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Mass Fraction

03rd Sep 2019 @ 13 min read

Analytical Chemistry

The mass fraction is a property of a mixture. A mixture is the physical combination of two or more different substances. The types of mixtures are solutions, suspensions, and colloids.

salt in water dirty water a glass of milk
Figure 1: Types of Mixtures: Solutions (salt and water), Suspensions (soil and water), and Colloids (milk) (from left to right)
[Image Sources: Wikimedia, Flickr, and Public Domain]

In chemistry, most mixtures are solutions, homogeneous mixture of a solute and a solvent. But the below definition of the mass fraction is also applicable to the other types.

Definition and Formula

The mass fraction of a substance in a mixture is the ratio of the mass of the substance to the total mass of the mixture. It is also known as mass per cent or percentage by mass when expressing in percentage. Since the mass fraction is a ratio of mass to mass, it is a dimensionless quantity.

For a mixture,


Here, wi is the mass fraction of the ith component, mi is the mass of the ith component, and mT is the total mass of the mixture.

The total mass of a mixture is the sum of the mass of each component.

m_\text{T} &=m_1+m_2+m_3+\ldots\\ &=\sum m_i

Using the above two equations,

w_i=\frac{m_i}{\sum m_i}

Note: The sum of all mass fractions is equal to unity.

w_1+w_2+w_3+\ldots=\sum w_i=1

Example 1: Wine

Statement: A glass of wine contains 24 g of alcohol and the remaining is water. The mass of water in the wine is 211 g. Calculate the mass fraction of the alcohol and the water?

a glass of wine
Figure 2: A Glass of Wine
[Image Source: PixaBay]

Solution: The mass of the water is 211 g and the alcohol, 24 g.

m_\text{water} =211 \,\text{g} && m_\text{alcohol} =24 \,\text{g}

The mass fraction of the water and the alcohol be wwater and walcohol.

For water,

w_\text{water} &=\frac{m_\text{water}}{m_\text{water}+m_\text{alcohol}}\\ &=\frac{24}{24+211}\\ &=\frac{24}{235}\\ &=0.102

For alcohol,

w_\text{alcohol} &=\frac{m_\text{alcohol}}{m_\text{water}+m_\text{alcohol}}\\ &=\frac{211}{24+211}\\ &=\frac{211}{235}\\ &=0.898

Note: The mass fraction of a substance is independent of pressure, temperature, and location; it is an intensive quantity. It remains constant unless there is a chemical change or a net physical transfer of any substance. So, the mass percentages of the wine (in the above example) is constant throughout the world.

The mass fraction of a solute in a solution is defined as the mass of the solute to the mass of the solution.

Let msolute be the mass of a solute and msolution, the mass of the solution. The mass fraction of the solute is as follows:


Example 2: Sodium Chloride Solution

Statement: 23 g of sodium chloride is dissolved in 125 g of water. Find the mass fraction of the solute.

a beaker of sodium chloride solution
Figure 3: A Beaker of Sodium Chloride Solution
[Image Source: University of California, Los Angeles]

Solution: The mass of sodium chloride (the solute) is 23 g. And the mass of water (the solvent) is 125 g.

m_\text{NaCl} =23 \,\text{g} && m_{\text{H}_2\text{O}} =125 \,\text{g}

The mass fraction of NaCl is wNaCl.

w_\text{NaCl} &=\frac{m_\text{NaCl}}{m_\text{NaCl}+m_{\text{H}_2\text{O}}}\\ &=\frac{23}{23+125}\\ &=\frac{23}{148}\\ &=0.155

Mass Percentage

The mass fraction is also known as the mass percentage, mass per cent, percentage by mass, or percentage by weight when expressed in the percentage. It is abbreviated mass % or wt %.

