Dr. Ray's Learning Center

Expressions of Concentration

General Chemistry | Dr. Ray’s Learning Center

Estimated study time: 20–25 minutes
Prerequisite: the mole concept, molar mass, and mass and volume conversions

In chemical and environmental engineering, we need to know not only what is in a solution but how much of it is there. The concentration of a solution expresses the amount of solute relative to a specified amount of solution or solvent.

Different units suit different jobs: molarity for reaction calculations, mass percent for formulations, and parts per million (ppm) for environmental measurements.

Learning Objectives

After this lesson, you should be able to:

  1. Distinguish among common concentration expressions.
  2. Calculate molarity, mass percent, and parts per million.
  3. Solve simple dilution problems.

Key Ideas

1. Molarity

Molarity is the amount of solute (mol) divided by the volume of the entire solution (L). Its unit is mol/L, written M.

C=nsoluteVsolutionC = \frac{n_{\mathrm{solute}}}{V_{\mathrm{solution}}}
  • CC = molarity (mol/L, or M)
  • nsoluten_{\mathrm{solute}} = amount of solute (mol)
  • VsolutionV_{\mathrm{solution}} = total solution volume (L)

Since n=m/Mn = m/M, where MM is the molar mass (as in Lesson 1), molarity can also be calculated directly from the solute mass:

C=msoluteMsoluteVsolutionC = \frac{m_{\mathrm{solute}}}{M_{\mathrm{solute}}\,V_{\mathrm{solution}}}

Note on symbols: many textbooks also use MM for molarity. Here we write molarity as CC so it is never confused with molar mass MM. The unit “M” (mol/L) is unchanged.

Important: use the final volume of the solution, not the volume of solvent added.

2. Mass Percent

Mass percent expresses the mass of solute relative to the total mass of solution:

mass percent=msolutemsolution×100%\text{mass percent} = \frac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 100\%

For a simple two-component solution:

msolution=msolute+msolventm_{\mathrm{solution}} = m_{\mathrm{solute}} + m_{\mathrm{solvent}}

3. Parts per Million

Parts per million (ppm) by mass is the mass fraction of a component multiplied by one million:

concentration in ppm=msolutemsolution×106\text{concentration in ppm} = \frac{m_{\mathrm{solute}}}{m_{\mathrm{solution}}} \times 10^{6}

For dilute aqueous solutions with a density close to 1.00 kg/L (1 L of solution ≈ 106 mg):

1mg/L≈1ppm1\,\mathrm{mg/L} \approx 1\,\mathrm{ppm}

This approximation is used constantly in water-quality work, but it is not exact: it depends on the solution density.

4. Dilution

Dilution adds solvent without changing the amount of solute. Because the moles of solute before and after are equal:

C1V1=C2V2C_1 V_1 = C_2 V_2
  • C1C_1, V1V_1 = molarity and volume of the initial (stock) solution
  • C2C_2, V2V_2 = molarity and volume of the final (diluted) solution

This assumes no solute is lost, consumed, or produced, and that V1V_1 and V2V_2 are in the same units. (You will often see this written as M1V1=M2V2M_1V_1 = M_2V_2.)

Worked Example 1 — Calculating Molarity

Problem: A solution is prepared by dissolving 5.85 g of sodium chloride in water and diluting to a final volume of 500.0 mL. Determine the molarity. Use MNaCl=58.44g/molM_{\mathrm{NaCl}} = 58.44\,\mathrm{g/mol}.

Step 1. List the given information.

  • mNaCl=5.85gm_{\mathrm{NaCl}} = 5.85\,\mathrm{g}
  • MNaCl=58.44g/molM_{\mathrm{NaCl}} = 58.44\,\mathrm{g/mol}
  • Vsolution=500.0mLV_{\mathrm{solution}} = 500.0\,\mathrm{mL}

Step 2. Convert the volume to liters.

V=500.0mL×1L1000mL=0.5000LV = 500.0\,\mathrm{mL} \times \frac{1\,\mathrm{L}}{1000\,\mathrm{mL}} = 0.5000\,\mathrm{L}

Step 3. Calculate the moles of NaCl.

nNaCl=5.85g58.44g/mol=0.1001moln_{\mathrm{NaCl}} = \frac{5.85\,\mathrm{g}}{58.44\,\mathrm{g/mol}} = 0.1001\,\mathrm{mol}

Step 4. Calculate the molarity.

C=0.1001mol0.5000L=0.2002mol/L≈0.200MC = \frac{0.1001\,\mathrm{mol}}{0.5000\,\mathrm{L}} = 0.2002\,\mathrm{mol/L} \approx 0.200\,\mathrm{M}

Final Answer: 0.200 M NaCl (three significant figures, set by 5.85 g)

Worked Example 2 — Mass Percent and ppm

Problem: A solution contains 10.0 g of sodium chloride and 190.0 g of water. Determine (a) the mass percent of NaCl and (b) the NaCl concentration in ppm by mass.

