Mole Fraction Calculator

Moles of component A	2 mol
Moles of component B	8 mol
Moles of component C	0 mol

Moles of component A

2 mol

Moles of component B

8 mol

Moles of component C

0 mol

Results

Mole fraction A	0
$\begin{aligned} x_A &= \dfrac{n_A}{n_A + n_B + n_C} \\ &= \dfrac{2\,\text{mol}}{?\,\text{mol}} \\ &= 0 \end{aligned}$
Mole fraction B	0
$\begin{aligned} x_B &= \dfrac{n_B}{n_A + n_B + n_C} \\ &= \dfrac{8\,\text{mol}}{?\,\text{mol}} \\ &= 0 \end{aligned}$
Mole fraction C	0
$\begin{aligned} x_C &= \dfrac{n_C}{n_A + n_B + n_C} \\ &= \dfrac{0\,\text{mol}}{?\,\text{mol}} \\ &= 0 \end{aligned}$
Mole percent A	0
Mole percent B	0
Mole percent C	0

Mole fraction A

\begin{aligned} x_A &= \dfrac{n_A}{n_A + n_B + n_C} \\ &= \dfrac{2\,\text{mol}}{?\,\text{mol}} \\ &= 0 \end{aligned}

Mole fraction B

\begin{aligned} x_B &= \dfrac{n_B}{n_A + n_B + n_C} \\ &= \dfrac{8\,\text{mol}}{?\,\text{mol}} \\ &= 0 \end{aligned}

Mole fraction C

\begin{aligned} x_C &= \dfrac{n_C}{n_A + n_B + n_C} \\ &= \dfrac{0\,\text{mol}}{?\,\text{mol}} \\ &= 0 \end{aligned}

Mole percent A

Mole percent B

Mole percent C

Last updated: 2026-07-01

Understanding mole fraction

Mole fraction expresses the composition of a mixture in terms of amount of substance. For any component, it is the number of moles of that component divided by the total number of moles of all components present. Because both the numerator and the denominator are measured in moles, the mole fraction is dimensionless, and it always falls between 0 and 1. It is the concentration measure that appears directly in Raoult's law for vapour pressure and in Dalton's law of partial pressures.

The mole fraction formula

x_i = \frac{n_i}{n_{\text{total}}} = \frac{n_i}{n_A + n_B + n_C + \cdots}

where $n_i$ is the amount of component $i$ in moles and $n_{\text{total}}$ is the sum of the moles of every component. The mole percent of the same component is simply the mole fraction times 100.

Term	Symbol	Meaning
Mole fraction	$x_i$	Share of the total amount, from 0 to 1
Component moles	$n_i$	Amount of one component
Total moles	$n_{\text{total}}$	Sum of the moles of all components

A defining property is that the mole fractions of all components add to exactly 1:

x_A + x_B + x_C + \cdots = 1

Equivalently, the mole percents sum to 100. This provides a quick consistency check on any composition calculation.

Worked example

A gas mixture contains 2 mol of oxygen and 8 mol of nitrogen. Find the mole fraction of each gas.

\begin{aligned} n_{\text{total}} &= 2 + 8 \\ &= 10\ \text{mol} \\ x_{\text{O}_2} &= \frac{n_{\text{O}_2}}{n_{\text{total}}} = \frac{2}{10} \\ &= 0.20 \\ x_{\text{N}_2} &= \frac{n_{\text{N}_2}}{n_{\text{total}}} = \frac{8}{10} \\ &= 0.80 \end{aligned}

The oxygen mole fraction is 0.20 (20 mol %) and the nitrogen mole fraction is 0.80 (80 mol %). As expected, the two fractions add to 1.

Adding a third component

Mole fraction extends to any number of components; only the total in the denominator grows. Suppose 1 mol of argon is added to the mixture above, giving 2 mol O₂, 8 mol N₂, and 1 mol Ar.

\begin{aligned} n_{\text{total}} &= 2 + 8 + 1 \\ &= 11\ \text{mol} \\ x_{\text{Ar}} &= \frac{1}{11} \\ &\approx 0.091 \end{aligned}

The oxygen fraction changes to $2 \div 11 \approx 0.182$ and the nitrogen fraction to $8 \div 11 \approx 0.727$ , and all three still sum to 1.

Mole fraction in colligative and gas laws

Mole fraction is the natural composition variable in several physical laws. In Raoult's law, the vapour pressure of a solvent above a solution equals the pure-solvent vapour pressure multiplied by the solvent's mole fraction. In Dalton's law, the partial pressure of a gas in a mixture equals its mole fraction multiplied by the total pressure. In both cases the composition must be expressed as a mole fraction rather than a mass fraction, because these laws count particles, not mass. The related Molarity Calculator, Raoult's Law Calculator, and Dalton's Law of Partial Pressures Calculator cover those applications in detail.

Converting from mass to moles

Compositions are often given as masses rather than moles. To find a mole fraction from masses, first convert each mass to moles using the molar mass $M$ :

n_i = \frac{m_i}{M_i}

then apply the mole fraction formula to those mole amounts. For example, 46 g of ethanol (molar mass 46 g/mol) is 1 mol, and 54 g of water (molar mass 18 g/mol) is 3 mol; the ethanol mole fraction is $1 \div (1 + 3) = 0.25$ .

Frequently Asked Questions (FAQ)

What is the mole fraction formula?

The mole fraction of a component is its amount in moles divided by the total moles of all components: xᵢ = nᵢ ÷ n(total). For a mixture of A and B, xₐ = nₐ ÷ (nₐ + n_b). Mole fraction is dimensionless and always lies between 0 and 1.

What is the difference between mole fraction and mole percent?

Mole fraction is a value between 0 and 1; mole percent is the same quantity multiplied by 100, so it runs from 0 to 100. A mole fraction of 0.20 is a mole percent of 20. Mole percent is often more intuitive to read, while mole fraction is the form used directly in Raoult’s law and Dalton’s law of partial pressures.

Do mole fractions always add up to 1?

Yes. Because each mole fraction is a share of the same total, the fractions of all components sum to exactly 1 (and the mole percents sum to 100). This is a useful check: if the computed fractions do not add to 1, an input amount was entered incorrectly.

How do I handle a third component?

Enter its amount in the component C field; the total moles and every mole fraction update to include it. Leaving component C at 0 gives a two-component mixture, where the third fraction is 0 and does not affect the others.

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