\text{Mass Percentage} &=\frac{m_i}{m_\text{T}} \times 100 \\ &=\frac{m_i}{\sum m_i} \times 100

Example 3: Steel

Statement: A high carbon steel block contains 104 g of carbon and 5100 g of iron. Calculate the mass fraction of each.

a high carbon steel knife
Figure 4: A High Carbon Steel Knife
[Image Source: Mercer kitchen knife]

Solution: The mass carbon is 104 g of carbon and iron, 5100 g.

m_\text{C} =104 \,\text{g} && m_\text{Fe} =510\,0 \,\text{g}

Now, the weight percentage of carbon and iron is calculated as:

\text{wt \% of carbon} &=\frac{m_\text{C}}{m_\text{C}+m_\text{Fe}} \times 100\\ &=\frac{104}{104+510\,0} \times 100\\ &=\frac{104}{520\,4} \times 100\\ &\approx 2.00 \,\% \text{wt \% of iron} &=\frac{m_\text{Fe}}{m_\text{C}+m_\text{Fe}} \times 100\\ &=\frac{510\,0}{104+510\,0} \times 100\\ &=\frac{510\,0}{520\,4} \times 100\\ & \approx 98.00 \,\%

For a solution,

\text{Mass Percentage} =\frac{m_\text{solute}}{m_\text{solution}}

Example 4: Manganese (II) Sulphate

Statement: A flask of manganese (II) sulphate solution contains 86 g of MnSO4 and the rest is H2O. The amount of the water is 74 g.

manganese (II) sulphate flask
Figure 5: Manganese (II) Sulphate
[Image Source: Wikimedia]

Solution: The mass of the manganese sulphate is 86 g and the water, 74 g.

m_{\text{MnSO}_4} =86 \,\text{g} && m_{\text{H}_2\text{O}} =74 \,\text{g}

The weight percentage of MnSO4 and H2O are calculated as follows:

\text{wt \% of manganese sulphate} &=\frac{m_{\text{MnSO}_4}}{m_{\text{H}_2\text{O}}+m_{\text{MnSO}_4}} \times 100\\ &=\frac{86}{74+86} \times 100\\ &=\frac{86}{160} \times 100\\ & \approx 54 \,\% \text{wt \% of water} &=\frac{m_{\text{H}_2\text{O}}}{m_{\text{H}_2\text{O}}+m_{\text{MnSO}_4}} times 100\\ &=\frac{74}{74+86} \times 100\\ &=\frac{74}{160} \times 100\\ & \approx 46 \,\%

Relationship with Other Quantities

The mass fraction is one of a quantity used for expressing the composition of a mixture. But there are also other quantities like mole fraction, molarity, mass concentration, among others. Some of them are discussed below.

Mixing Ratio

The mixing ratio of a component is the abundance of that component of a mixture relative to other components. Let m1 be the mass of pure substance 1 and m2, of pure substance 2. The mixture ratio is rm.

r_\text{m} =\frac{m_2}{m_1}

Now, we can express w1 and w2 in terms of rm.

w_1 &=\frac{m_1}{m_1+m_2} \\ &=\frac{1}{1+m_2 \big/ m_1}\\ &=\frac{1}{1+r_\text{m}} w_2 &=\frac{m_2}{m_1+m_2} \\ &=\frac{m_2 \big/ m_1}{m_2 \big/ m_1+1}\\ &=\frac{r_\text{m}}{1+r_\text{m}}

Mole Fraction

The mole fraction is widely used quantity in chemistry and engineering. It is also a dimensionless quantity and is the ratio of the mole of a component to the total mole of the mixture. It is usually denoted by xi.

