Molality and Mole Fraction

General Chemistry · Expressions of Concentration

Dr. Ray’s Learning Center · covers: Molality and Mole Fraction

25 to 35 min
7 figures
1 simulator
4 worked examples
2 practice types: molality, mole fraction

Before this lesson:

The Mole and Molar Mass

Molarity counts the moles of solute in each liter of solution. A liter of solution is not a fixed amount of matter, though: warm the solution and it expands, so the same solute is spread through a larger volume and the molarity drops a little. Freezing points and boiling points are measured at changing temperatures, so chemists need concentrations that do not depend on temperature.

This lesson introduces two of them, both built only from masses and moles. The molality divides the moles of solute by the mass of the solvent in kg. The mole fraction gives the share of all the moles that belongs to one component (Figure 1).

Massesof solute and solvent, in g
divide by Meach component
Molesof each component
divideby kg of solvent, or by total moles
Molality or mole fractionmol/kg, or no unit

Molality: moles of solute over kilograms of solvent.Mole fraction: moles of one component over the total moles.

Figure 1. From masses to moles, then one division: by the kilograms of solvent for a molality, by the total moles for a mole fraction.

Learning Objectives

After this lesson, you should be able to:

Calculate molality (mol solute per kg solvent).
Calculate mole fractions of the components of a mixture.

Key Ideas

1. Molality: Moles of Solute for Each Kilogram of Solvent

The molality of a solution is the amount of solute (in mol) divided by the mass of the solvent (in kg):

m = \dfrac{n_{\text{solute}}}{m_{\text{solvent}}}

Its unit is mol/kg. A solution with 1.00 mol of solute for every 1 kg of solvent is “1.00 molal”; many textbooks write this as 1.00 m. Double the solvent and double the solute, and the molality stays the same (Figure 2).

Notation: this lesson follows Whitten and writes molality as an italic $m$ . A mass always carries a subscript, $m_{\text{solute}}$ or $m_{\text{solvent}}$ , so the two never get mixed up. (IUPAC writes molality as $b$ .)

Figure 2. Twice the solvent with twice the solute: the moles of solute for each kilogram of solvent stay the same, so both samples have a molality of 1.00 mol/kg.

Key idea: molality divides by the mass of the solvent in kg, not by the volume of the solution in L, and not by the mass of the solution.

2. Molality and Molarity Compared

Molarity uses the volume of the whole solution; molality uses the mass of the solvent alone (Figure 3):

\text{molarity} = \dfrac{n_{\text{solute}}}{V_{\text{solution}}}

m = \dfrac{n_{\text{solute}}}{m_{\text{solvent}}}

You weigh the solvent on a balance; you do not need a volumetric flask.
Masses do not change with temperature, so a molality does not either. A volume, and so a molarity, changes a little when the solution warms up or cools down.
The denominator of a molality leaves the solute out: the solvent is only part of the solution.

Figure 3. Molarity and molality have the same numerator. Molarity divides by the volume of the solution, molality by the mass of the solvent.

For a dilute solution in water the two numbers are close, but they are not the same. The CRC Handbook lists a glucose solution that is 10.0% glucose by mass, at 20 °C, as 0.617 mol/kg and 0.576 mol/L. The lesson Converting Between Concentration Units shows how to get one from the other with the density of the solution.

Key idea: molarity: mol of solute and L of solution. Molality: mol of solute and kg of solvent.

3. The Mole Fraction: a Share of All the Moles

The mole fraction of a component A, written $\chi_{\text{A}}$ (the Greek letter chi), is its amount (in mol) divided by the total amount of all the components:

\chi_{\text{A}} = \dfrac{n_{\text{A}}}{n_{\text{total}}}

For a mixture of three components A, B and C, the total is the sum of all three (add one more term for each extra component):

n_{\text{total}} = n_{\text{A}} + n_{\text{B}} + n_{\text{C}}

A mole fraction has no unit, because mol cancels. It lies between 0 and 1, and the mole fractions of all the components add up to exactly 1. For a solution of ethanol in water (Figure 4):

\chi_{\mathrm{C_2H_5OH}} + \chi_{\mathrm{H_2O}} = 1

Figure 4. In Worked Example 3, ethanol has 0.111 of all the moles and water has the rest, 0.889. As shares, the mole fractions fill the whole bar.

