Raoult's Law Calculator
Calculate the total vapor pressure and vapor-phase composition of an ideal two-component liquid mixture using Raoult's law. Enter the mole fractions and the pure vapor pressures of both components.
Vapor Pressure of an Ideal Liquid Mixture
Raoult's Law Formula
Where:
- \(p_1\), \(p_2\) — partial vapor pressures of components 1 and 2 above the mixture
- \(x_1\), \(x_2\) — mole fractions of the components in the liquid phase (\(x_1 + x_2 = 1\))
- \(P_1\degree\), \(P_2\degree\) — vapor pressures of the pure components at the mixture temperature
- \(P\) — total vapor pressure above the solution
- \(y_1\), \(y_2\) — mole fractions of the components in the vapor phase
Raoult's law states that each component's partial pressure equals its mole fraction in the liquid multiplied by the vapor pressure of the pure component at the same temperature. The total pressure is the sum of the partial pressures, and the vapor composition follows from Dalton's law.
Raoult's Law – Calculation Example
Example: a liquid mixture of benzene and toluene at 25 °C contains 40 mol% benzene and 60 mol% toluene. At 25 °C, pure benzene has a vapor pressure of 95.1 mmHg and pure toluene 28.4 mmHg. Find the total vapor pressure and the composition of the vapor.
\[ p_{\text{benzene}} = x_1 P_1\degree = 0.4 \times 95.1 = 38.04 \text{ mmHg} \\ p_{\text{toluene}} = x_2 P_2\degree = 0.6 \times 28.4 = 17.04 \text{ mmHg} \\ P = 38.04 + 17.04 = 55.08 \text{ mmHg} \\ y_1 = \frac{38.04}{55.08} = 0.6905, \quad y_2 = 0.3095 \
Result: the total vapor pressure is 55.08 mmHg (7.34 kPa), and the vapor contains 69.05 mol% benzene and 30.95 mol% toluene. The vapor is richer in benzene than the liquid (40 mol%) because benzene is the more volatile component.
How to Use the Calculator
- Enter the Mole fraction of component 1 (\(x_1\)) as a number between 0 and 1. In the default example this is benzene with \(x_1 = 0.4\).
- Enter the Mole fraction of component 2 (\(x_2\)). The two fractions must sum to 1; if you enter 0.4 for component 1, use 0.6 here.
- Enter the Pure vapor pressure of component 1 (\(P_1\degree\)) and pick its unit (mmHg, kPa, atm, bar, or psi) from the dropdown.
- Enter the Pure vapor pressure of component 2 (\(P_2\degree\)) at the same temperature, with its own unit selection.
- Read the results: the partial pressures \(p_1\) and \(p_2\), the total vapor pressure \(P\) (also converted to other units), and the vapor-phase mole fractions \(y_1\) and \(y_2\). The calculation details show every step.
What Is Raoult's Law?
Raoult's law, formulated by French chemist François-Marie Raoult in 1887, describes the vapor pressure of an ideal solution. When two liquids mix to form an ideal solution, the presence of the other component dilutes each one, lowering its partial vapor pressure in direct proportion to its mole fraction in the liquid. This happens because fewer molecules of each component sit at the surface of the liquid, so fewer escape into the vapor phase per unit time.
The law is the foundation of much of vapor–liquid equilibrium chemistry. It explains why adding a solute lowers the vapor pressure of a solvent — the basis of boiling point elevation and freezing point depression — and it underlies the design of distillation columns, where mixtures are separated by repeatedly vaporizing and condensing them.
A key insight from the results is the difference between liquid and vapor composition. The vapor above a mixture is always richer in the more volatile component (the one with the higher pure vapor pressure). In the benzene–toluene example, the liquid is 40% benzene but the vapor is about 69% benzene. This enrichment is exactly what makes distillation work: each vaporization step concentrates the more volatile component further.
Note that the pure vapor pressures themselves depend strongly on temperature. If your mixture is not at 25 °C, look up (or estimate with the Clausius–Clapeyron equation) the vapor pressures of each pure component at your actual temperature before using this calculator.
Assumptions, Limitations and Tips
- Ideal solution: Raoult's law assumes the two components interact with each other exactly as strongly as with themselves. It works best for chemically similar liquids, such as benzene/toluene, hexane/heptane, or close isomers.
