6.1 Convert between mole and mass fractions | University of Alberta MEC E 340: Applied Thermodynamics

University of Alberta MEC E 340: Applied Thermodynamics

Gas Mixtures and Real Gases

MEC E 340 Applied Thermodynamics — Study topic 6.1

Consider a gas mixture in a control volume at state 1. Its components are numbered by species, such as species 1 and species 2; these numbers identify components, not successive thermodynamic states. The mixture may enter or leave a device, but this topic concerns how its composition is reported, not the device’s energy performance. Assume the composition describes the same mixture sample and that each component’s molar mass is known. No property tables or energy balances are needed for this conversion. Mole fraction counts each species by amount of substance; mass fraction counts it by mass. The two descriptions are related through molar mass, so they are generally not numerically equal.

What you will learn

Distinguish mole fraction from mass fraction and identify the basis of each composition.
Convert known mole fractions to mass fractions using component molar masses.
Convert known mass fractions to mole fractions using component molar masses.
Check that calculated fractions are dimensionless, nonnegative, and sum to one.

1. Identify the composition basis

A mixture contains several chemical species. For species i, its mole fraction is the number of moles of that species divided by the total number of moles in the mixture. Its mass fraction is the mass of that species divided by the total mixture mass. Both are dimensionless.

Use n_i for the amount of species i, m_i for its mass, and M_i for its molar mass. Since m_i=n_iM_i, a species with a larger molar mass contributes more mass per mole. That is why a mixture’s mass fractions can differ substantially from its mole fractions.

Fractions refer to the same mixture sample. On either basis, all species must be included in the denominator. If the composition is given in percent, divide each percentage by 100 before using it as a fraction.

x_i=\frac{n_i}{\sum_j n_j},\quad w_i=\frac{m_i}{\sum_j m_j},\quad m_i=n_iM_i

Mole fraction is based on moles; mass fraction is based on mass.
Molar mass connects the two descriptions.
The fractions for all species sum to one on either basis.

2. Convert mole fractions to mass fractions

When mole fractions are known, imagine a convenient total amount of mixture, such as one mole. Species i then contributes x_i moles and has mass x_iM_i. The total mixture mass on this basis is the sum of x_jM_j over all species. Dividing each species mass by this total gives its mass fraction.

The result is independent of the imagined total amount: choosing one mole is only a convenient calculation basis. Molar masses must use consistent units, for example kilograms per kilomole for amounts in kilomoles. Since each term in the numerator and denominator has the same units, the resulting mass fraction is dimensionless.

For a binary mixture, the second mole fraction is one minus the first. This can reduce arithmetic, but the weighted molar-mass denominator must still include both components.

w_i=\frac{x_iM_i}{\sum_j x_jM_j}

Multiply each mole fraction by its molar mass to obtain a relative mass contribution.
Normalize by the sum of all relative mass contributions.
Use the molar mass for the stated species and keep units consistent.

3. Convert mass fractions to mole fractions

When mass fractions are known, take a convenient total mixture mass, such as one kilogram. Species i then has mass w_i kilograms. Dividing by its molar mass gives its amount of substance, proportional to w_i/M_i. Normalize these amounts by their sum to obtain mole fractions.

This direction uses division by molar mass, not multiplication. For a fixed mass, a species with smaller molar mass represents more moles. The normalization step is essential: the values w_i/M_i are proportional amounts, not usually mole fractions yet.

The formulas work for any number of components. Include every species in the sum, and check the result by confirming that the mole fractions total one within rounding.

x_i=\frac{w_i/M_i}{\sum_j w_j/M_j}

Divide each mass fraction by its molar mass to get a relative mole amount.
Normalize the relative amounts by their sum.
A lower molar mass can correspond to a larger mole fraction than mass fraction.

4. Checks and interpretation

Before interpreting a converted composition, check that every fraction is between zero and one and that the fractions sum to one. Small departures from one can result from rounding; large departures usually indicate a missing species, an incorrect denominator, or a mix-up between the two conversion directions.

A useful physical check is to compare molar masses. If species i is heavier per mole than the mixture’s mole-weighted average molar mass, it will have a larger mass fraction than mole fraction. If it is lighter, its mass fraction will be smaller. This check helps catch swapped molar masses or an inverted conversion formula.

These conversions describe composition only. They do not require assumptions about pressure, temperature, phase, ideal-gas behavior, or energy transfer. Use a molar mass appropriate to each named species and the supplied composition basis.

\sum_i x_i=1,\quad \sum_i w_i=1

Check normalization and nonnegative values.
Compare each species’ molar mass with the mixture’s mole-weighted average as a plausibility check.
Do not treat mole percent and mass percent as interchangeable.

Worked example

Mole fractions of nitrogen and oxygen to mass fractions

A two-species gas mixture has mole fractions

x_{\mathrm{N_2}}=0.79

and

x_{\mathrm{O_2}}=0.21

. Use molar masses

M_{\mathrm{N_2}}=28.0134\ \mathrm{kg/kmol}

and

M_{\mathrm{O_2}}=31.998\ \mathrm{kg/kmol}

. Find the mass fractions.