Key idea: a mole fraction counts every component, the solvent too, in the denominator. It has no unit.

Try it: molality and mole fraction

Pick a solute and set the masses of solute and of water. Load the settings of a worked example, or open a challenge: make a 1.00 mol/kg solution, then double the water and compare. The molality halves exactly, but the mole fraction does not halve, because the water is in its denominator too.

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4. Masses to Moles First

A problem usually gives masses. Both concentrations need moles, so the first step is always the same: divide the mass of each component by its molar mass. For a molality, also convert the mass of the solvent from g to kg (Figure 5):

m_{\text{solvent}}\ \text{in kg} = m_{\text{solvent}}\ \text{in g} \times \dfrac{1\,\text{kg}}{1000\,\text{g}}

Mass of solutein g
$\times \frac{1}{M}$ molar mass of the solute
Moles of solutein mol
$\times \frac{1}{m_{\text{solvent}}}$ mass of solvent in kg
Molalityin mol/kg

Figure 5. Find the moles of solute first, then divide by the mass of the solvent in kg. Convert the mass of the solvent with 1 kg in every 1000 g.

Key idea: grams to moles for each component, grams of solvent to kilograms; only then divide.

Worked Example 1: Molality from Two Masses

Problem: A solution is made by dissolving 45.0 g of glucose (C₆H₁₂O₆) in 375.0 g of water. What is the molality of the solution?

Step 1. Convert the mass of glucose to moles.

M_{\mathrm{C_6H_{12}O_6}} = 180.16\,\text{g/mol}

n_{\mathrm{C_6H_{12}O_6}} = \dfrac{45.0\,\text{g}}{180.16\,\text{g/mol}} \approx 0.2498\,\text{mol}

Step 2. Convert the mass of the solvent, water, to kilograms.

m_{\text{solvent}} = 375.0\,\text{g} \times \dfrac{1\,\text{kg}}{1000\,\text{g}} = 0.3750\,\text{kg}

Step 3. Divide the moles of solute by the kilograms of solvent.

\begin{aligned} m &= \dfrac{0.2498\,\text{mol}}{0.3750\,\text{kg}} \\ &\approx 0.6661\,\text{mol/kg} \approx 0.666\,\text{mol/kg} \end{aligned}

Final Answer: 0.666 mol/kg

Glucose $45.0\,\text{g}\ \mathrm{C_6H_{12}O_6}$ given
$\times \frac{1\,\text{mol}}{180.16\,\text{g}}$ g to mol
Moles of glucose $0.2498\,\text{mol}$ 4 significant figures
$\times \frac{1}{0.3750\,\text{kg}}$ 375.0 g of water is 0.3750 kg
Molality $0.666\,\text{mol/kg}$ Final answer to 3 significant figures

Figure 6. Worked Example 1 one step at a time: moles of glucose, then divide by the kilograms of water.

Rounded operand: Step 3 uses 0.2498 mol, the amount of glucose rounded to 4 significant figures. The unrounded amount gives the same final answer, 0.666 mol/kg.

Your turn: molality from two masses

Same three steps as Worked Example 1: moles of solute, kilograms of water, then divide.

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Worked Example 2: Using a Molality to Find a Mass

Problem (a): What mass of potassium chloride (KCl) must be dissolved in 325 g of water to make a solution with a molality of 0.500 mol/kg?

Step 1. Convert the mass of water to kilograms, then multiply by the molality. kg cancels:

m_{\text{solvent}} = 325\,\text{g} \times \dfrac{1\,\text{kg}}{1000\,\text{g}} = 0.325\,\text{kg}

n_{\mathrm{KCl}} = 0.500\,\dfrac{\text{mol}}{\text{kg}} \times 0.325\,\text{kg} = 0.1625\,\text{mol}

Step 2. Convert the moles of KCl to grams.

\begin{aligned} m_{\mathrm{KCl}} &= 0.1625\,\text{mol} \times 74.55\,\text{g/mol} \\ &\approx 12.11\,\text{g} \approx 12.1\,\text{g} \end{aligned}

Final Answer: 12.1 g KCl

Problem (b): What mass of water is needed to dissolve 10.0 g of KCl and make a solution with a molality of 0.750 mol/kg?

