Product labels and water reports rarely use molarity. Household vinegar is about 5% acetic acid by mass; a water-quality report lists nitrate, lead or fluoride in parts per million (ppm) or in mg/L.

Both units compare the mass of the solute with the mass of the whole solution. Percent scales that fraction by 100; ppm scales it by one million, which suits very small amounts (Figure 1). No molar masses are needed.

1. startMassessolute and solution, same unit
2. $\frac{m_{\text{solute}}}{m_{\text{solution}}}$mass fraction
3. Mass fractiona number below 1
4. $\times 100\%$ or $\times 10^{6}$scale
5. answerPercent or ppm% or parts per million

The solution mass includes the solute.For dilute water, $1\,\text{mg/L} \approx 1\,\text{ppm}$.

Learning Objectives

After this lesson, you should be able to:

1. Calculate mass percent and ppm from solute and solution masses.
2. Relate mg/L to ppm for dilute aqueous solutions.

Key Ideas

1. Mass Percent

The solution mass is the solute plus the solvent: $m_{\text{solution}} = m_{\text{solute}} + m_{\text{solvent}}$. Dissolving 10.0 g of salt in 190.0 g of water gives 200.0 g of solution, and the salt is 5.00% of it (Figure 2).

Key idea: divide by the mass of the solution, not by the mass of the water.

2. Parts per Million

One part per million is 1 g of solute in 10⁶ g of solution, which is the same as 1 mg in 1 kg. Percent would need many zeros for amounts this small: 1 ppm is 0.0001% (Figure 3).

3. Concentration in mg/L and ppm

Dilute water samples have a density of about 1.00 g/mL, so 1 L of the sample has a mass of about 1000 g = 10⁶ mg. A concentration of 1 mg/L is therefore about 1 mg in every 10⁶ mg, which is 1 ppm (Figure 4).

Key idea: for dilute aqueous solutions, mg/L and ppm are the same number. This only works when the density is close to 1.00 g/mL.

Worked Example 1: Mass Percent of a Solution

Problem: A solution is made by dissolving 10.0 g of sodium chloride ($\mathrm{NaCl}$) in 190.0 g of water. What is the mass percent of NaCl?

Step 1. Add the solute and solvent masses.

Step 2. Divide by the solution mass and multiply by 100%.

Final Answer: 5.00% NaCl by mass

1. Masses given$10.0\,\text{g} + 190.0\,\text{g}$
2. Step 1+add
3. Solution$200.0\,\text{g}$
4. Step 2$\frac{10.0\,\text{g}}{200.0\,\text{g}}$ $\times 100\%$fraction times 100
5. Mass percent$5.000\% \approx 5.00\%$Final answer to 3 significant figures

Your turn: mass percent

Same method as Worked Example 1: add the masses first, then divide the solute mass by the solution mass.

Worked Example 2: Mass of Solute from a Percent

Problem: A bottle of vinegar holds 365 g of solution that is 4.80% acetic acid ($\mathrm{CH_3\!COOH}$) by mass. What mass of acetic acid does it contain?

Step 1. Write the percent as a fraction. 4.80% means 4.80 g of acetic acid in every 100 g of vinegar: $\dfrac{4.80}{100}$.

Step 2. Multiply the solution mass by the fraction.

Final Answer: 17.5 g of acetic acid

1. Vinegar$365\,\text{g}$
2. Steps 1 and 2$\times \dfrac{4.80}{100}$percent as a fraction
3. Acetic acid$17.52\,\text{g} \approx 17.5\,\text{g}$Final answer to 3 significant figures

Your turn: mass of solute from a percent

Same method as Worked Example 2: turn the percent into a fraction and multiply by the solution mass.

Worked Example 3: Parts per Million in Water

Problem: A 625 g sample of well water contains 3.40 mg of dissolved nitrate. (a) What is the nitrate concentration in ppm? (b) Express it in mg/L.

Step 1. Convert the solute mass to grams, so both masses have the same unit.

Step 2. Divide by the sample mass and multiply by $10^{6}$.

Step 3. (b) Use the density of dilute water. 625 g of water is about 625 mL = 0.625 L.

Final Answer: (a) 5.44 ppm nitrate; (b) 5.44 mg/L

1. Nitrate$3.40\,\text{mg}$
2. Step 1$\times \dfrac{1\,\text{g}}{1000\,\text{mg}}$mg to g
3. Nitrate$0.00340\,\text{g}$
4. Step 2$\times \dfrac{10^{6}}{625\,\text{g}}$fraction times one million
5. Concentration$5.440\,\text{ppm} \approx 5.44\,\text{ppm}$Final answer to 3 significant figures

Exam tip: the two answers match because the sample is dilute water (Key Idea 3). For seawater or a syrup, mg/L and ppm differ.

Your turn: parts per million

Same method as Worked Example 3: by mass (convert mg to g first) or by volume for a dilute water sample (in mg/L).

Common Mistakes

Common mistake: Dividing by the mass of water. In Worked Example 1, $\frac{10.0\,\text{g}}{190.0\,\text{g}} \times 100\% = 5.26\%$, not 5.00%. The denominator is the whole solution, solute included.

Common mistake: Forgetting the factor of 100. 0.0500 is the mass fraction; the mass percent is 5.00%.

Common mistake: Using the percent as a whole number. 365 g × 4.80 = 1752 g, more than the whole bottle. Divide the percent by 100 first.

Common mistake: Mixing mg and g in ppm. $\frac{3.40\,\text{mg}}{625\,\text{g}} \times 10^{6}$ gives 5440, a thousand times too large. Put both masses in g (or use mg/kg).

Common mistake: Rounding too early. Keep 4 significant figures in intermediate values and round only the final answer to 3.

Key Takeaway

Mass percent and ppm are the same fraction, solute mass over solution mass, scaled by 100 or by one million. For dilute water samples, 1 mg/L is 1 ppm.

Practice: randomized problems

Each problem below uses fresh numbers every time: work it out, check your answer, then open the worked solution to compare your steps.

Your results

This summary collects every practice problem you try on this page and suggests what to review next.

Figures: original diagrams by Dr. Ray’s Learning Center, licensed CC BY 4.0. The practice problems are generated and checked by an audited engine; see the lesson audit log.