Step 1. Calculate the total mass of solution.

msolution=10.0g+190.0g=200.0gm_{\mathrm{solution}} = 10.0\,\mathrm{g} + 190.0\,\mathrm{g} = 200.0\,\mathrm{g}

Step 2. Calculate the mass percent.

mass percent=10.0g200.0g×100%=5.00%\text{mass percent} = \frac{10.0\,\mathrm{g}}{200.0\,\mathrm{g}} \times 100\% = 5.00\%

Step 3. Calculate the ppm.

10.0g200.0g×106=5.00×104ppm\frac{10.0\,\mathrm{g}}{200.0\,\mathrm{g}} \times 10^{6} = 5.00 \times 10^{4}\,\mathrm{ppm}

Final Answers: (a) 5.00% NaCl by mass; (b) 5.00 × 104 ppm NaCl by mass

The ppm definition is still valid here, but ppm is meant for trace amounts. For a solution this concentrated, mass percent is the sensible unit to report.

Worked Example 3 — Dilution

Problem: What volume of 2.00 M hydrochloric acid is required to prepare 250.0 mL of 0.200 M hydrochloric acid? Assume no solute is lost.

Step 1. List the known quantities.

  • C1=2.00MC_1 = 2.00\,\mathrm{M}
  • C2=0.200MC_2 = 0.200\,\mathrm{M}
  • V2=250.0mLV_2 = 250.0\,\mathrm{mL}
  • V1=?V_1 = ?

Step 2. Rearrange the dilution equation for the unknown.

V1=C2V2C1V_1 = \frac{C_2 V_2}{C_1}

Step 3. Substitute and calculate.

V1=(0.200M)(250.0mL)2.00M=25.0mLV_1 = \frac{(0.200\,\mathrm{M})(250.0\,\mathrm{mL})}{2.00\,\mathrm{M}} = 25.0\,\mathrm{mL}

Final Answer: 25.0 mL of 2.00 M HCl

In the lab: measure 25.0 mL of the stock acid and dilute with water to a total volume of 250.0 mL in a volumetric flask (add acid to water, never the reverse). Do not simply add 225.0 mL of water — volumes are not strictly additive, and the target is the final solution volume.

Practice Problems

Problem 1 — Molarity. Calculate the molarity of a solution prepared by dissolving 9.80 g of sulfuric acid in water and diluting to a final volume of 500.0 mL. Use MH2SO4=98.08g/molM_{\mathrm{H_2SO_4}} = 98.08\,\mathrm{g/mol}.

Problem 2 — Mass Percent. A solution contains 15.0 g of glucose and 135.0 g of water. Determine the mass percent of glucose.

Problem 3 — ppm. A water sample contains 2.50 mg of dissolved solute in 500.0 g of solution. Calculate the concentration in ppm by mass.

Problem 4 — Dilution. Calculate the volume of 1.50 M sodium hydroxide needed to prepare 300.0 mL of 0.100 M sodium hydroxide. Assume no reaction or loss of solute.

Show answers

1. 0.200 M H2SO4

C=9.80g(98.08g/mol)(0.5000L)=0.1998M≈0.200MC = \frac{9.80\,\mathrm{g}}{(98.08\,\mathrm{g/mol})(0.5000\,\mathrm{L})} = 0.1998\,\mathrm{M} \approx 0.200\,\mathrm{M}

2. 10.0% glucose by mass

mass percent=15.0g15.0g+135.0g×100%=10.0%\text{mass percent} = \frac{15.0\,\mathrm{g}}{15.0\,\mathrm{g} + 135.0\,\mathrm{g}} \times 100\% = 10.0\%

3. 5.00 ppm by mass

2.50×10−3g500.0g×106=5.00ppm\frac{2.50 \times 10^{-3}\,\mathrm{g}}{500.0\,\mathrm{g}} \times 10^{6} = 5.00\,\mathrm{ppm}

4. 20.0 mL of 1.50 M NaOH

V1=(0.100M)(300.0mL)1.50M=20.0mLV_1 = \frac{(0.100\,\mathrm{M})(300.0\,\mathrm{mL})}{1.50\,\mathrm{M}} = 20.0\,\mathrm{mL}

Common Pitfalls

  • Confusing solution volume with solvent volume. Molarity uses the total volume of solution.
  • Using the wrong denominator for mass percent. Divide by the total solution mass, not the solvent mass.
  • Assuming ppm always equals mg/L. That holds only for dilute aqueous solutions with density near 1.00 kg/L.
  • Using the dilution equation when a reaction occurs. If the solute is consumed or formed, you need reaction stoichiometry as well.
  • Treating molarity and molality as the same. Molarity is per liter of solution; molality is per kilogram of solvent.

Key Takeaway

Concentration is always “how much solute per how much of something.” Before you calculate, identify the basis the problem asks for — moles, mass, or volume — and make sure the denominator matches it.

Next Lesson: Lesson 3. Limiting Reactants and Percent Yield (coming soon)
Previous Lesson: Lesson 1. Reaction Stoichiometry
← Back to General Chemistry