Step 1. Convert the mass of KCl to moles.

n_{\mathrm{KCl}} = \dfrac{10.0\,\text{g}}{74.55\,\text{g/mol}} \approx 0.1341\,\text{mol}

Step 2. Rearrange the molality equation for the mass of solvent, then convert kg to g.

m_{\text{solvent}} = \dfrac{n_{\text{solute}}}{m}

\begin{aligned} m_{\text{solvent}} &= \dfrac{0.1341\,\text{mol}}{0.750\,\text{mol/kg}} \\ &= 0.1788\,\text{kg} \end{aligned}

\begin{aligned} m_{\text{solvent}} &= 0.1788\,\text{kg} \times \dfrac{1000\,\text{g}}{1\,\text{kg}} \\ &= 178.8\,\text{g} \approx 179\,\text{g} \end{aligned}

Final Answer: 179 g of water

Rounded operands: part (b) carries 0.1341 mol and 0.1788 kg, each rounded to 4 significant figures. The unrounded values give the same final answer, 179 g. Part (a) has no rounded operand: 0.1625 mol is exact.

Your turn: using a molality to find a mass

Same steps as Worked Example 2: kilograms of water times the molality gives moles of solute; moles of solute divided by the molality gives kilograms of water.

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Worked Example 3: Mole Fraction in a Two-Component Solution

Problem: A solution contains 23.0 g of ethanol (C₂H₅OH) dissolved in 72.0 g of water. What is the mole fraction of ethanol in the solution?

Step 1. Convert both masses to moles.

M_{\mathrm{C_2H_5OH}} = 46.07\,\text{g/mol} \text{ and } M_{\mathrm{H_2O}} = 18.02\,\text{g/mol}

n_{\mathrm{C_2H_5OH}} = \dfrac{23.0\,\text{g}}{46.07\,\text{g/mol}} \approx 0.4992\,\text{mol}

n_{\mathrm{H_2O}} = \dfrac{72.0\,\text{g}}{18.02\,\text{g/mol}} \approx 3.996\,\text{mol}

Step 2. Add the moles of all the components.

n_{\text{total}} = 0.4992\,\text{mol} + 3.996\,\text{mol} = 4.4952\,\text{mol}

Step 3. Divide the moles of ethanol by the total. mol cancels, so there is no unit:

\begin{aligned} \chi_{\mathrm{C_2H_5OH}} &= \dfrac{0.4992\,\text{mol}}{4.4952\,\text{mol}} \\ &\approx 0.1111 \approx 0.111 \end{aligned}

Final Answer: 0.111

Check: $\chi_{\mathrm{H_2O}} = \dfrac{3.996\,\text{mol}}{4.4952\,\text{mol}} \approx 0.8889$ , and $0.1111 + 0.8889 = 1$ , as it must be.

Rounded operands: the steps carry 0.4992 mol and 3.996 mol, each rounded to 4 significant figures. The unrounded amounts give the same final answer, 0.111.

Your turn: mole fraction in a two-component solution

Same three steps as Worked Example 3: moles of each component, the total, then divide.

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Worked Example 4: Mole Fractions in a Three-Component Mixture

Problem: A mixture contains 16.0 g of methanol (CH₃OH), 46.0 g of glycerol (C₃H₈O₃) and 90.0 g of water. What is the mole fraction of water in the mixture?

Step 1. Convert each mass to moles.

