Concentration Units

Overview

Concentration describes the amount of solute dissolved in a given amount of solution or solvent. Different units are useful for different applications.

Common Concentration Units

Unit	Formula	When to Use
Molarity (M)	mol solute / L solution	Most lab work
Molality (m)	mol solute / kg solvent	Colligative properties
Mole Fraction (X)	mol component / total mol	Gas mixtures, Raoult's law
Mass Percent	(mass solute / mass solution) × 100	Consumer products
Parts per million (ppm)	(mass solute / mass solution) × 10⁶	Trace amounts

Molarity (M)

Moles of solute per liter of solution.

$$M = \frac{n}{V} = \frac{\text{mol solute}}{\text{L solution}}$$

Example

Dissolve 58.5 g NaCl in water to make 500 mL solution.

$$\text{mol NaCl} = \frac{58.5 \text{ g}}{58.5 \text{ g/mol}} = 1.00 \text{ mol}$$ $$M = \frac{1.00 \text{ mol}}{0.500 \text{ L}} = 2.00 \text{ M}$$

Important: Volume of Solution, Not Solvent!

Molality (m)

Moles of solute per kilogram of solvent.

$$m = \frac{\text{mol solute}}{\text{kg solvent}}$$

Example

Dissolve 36.0 g glucose (C₆H₁₂O₆) in 500 g water.

$$\text{mol glucose} = \frac{36.0}{180} = 0.200 \text{ mol}$$ $$m = \frac{0.200 \text{ mol}}{0.500 \text{ kg}} = 0.400 \text{ m}$$

Advantage

• Temperature independent (mass doesn't change with T)
• Useful for colligative properties

Mole Fraction (X)

Ratio of moles of component to total moles.

$$X_A = \frac{n_A}{n_A + n_B + n_C + \cdots}$$

For a two-component system:

$$X_A + X_B = 1$$

Example

Mix 2 mol ethanol with 8 mol water.

$$X_{\text{ethanol}} = \frac{2}{2 + 8} = 0.2$$ $$X_{\text{water}} = \frac{8}{2 + 8} = 0.8$$

Mass Percent (% w/w)

Mass of solute per 100 g of solution.

$$\text{Mass \%} = \frac{\text{mass solute}}{\text{mass solution}} \times 100\%$$

Example

20 g NaCl dissolved in 80 g water.

$$\text{Mass \%} = \frac{20}{100} \times 100\% = 20\%$$

Parts per Million (ppm)

Used for very dilute solutions.

$$\text{ppm} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^6$$

For aqueous solutions (density ≈ 1 g/mL):

$$1 \text{ ppm} \approx 1 \text{ mg/L} = 1 \text{ mg/kg}$$

Example

Drinking water contains 0.002 g fluoride per liter.

$$\text{ppm} = \frac{0.002 \text{ g}}{1000 \text{ g}} \times 10^6 = 2 \text{ ppm}$$

Parts per Billion (ppb)

$$\text{ppb} = \frac{\text{mass solute}}{\text{mass solution}} \times 10^9$$

Converting Between Units

Molarity ↔ Molality

$$M = \frac{m \times \rho}{1 + \frac{mM_{\text{solute}}}{1000}}$$

For dilute solutions:

$$M \approx m \times \rho \quad \text{(when } mM \ll 1000\text{)}$$

Molarity from Mass Percent

$$M = \frac{\% \times \rho \times 10}{M_{\text{solute}}}$$

Where:

• $\%$ = mass percent
• $\rho$ = solution density (g/mL)
• $M_{\text{solute}}$ = molar mass of solute

Example

Find molarity of 36.5% HCl ($\rho$ = 1.18 g/mL).

$$M = \frac{36.5 \times 1.18 \times 10}{36.5} = 11.8 \text{ M}$$

Dilution

When diluting solutions:

$$M_1V_1 = M_2V_2$$

Moles of solute remain constant.

Example

Dilute 50 mL of 6.0 M HCl to make 0.50 M solution.

$$V_2 = \frac{M_1V_1}{M_2} = \frac{(6.0)(50)}{0.50} = 600 \text{ mL}$$

Add 550 mL water to original 50 mL.

Summary Table

Unit	Symbol	Formula	Units
Molarity	M	mol/L solution	mol/L
Molality	m	mol/kg solvent	mol/kg
Mole Fraction	X	$n/n_{\text{total}}$	unitless
Mass Percent	% w/w	$(m_{\text{solute}}/m_{\text{solution}}) \times 100$	%
Volume Percent	% v/v	$(V_{\text{solute}}/V_{\text{solution}}) \times 100$	%
ppm	ppm	$(m_{\text{solute}}/m_{\text{solution}}) \times 10^6$	mg/kg

Stoichiometry with Solutions

Using Molarity

$$\text{mol} = M \times V \text{ (in liters)}$$

Example

How many moles in 250 mL of 0.100 M NaOH?

$$\text{mol} = 0.100 \text{ M} \times 0.250 \text{ L} = 0.0250 \text{ mol}$$

Titration Calculation

$$\frac{M_AV_A}{\text{coeff}_A} = \frac{M_BV_B}{\text{coeff}_B}$$

For 1:1 reactions:

$$M_AV_A = M_BV_B$$