Let ni be the mole of the ith component of a mixture. The total mole in the mixture is \sum n_i. The mole fraction of the ith component is xi.

x_i =\frac{n_i}{\sum n_i}

We also know, the mole of a substance is a ratio of the mass (mi) to the molar mass (Mi).

n_i =\frac{m_i}{M_i}

Using the above two equations,

x_i &=\frac{\frac{m_i}{M_i}}{\sum \frac{m_i}{M_i}}\\ &=\frac{m_i}{M_i} \times \frac{1}{\sum \frac{m_i}{M_i}}\\ &=\frac{m_i}{\sum m_i} \times \frac{\sum m_i}{M_i} \times \frac{1}{\sum \frac{m_i}{M_i}}\\ &=w_i \times \frac{1}{M_i} \times \frac{\sum m_i}{\sum \frac{m_i}{M_i}}\\ &=\frac{w_i}{M_i} \times M

Here, M=\frac{\sum m_i}{\sum \frac{m_i}{M_i}} is the average molar mass of the mixture. But we also know that M=\sum \frac{M_i}{w_i}.

Substituting M=\sum \frac{M_i}{w_i},

x_i =\frac{w_i}{M_i} \frac{1}{\sum \frac{M_i}{w_i}}

Thus, we have established the relationship between the mole fraction and the mass fraction.

Volume Fraction

The volume fraction (vi) is the volume of a component (Vi) to the total volume of the mixture (Vmix).

v_i =\frac{V_i}{V_\text{mix}}

Volume is mass divided by density.

v_i &=\frac{m_i \big/ \rho_i}{m_\text{mix} \big/ \rho_\text{mix}} \\ &=\frac{m_i}{m_\text{mix}} \times \frac{\rho_\text{mix}}{\rho_i} \\ &=\frac{w_i \rho_\text{mix}}{\rho_i}

Mass Concentration

The mass concentration of a solute (ρi) is the ratio of the mass of a solute (mi) to the volume of the solution (V).

\rho_i &=\frac{m_i}{V}\\ &=\frac{m_i}{\sum m_i} \times \frac{\sum m_i}{V}\\ &=w_i \rho

Here, \frac{\sum m_i}{V} is the density of solution (ρ).

From the above equation, the mass concentration of a solute is the mass fraction of the solute times the density of the solution.

Molarity or Molar Concentration

The molarity or molar concentration (Ci) is the mole of the solute (ni) per unit volume of the solution (V). The volume of solution is typically expressed in decimetre cube (dm3).

C_i =\frac{n_i}{V}

The mole of a substance is the mass divided by the molar mass. And density is mass divided by volume.

C_i =\frac{m_i \big/ M_i}{{\sum m} \big/ \rho} =w_i \frac{\rho}{M_i}

Percentage Composition

A compound consists of different elements. When the mass fraction of an element is calculated, it is called the percentage composition of that element. In short, the mass fraction and the percentage composition are the same when it comes to elements of a compound.

Solved Problems

Problem 1: Air

Statement: Air contains 21 % of oxygen and 79 % of nitrogen. The molar mass of oxygen and nitrogen is 15.999 g mol−1 and 14.007 g mol−1.

Solution: The above percentages are not in mass but in the mole. So, 100 mol of air contains 21 mol of oxygen and 79 mol of nitrogen.

Converting the mole to the gram,

m_{\text{O}_2} =21 \times 2 \times 15.999 =672 \,\text{g} m_{\text{N}_2} =79 \times 2 \times 14.007 =221\,0 \,\text{g}

Let wO2 and wN2 be the mass fraction of oxygen and nitrogen respectively.

w_{\text{O}_2} &=\frac{m_{\text{O}_2}}{m_{\text{O}_2}+m_{\text{N}_2}} \\ &=\frac{672}{672+221\,0} \\ & \approx 0.23 w_{\text{N}_2} &=\frac{m_{\text{N}_2}}{m_{\text{O}_2}+m_{\text{N}_2}} \\ &=\frac{2210}{672+221\,0} \\ & \approx 0.77

Thus, the mass percentage of oxygen and nitrogen is 23 % and 77 %.

Problem 2: Gas Pipe

Statement: A gas pipe contains 2 mol of carbon dioxide, 10 mol of oxygen, 52 mol of nitrogen, and 1.3 mol of methane. The molar mass of hydrogen, carbon, nitrogen, and oxygen is 1.008 g mol−1, 12.011 g mol−1, 14.007 g mol−1, and 15.999 g mol−1.

