Expressing Concentration of Solutions

Saying a solution is “dilute” or “concentrated” gives a rough idea, but it is far too vague for real work. A doctor preparing an IV drip, an environmental scientist testing river water, or a chemist running a reaction all need an exact number that says precisely how much solute is present. Chemistry provides seven standard ways to express that number, each suited to different situations.

Percentage-Based Units: Ratios Out of a Hundred

The simplest concentration units compare some measure of the solute to the total solution and express the result as a percentage.

Mass Percentage (w/w)

Mass percentage (the mass of a component as a fraction of the total solution mass, times 100) is defined as:

$\text{Mass \% of a component} = \frac{\text{Mass of the component in the solution}}{\text{Total mass of the solution}} \times 100 \qquad \text{(1.1)}$

If someone tells you a solution is “10% glucose by mass”, that means every 100 g of solution contains 10 g of glucose and 90 g of water. Notice that the denominator is the total solution mass (solute + solvent together), not the solvent mass alone.

Mass percentage is widely used in industrial chemistry. For instance, commercial bleaching solution contains 3.62 mass percent sodium hypochlorite ($NaClO$) in water.

Volume Percentage (V/V)

When both solute and solvent are liquids, it is often more practical to measure volumes rather than masses. Volume percentage is defined as:

$\text{Volume \% of a component} = \frac{\text{Volume of the component}}{\text{Total volume of solution}} \times 100 \qquad \text{(1.2)}$

A “10% ethanol solution in water” means 10 mL of ethanol is present in enough water to make the total volume 100 mL.

A practical example: a 35% (v/v) ethylene glycol solution is used as antifreeze in car engines. At this concentration, the solution lowers the freezing point of water all the way down to 255.4 K ($-17.6\degree C$), preventing the coolant from freezing in winter.

Mass by Volume Percentage (w/V)

Mass by volume percentage tells you the mass of solute dissolved in 100 mL of solution. This unit is a favourite in medicine and pharmacy because liquid medicines are dispensed by volume, and knowing the drug mass per fixed volume makes dosing straightforward.

Tracking Tiny Amounts: Parts Per Million

Some solutes are present in such small quantities that percentages become inconveniently tiny numbers. For these trace (extremely small) quantities, we use parts per million (ppm):

$\text{Parts per million} = \frac{\text{Number of parts of the component}}{\text{Total number of parts of all components}} \times 10^6 \qquad \text{(1.3)}$

Just like percentage, ppm can be expressed on a mass-to-mass, volume-to-volume, or mass-to-volume basis.

Here is what ppm looks like in practice: one litre of sea water weighs about 1030 g and contains roughly $6 \times 10^{-3}$ g of dissolved $O_2$. That tiny mass works out to about 5.8 g of oxygen per $10^6$ g of sea water, or 5.8 ppm. Pollution levels in drinking water and trace gases in the atmosphere are routinely reported in ppm or the related unit $\mu g\;mL^{-1}$.

Counting Molecules Instead of Mass: Mole Fraction

All the units above measure concentration using masses or volumes. Mole fraction takes a different approach: it counts the actual number of moles of each component relative to the total.

The symbol for mole fraction is $x$, with a subscript showing which component you mean. For any component in a solution:

$\text{Mole fraction of a component} = \frac{\text{Number of moles of the component}}{\text{Total number of moles of all the components}} \qquad \text{(1.4)}$

For a binary solution (two components A and B) with $n_A$ moles of A and $n_B$ moles of B:

$x_A = \frac{n_A}{n_A + n_B} \qquad \text{(1.5)}$

If the solution has many components (labelled 1, 2, … i), the general formula is:

$x_i = \frac{n_i}{n_1 + n_2 + \ldots + n_i} = \frac{n_i}{\sum n_i} \qquad \text{(1.6)}$

An Important Property: Mole Fractions Always Add Up to One

No matter how many components a solution has, the sum of all their mole fractions is exactly 1:

$x_1 + x_2 + \ldots + x_i = 1 \qquad \text{(1.7)}$

This is a handy shortcut. In a binary solution, once you know one mole fraction, the other is simply $1$ minus that value.

Why Mole Fraction Matters

Mole fraction connects directly to several physical properties of solutions, most notably vapour pressure. It is also the go-to unit for calculations involving gas mixtures, where counting molecules is more natural than weighing them.

Worked Example 1.1: Mole Fraction of Ethylene Glycol

Problem: Find the mole fraction of ethylene glycol ($C_2H_6O_2$) in a solution that contains 20% $C_2H_6O_2$ by mass.

Solution:

Step 1: Pick a convenient total mass. Assume 100 g of solution. (Any amount works because mole fraction is a ratio; the answer will be the same.)

With 20% by mass, this gives:

• Mass of $C_2H_6O_2$ = 20 g
• Mass of $H_2O$ = 100 g $-$ 20 g = 80 g

Step 2: Convert masses to moles.

