Chemistry · General chemistry II · Worked example

Express a trace concentration in ppm and ppb

A 250.0 g sample of well water contains 1.2 mg of calcium ions. Express the calcium concentration in ppm, in ppb and as a mass percent.

Use one mass unit

Convert the solute’s mass to grams.

1.2 mg = 1.2 \times 10^{- 3} g

Find the mass fraction

Divide by the solution’s mass. The calcium adds only 0.0012 g to the 250.0 g, too little to change the total.

\frac{1.2 \times 10^{- 3} g}{250.0 g} = 4.8 \times 10^{- 6}

Scale it

Multiply by 10⁶ for ppm, by 10⁹ for ppb and by 100 for percent.

\begin{array}{l} 4.8 \times 10^{- 6} \times 10^{6} = 4.8 \\ 4.8 \times 10^{- 6} \times 10^{9} = 4800 \\ 4.8 \times 10^{- 6} \times 100 = 4.8 \times 10^{- 4} \end{array}

Check with milligrams per liter

The sample is about 250 mL of water, and 1.2 mg ÷ 0.250 L = 4.8 mg/L: the same number as ppm, as expected for a dilute water solution.

Result

4.8 ppm, 4800 ppb, or 4.8 × 10⁻⁴ % by mass.

Your turn

A 2.0 kg sample of water contains 0.050 mg of lead. Express the concentration in ppb.

Show the answer and explanation

25 ppb.

0.050 mg = 5.0 × 10⁻⁵ g, and (5.0 × 10⁻⁵ g ÷ 2.0 × 10³ g) × 10⁹ = 25 ppb.

\frac{5.0 \times 10^{- 5}}{2.0 \times 10^{3}} \times 10^{9} = 25

Keep exploring

Solutions & concentration opens in mass percent with 1.2 × 10⁻³ g of calcium in 250.0 g of solution and gives 4.8 × 10⁻⁴ %. Multiply by 10⁴ to get ppm.

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