- Positive deviation: if unlike molecules attract each other less than like molecules (e.g. ethanol/water), the actual total pressure is higher than Raoult's law predicts, and the mixture may form a minimum-boiling azeotrope.
- Negative deviation: if unlike molecules attract each other more strongly (e.g. acetone/chloroform), the actual pressure is lower than predicted, possibly giving a maximum-boiling azeotrope.
- Same temperature: both pure vapor pressures must be taken at the temperature of the mixture. Mixing values measured at different temperatures gives meaningless results.
- Nonvolatile solutes: if component 2 is nonvolatile (\(P_2\degree \approx 0\)), the total pressure reduces to \(P = x_1 P_1\degree\) — the dilution form used in colligative property calculations.
- Extension to multicomponent mixtures: for more than two components, simply sum \(x_i P_i\degree\) over all of them; the two-component form here is the most common case.
For related equilibrium calculations, see the partition coefficient calculator or the equilibrium constant calculator.
Frequently Asked Questions
How do I calculate Raoult's law by hand without a calculator?
Multiply each component's mole fraction in the liquid by its pure vapor pressure to get partial pressures, then add them: \(P = x_1 P_1^\degree + x_2 P_2^\degree\). For the vapor composition, divide each partial pressure by the total: \(y_1 = p_1/P\). A quick check: if \(x_1 = 0.5\), the total pressure is simply the average of the two pure vapor pressures.
How do I convert mole fractions from a mass (weight) fraction mixture?
Convert each mass fraction to moles first: divide the mass of each component by its molar mass (g/mol), then divide each by the total moles. Example: 50 g of benzene (M = 78.11) is 0.640 mol; 50 g of toluene (M = 92.14) is 0.543 mol, so \(x_{benzene} = 0.640/(0.640+0.543) = 0.541\). Raoult's law always requires mole fractions, not mass fractions — a very common mistake.
How do I convert vapor pressure between mmHg, kPa, atm, and psi?
Use these fixed relations: 1 atm = 101.325 kPa = 760 mmHg = 1.01325 bar = 14.696 psi; 1 mmHg ≈ 0.1333 kPa. So 95.1 mmHg of benzene equals 12.68 kPa or 0.125 atm. The calculator converts the result automatically, but both pure vapor pressures must still be looked up at the same temperature before you enter them.
What if my mixture is not at 25 °C — how do I get vapor pressures at another temperature?
Vapor pressure rises steeply with temperature, so you must use values at your actual temperature. Look up tabulated values (e.g., Antoine equation constants) or estimate them with the Clausius–Clapeyron equation: \(\ln(P_2/P_1) = -\Delta H_{vap}/R \cdot (1/T_2 - 1/T_1)\), using one known vapor pressure and the enthalpy of vaporization. Using 25 °C values for a hot mixture gives seriously wrong results.
When is Raoult's law accurate, and when does it fail?
It works well for chemically similar liquids — benzene/toluene, hexane/heptane, close isomers — where unlike and like molecular attractions are nearly equal. It fails for mixtures with hydrogen bonding or strong specific interactions: ethanol/water shows positive deviation (real pressure higher than predicted), while acetone/chloroform shows negative deviation (lower). Strongly deviating mixtures can form azeotropes, which Raoult's law cannot predict.
What is the difference between Raoult's law and Henry's law?
Both give partial pressure as a proportionality constant times concentration, but they apply at opposite ends. Raoult's law uses the pure-component vapor pressure as the constant and is most accurate for the component in large excess (the solvent). Henry's law uses an empirical constant and applies to dilute solutes, especially gases dissolved in liquids. For a two-liquid ideal mixture, Raoult's law applies to both components across the whole composition range.
Why is the vapor richer in the more volatile component than the liquid?
The more volatile component has the higher pure vapor pressure, so it contributes disproportionately to the gas phase. Its vapor mole fraction is \(y_1 = x_1 P_1^\degree/P\), which exceeds \(x_1\) whenever \(P_1^\degree > P\). This enrichment is the physical basis of distillation: each vaporization–condensation cycle pushes the vapor composition further toward the more volatile component.