding solvent without adding solute increases the volume and lowers the concentration: the same mass of solute is spread through more solution.

Matching the volume units

One cubic decimetre is one litre, and:

$$1\text{ dm}^3 = 1000\text{ cm}^3$$

Divide a volume in cm³ by 1000 to express it in dm³. Multiply by 1000 to convert back.

For example, consider a solution containing 10 g of sodium hydroxide with a final volume of 2000 cm³. Its volume is $2000 \div 1000 = 2\text{ dm}^3$, so its concentration is $10 \div 2 = 5\text{ g dm}^{-3}$. Each cubic decimetre of this solution contains 5 g of dissolved sodium hydroxide.

The concentration equation can also be rearranged to find either mass or volume:

$$\text{mass} = \text{concentration} \times \text{volume}$$ $$\text{volume} = \frac{\text{mass}}{\text{concentration}}$$

With concentration in g dm⁻³, use mass in grams and volume in dm³.

$$\text{percentage by mass} = \frac{A_r\text{ of element} \times \text{number of its atoms}}{M_r\text{ of compound}} \times 100$$

Compare the element’s mass contribution with the whole compound’s formula mass, not its fraction of the atom count.

Solution concentration

$$c = \frac{m}{V}, \qquad m = cV, \qquad V = \frac{m}{c}$$

Use $c$ in g dm⁻³, solute mass $m$ in g and final solution volume $V$ in dm³.

• $1\text{ dm}^3 = 1000\text{ cm}^3 = 1\text{ litre}$.
• cm³ → dm³: divide by 1000.
• More solute at fixed volume increases concentration; more solvent at fixed solute mass decreases it.