Choose a calculation basis
Take one kilomole of mixture. The component amounts are therefore 0.79 kmol of nitrogen and 0.21 kmol of oxygen.
$n_{\mathrm{N_2}}=0.79\ \mathrm{kmol},\quad n_{\mathrm{O_2}}=0.21\ \mathrm{kmol}$
Find component masses
Multiply each component amount by its molar mass. The products are in kilograms because kmol multiplied by kg/kmol gives kg.
$m_{\mathrm{N_2}}=22.1306\ \mathrm{kg},\quad m_{\mathrm{O_2}}=6.71958\ \mathrm{kg}$
Normalize by total mass
The total mass is 28.85018 kg. Divide each component mass by this total and round the fractions to four decimal places.
$w_{\mathrm{N_2}}=0.7671,\quad w_{\mathrm{O_2}}=0.2329$

Answer: The mass fractions are approximately 0.7671 nitrogen and 0.2329 oxygen, or 76.71% and 23.29% by mass.

Check: The fractions sum to 1.0000. Nitrogen has a lower molar mass than oxygen, so its mass fraction is below its mole fraction, as expected.

Worked example

Component masses to mole fractions

A sample contains 2.000 kg of methane and 8.000 kg of carbon dioxide. Use

M_{\mathrm{CH_4}}=16.043\ \mathrm{kg/kmol}

and

M_{\mathrm{CO_2}}=44.01\ \mathrm{kg/kmol}

. Find the mole fractions.

Calculate component amounts
Divide each component mass by its molar mass. The resulting amounts are in kilomoles.
$n_{\mathrm{CH_4}}=0.1247\ \mathrm{kmol},\quad n_{\mathrm{CO_2}}=0.1818\ \mathrm{kmol}$
Normalize the amounts
The total amount is approximately 0.3065 kmol. Divide each amount by this total to obtain the mole fractions.
$x_{\mathrm{CH_4}}\approx0.4069,\quad x_{\mathrm{CO_2}}\approx0.5931$

Answer: The mole fractions are approximately 0.4069 methane and 0.5931 carbon dioxide.

Check: The fractions sum to 1.0000. Although methane is only 20% of the sample mass, its much smaller molar mass gives it a larger share of the total moles.

Worked example

Mass fractions to mole fractions for a binary mixture

A binary mixture has mass fractions

w_A=0.20

and

w_B=0.80

. The supplied molar masses are

M_A=20.0\ \mathrm{kg/kmol}

and

M_B=40.0\ \mathrm{kg/kmol}

. Find the mole fractions.

Choose a mixture mass
Take one kilogram of mixture. The component masses are 0.20 kg and 0.80 kg. Divide each mass by its molar mass to find the corresponding amounts.
$n_A=0.0100\ \mathrm{kmol},\quad n_B=0.0200\ \mathrm{kmol}$
Find mole fractions
The total amount is 0.0300 kmol. Normalize each component amount by this total.
$x_A=\frac{1}{3},\quad x_B=\frac{2}{3}$

Answer: The mole fractions are

x_A=0.3333

and

x_B=0.6667

.

Check: The mole fractions sum to one. Species B has twice the mass fraction of species A, but its larger molar mass means the mole-fraction ratio is also influenced by the mass-per-mole difference.

Common mistakes and how to avoid them

Using mole fractions directly as mass fractions.

Correction: Convert using each species’ molar mass, then normalize by the total weighted amount.

Multiplying mass fractions by molar mass to find relative moles.

Correction: Divide each mass fraction by its molar mass, then normalize.

Stopping at values proportional to the desired fractions.

Correction: Divide each relative contribution by the sum of all contributions so the final fractions total one.

Using inconsistent molar-mass units or omitting a species from the sum.

Correction: Use consistent units throughout and include every mixture component in the normalization.

Lesson summary

Mole fractions describe shares of total moles; mass fractions describe shares of total mass.
Convert mole fractions to mass fractions by weighting each mole fraction with its molar mass and normalizing.
Convert mass fractions to mole fractions by dividing each mass fraction by its molar mass and normalizing.
Check that fractions are nonnegative and sum to one, allowing for rounding.

Check your understanding

Question 1

A mixture contains equal moles of species A and B. If A has the greater molar mass, which species has the greater mass fraction?

A
B
They must have equal mass fractions
It cannot be determined even if both molar masses are known

Show answer and explanation

A

Equal mole amounts have masses proportional to their molar masses, so the heavier species contributes more mass.

Question 2

For a mixture with known mass fractions, which quantity is proportional to the number of moles of species i?

$w_iM_i$
$w_i/M_i$
$M_i/w_i$
$1-w_i$

Show answer and explanation

w_i/M_i

For a chosen total mixture mass, species i has mass proportional to

w_i

. Dividing by

M_i

gives an amount proportional to its moles.

Question 3

After converting mole fractions to relative mass contributions, what should be done next?

Multiply each contribution by the total number of species
Divide each contribution by the sum of all contributions
Subtract the smallest contribution from each value
Use the same numerical values as the mole fractions

Show answer and explanation

Divide each contribution by the sum of all contributions

Normalization makes the mass fractions sum to one.

Key terms

Mole fraction: The amount of a component divided by the total amount of mixture, on a mole basis.
Mass fraction: The mass of a component divided by the total mass of the mixture.
Molar mass: The mass per amount of substance for a species, commonly expressed in kg/kmol.
Normalize: Divide each relative component value by the sum of all such values so the resulting fractions total one.

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6.1 · Convert between mole and mass fractions

University of Alberta MEC E 340: Applied Thermodynamics

MEC E 340 Applied Thermodynamics — Study topic 6.1

What you will learn

1. Identify the composition basis

2. Convert mole fractions to mass fractions

3. Convert mass fractions to mole fractions

4. Checks and interpretation

Mole fractions of nitrogen and oxygen to mass fractions

Component masses to mole fractions

Mass fractions to mole fractions for a binary mixture

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through MEC E 340

About this lesson and its review