M_{\mathrm{CH_3OH}} = 32.04\,\text{g/mol},\ M_{\mathrm{C_3H_8O_3}} = 92.09\,\text{g/mol}

n_{\mathrm{CH_3OH}} = \dfrac{16.0\,\text{g}}{32.04\,\text{g/mol}} \approx 0.4994\,\text{mol}

n_{\mathrm{C_3H_8O_3}} = \dfrac{46.0\,\text{g}}{92.09\,\text{g/mol}} \approx 0.4995\,\text{mol}

n_{\mathrm{H_2O}} = \dfrac{90.0\,\text{g}}{18.02\,\text{g/mol}} \approx 4.994\,\text{mol}

Step 2. Add the moles of all three components.

\begin{aligned} n_{\text{total}} &= 0.4994\,\text{mol} + 0.4995\,\text{mol} \\ &\;\; + 4.994\,\text{mol} \\ &= 5.9929\,\text{mol} \end{aligned}

Step 3. Divide the moles of water by the total.

\begin{aligned} \chi_{\mathrm{H_2O}} &= \dfrac{4.994\,\text{mol}}{5.9929\,\text{mol}} \\ &\approx 0.8333 \approx 0.833 \end{aligned}

Final Answer: 0.833

Figure 7. In Worked Example 4, water has 0.833 of all the moles. Methanol and glycerol have about one twelfth each.

Check: the other two mole fractions are $\chi_{\mathrm{CH_3OH}} = 0.08333$ and $\chi_{\mathrm{C_3H_8O_3}} = 0.08335$ . The three add up to 0.99998, not exactly 1, only because each one was rounded. Subtracting the other two from 1 gives 0.83332, the same final answer.

Rounded operands: the steps carry 0.4994, 0.4995 and 4.994 mol, each rounded to 4 significant figures. The unrounded amounts give the same final answer, 0.833.

Your turn: mole fractions in a three-component mixture

Same three steps as Worked Example 4: every component goes into the total.

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Common Mistakes

Common mistake: Dividing by the mass of the solution. In Worked Example 1, dividing by 45.0 g + 375.0 g, that is 0.4200 kg of solution, gives 0.595 mol/kg. Molality uses the solvent alone: 0.3750 kg of water.

Common mistake: Leaving the mass of solvent in g. Dividing 0.2498 mol by 375.0 instead of 0.3750 gives 0.000666 mol/kg, a thousand times too small. Convert g to kg first.

Common mistake: Using the mass of solute instead of its moles. 45.0 g divided by 0.3750 kg is 120 g/kg, which is not a molality. Divide the mass by the molar mass first.

Common mistake: Dividing by the other component instead of the total. In Worked Example 3, 0.4992 mol of ethanol divided by 3.996 mol of water alone gives 0.125. That is a mole ratio; the mole fraction divides by the total, 4.4952 mol.

Common mistake: Using masses instead of moles. 23.0 g of ethanol in 95.0 g of solution gives 0.242, the mass fraction. A mole fraction needs moles of every component.

Common mistake: Leaving a component out of the total. In Worked Example 4, leaving out the glycerol gives a mole fraction of water of 0.909. Every component goes into the total.

Exam tip: a mole fraction has no unit, lies between 0 and 1, and all the mole fractions of a mixture add up to 1. A molality is in mol/kg: check that you divided by the solvent alone, in kg.

Key Takeaway

Molality and mole fraction are built from masses and moles only, so they do not change with temperature. Convert every mass to moles first. Molality divides the moles of solute by the kilograms of solvent and has the unit mol/kg. A mole fraction divides the moles of one component by the total moles of all the components; it has no unit, and all the mole fractions add up to 1.

The relations to keep at hand:

m = \dfrac{n_{\text{solute}}}{m_{\text{solvent}}}

\chi_{\text{A}} = \dfrac{n_{\text{A}}}{n_{\text{total}}}

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.

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Your results

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

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Figures and simulator: original work by Dr. Ray’s Learning Center, licensed CC BY 4.0. Atomic weights: IUPAC Commission on Isotopic Abundances and Atomic Weights, standard atomic weights 2021 (abridged values), the same table as the other stoichiometry lessons. Glucose solution data in the simulator: CRC Handbook of Chemistry and Physics, 97th ed. (2016), Section 5, Concentrative Properties of Aqueous Solutions. Molality and mole fraction definitions: K. Whitten, R. Davis, M. Peck and G. Stanley, General Chemistry, 9th ed., section 14-8, and J. Moore, C. Stanitski and P. Jurs, Chemistry: The Molecular Science, 5th ed., section 13-6. The simulator and the practice problems are audited; see the lesson audit log.