Molar mass of $C_2H_6O_2$:

$= (12 \times 2) + (1 \times 6) + (16 \times 2) = 24 + 6 + 32 = 62\;\text{g mol}^{-1}$

Moles of $C_2H_6O_2$:

$= \frac{20\;\text{g}}{62\;\text{g mol}^{-1}} = 0.322\;\text{mol}$

Moles of $H_2O$:

$= \frac{80\;\text{g}}{18\;\text{g mol}^{-1}} = 4.444\;\text{mol}$

Step 3: Apply the mole fraction formula.

$x_{\text{glycol}} = \frac{0.322}{0.322 + 4.444} = \frac{0.322}{4.766} = 0.068$

Step 4: Find the mole fraction of water using the sum rule.

$x_{\text{water}} = 1 - 0.068 = 0.932$

You can verify this independently:

$x_{\text{water}} = \frac{4.444}{0.322 + 4.444} = \frac{4.444}{4.766} = 0.932 \;\checkmark$

Molarity: Moles of Solute Per Litre of Solution

Molarity (symbol M) is probably the most frequently used concentration unit in the chemistry lab. It tells you how many moles of solute are dissolved in one litre (one cubic decimetre) of the total solution:

$\text{Molarity} = \frac{\text{Moles of solute}}{\text{Volume of solution in litre}} \qquad \text{(1.8)}$

Saying a solution is “0.25 M $NaOH$” (read as “0.25 molar sodium hydroxide”) means 0.25 mol of $NaOH$ is present in every litre of that solution.

Worked Example 1.2: Molarity of a NaOH Solution

Problem: Find the molarity of a solution that contains 5 g of $NaOH$ in 450 mL of solution.

Solution:

Step 1: Convert grams to moles.

Molar mass of $NaOH$ = 23 + 16 + 1 = 40 g mol$^{-1}$

$\text{Moles of } NaOH = \frac{5\;\text{g}}{40\;\text{g mol}^{-1}} = 0.125\;\text{mol}$

Step 2: Convert mL to L.

$\text{Volume} = \frac{450\;\text{mL}}{1000\;\text{mL L}^{-1}} = 0.450\;\text{L}$

Step 3: Apply the molarity formula (Equation 1.8).

$\text{Molarity} = \frac{0.125\;\text{mol}}{0.450\;\text{L}} = 0.278\;\text{M}$

This can also be written as $0.278\;\text{mol L}^{-1}$ or $0.278\;\text{mol dm}^{-3}$. All three notations mean exactly the same thing.

Molality: Moles of Solute Per Kilogram of Solvent

Molality (symbol m) looks similar to molarity but measures something subtly different. Instead of dividing by the volume of the solution, you divide by the mass of the solvent in kilograms:

$\text{Molality (m)} = \frac{\text{Moles of solute}}{\text{Mass of solvent in kg}} \qquad \text{(1.9)}$

A “1.00 m $KCl$ solution” means 1 mol of $KCl$ (74.5 g) is dissolved in 1 kg of water. Notice that the denominator is the solvent mass, not the total solution mass or volume.

Worked Example 1.3: Molality of Ethanoic Acid in Benzene

Problem: Calculate the molality when 2.5 g of ethanoic acid ($CH_3COOH$) is dissolved in 75 g of benzene.

Solution:

Step 1: Find the molar mass of ethanoic acid.

The molecular formula can also be written as $C_2H_4O_2$:

$\text{Molar mass} = (12 \times 2) + (1 \times 4) + (16 \times 2) = 24 + 4 + 32 = 60\;\text{g mol}^{-1}$

Step 2: Convert grams to moles.

$\text{Moles of } CH_3COOH = \frac{2.5\;\text{g}}{60\;\text{g mol}^{-1}} = 0.0417\;\text{mol}$

Step 3: Convert the solvent mass to kg.

$\text{Mass of benzene} = \frac{75\;\text{g}}{1000\;\text{g kg}^{-1}} = 0.075\;\text{kg}$

Step 4: Apply the molality formula (Equation 1.9).

$\text{Molality} = \frac{0.0417\;\text{mol}}{0.075\;\text{kg}} = 0.556\;\text{mol kg}^{-1}$

Choosing the Right Unit: The Temperature Question

Each of these seven methods has strengths and weaknesses, but there is one practical difference that stands out: temperature dependence.

• Mass percentage, ppm, mole fraction, and molality all rely on masses (or moles, which are derived from mass). Since mass does not change when you heat or cool a substance, these four units stay constant at any temperature.
• Molarity relies on the volume of the solution. Liquids expand when heated and contract when cooled, so the volume changes with temperature. That means the numerical value of molarity shifts with temperature even though no solute or solvent has been added or removed.

This is why molality is often preferred over molarity in precise thermodynamic work: it gives the same number regardless of whether the lab is at $20\degree C$ or $35\degree C$.