Mole Fraction Calculator

The mole fraction of a component $A$ is the share of the mixture's particles that are $A$:

$x_A = \frac{n_A}{n_{\text{total}}} = \frac{n_A}{n_A + n_B + n_C + \cdots}$

• $n_A$ — moles of component $A$.
• $n_{\text{total}}$ — moles of every component, solvent included.
• $x_A$ — a pure number between 0 and 1, with no units, because moles cancel.

Two consequences follow immediately. The fractions of all components sum to exactly one,

$\sum_i x_i = 1$

which gives a free check on any answer, and the relation inverts to $n_A = x_A n_{\text{total}}$.

Why it is used. Because it is a ratio of amounts rather than of volumes, mole fraction is independent of temperature — unlike molarity, which shifts as a solution expands. That makes it the natural composition variable in the laws that treat all particles alike: Dalton's law of partial pressures, $p_A = x_A P_{\text{total}}$, and Raoult's law for an ideal solution, $p_A = x_A p_A^{\circ}$. Both assume ideal behaviour: for gases, no intermolecular forces; for solutions, that A-B interactions resemble A-A and B-B interactions.

From moles

1. Add up the moles of every component to get $n_{\text{total}}$.
2. Divide each component's moles by that total.
3. Check that the fractions add to 1.000.

From masses

Masses cannot be added directly — convert each one to moles first:

$n_i = \frac{m_i}{\mathcal{M}_i}$

A 50:50 mixture by mass is nowhere near 50:50 by mole unless the two molar masses happen to match.

From a volume of solution

A volume on its own is not an amount. Convert it with the concentration, $n = cV$, or for a pure liquid with the density and molar mass, $n = \rho V / \mathcal{M}$. Only then can it enter the sum.

Back to moles

$n_A = x_A\, n_{\text{total}}$

If only $x_A$ and $n_B$ are known for a two-component mixture, use $x_B = 1 - x_A$ and $n_{\text{total}} = n_B/x_B$.

Significant figures

A mole fraction is a quotient, so it takes the fewest significant figures among the inputs. Because the fractions must sum to 1, rounding each one independently can leave a sum such as 1.001 — quote the components consistently and note the total is exact.

• Leaving the solvent out of the total. $n_{\text{total}}$ includes the solvent. Dividing solute moles by solute moles alone always gives 1.
• Adding masses instead of moles. 46.0 g of ethanol and 100.0 g of water is not a mole fraction of 0.315; converting first gives 0.152.
• Attaching units. A mole fraction is dimensionless. Writing 0.300 mol for $x$ confuses it with an amount.
• Using mole fraction where molarity is required. They answer different questions; only molarity carries a per-litre meaning.
• Forgetting that a dissolved salt splits into ions when the question asks about particles rather than formula units.
• Skipping the sum check. If the fractions do not add to 1, a component has been left out of the denominator.