Solution: The molar mass of carbon dioxide, oxygen, nitrogen, and methane be MCO2, MO2, MN2, and MCH4.

M_{\text{CO}_2} &=12.011+2 \times 15.999\\ &=44.009 \,\text{g}\,\text{mol}^{-1} M_{\text{O}_2} &=2 \times 15.999\\ &=31.998 \,\text{g}\,\text{mol}^{-1} M_{\text{N}_2} &=2 \times 14.007\\ &=28.014 \,\text{g}\,\text{mol}^{-1} M_{\text{CH}_4} &=12.011+4 \times 1.008\\ &=16.043 \,\text{g}\,\text{mol}^{-1}

Converting the mole to the gram,

m_{\text{CO}_2} &= 44.009 \,\text{g}\,\text{mol}^{-1} \times 2\,\text{mol} \\ &=88.02 \,\text{g} m_{\text{O}_2} &= 31.998 \,\text{g}\,\text{mol}^{-1} \times 10\,\text{mol} \\ &=319.98 \,\text{g} m_{\text{N}_2} &= 28.014 \,\text{g}\,\text{mol}^{-1} \times 52\,\text{mol} \\ &=1\,456.73 \,\text{g} m_{\text{CH}_4} &= 16.043 \,\text{g}\,\text{mol}^{-1} \times 1.3\,\text{mol} \\ &=20.86 \,\text{g}

The mass fraction is calculated as:

w_{\text{CO}_2} &=\frac{m_{\text{CO}_2}}{m_{\text{CO}_2}+m_{\text{O}_2}+m_{\text{N}_2}+m_{\text{CH}_4}}\\ &=\frac{88.02}{88.02+319.98+1\,456.73+20.86}\\ &=0.05 w_{\text{O}_2} &=\frac{m_{\text{O}_2}}{m_{\text{CO}_2}+m_{\text{O}_2}+m_{\text{N}_2}+m_{\text{CH}_4}}\\ &=\frac{319.98}{88.02+319.98+1\,456.73+20.86}\\ &=0.17 w_{\text{N}_2} &=\frac{m_{\text{N}_2}}{m_{\text{CO}_2}+m_{\text{O}_2}+m_{\text{N}_2}+m_{\text{CH}_4}}\\ &=\frac{1\,456.73}{88.02+319.98+1\,456.73+20.86}\\ &=0.77 w_{\text{CH}_4} &=\frac{m_{\text{CH}_4}}{m_{\text{CO}_2}+m_{\text{O}_2}+m_{\text{N}_2}+m_{\text{CH}_4}}\\ &=\frac{20.86}{88.02+319.98+1\,456.73+20.86}\\ &=0.01

Problem 3: Sulphuric Acid

Statement: The molar concentration of sulphuric acid is 2.0 M or 2 mol dm−3. The molar mass of sulphuric acid is 98.08 g mol−1. The density of the solution is 1.19 g cm−3.

Solution: Consider 1 L (or 1 dm−3) of the solution.

The mole of sulphuric acid in 1 L of the solution is

2.0\,\frac{\text{mol}}{\text{L}} \times 1 \,\text{L} =2.0 \,\text{mol}

Converting the mole to the gram,

2.0 \,\text{mol} \times 98.08 \,\text{g}\,\text{mol}^{-1} =196.16\,\text{g}

Thus, 1 L of the solution contains 196.16 g of sulphuric acid.

The density of the solution is 1.19 g cm−3. For 1 L of the solution, the mass of the solution is

1.19\,\frac{\text{g}}{\text{cm}^{3}} \times 1000 \,\frac{\text{cm}^3}{1\,\text{L}} \times 1\,\text{L} =1190\,\text{g}

The mass fraction of sulphuric acid in 119 0 g of the solution is

m_{\text{H}_2\text{SO}_4} &=\frac{m_{\text{H}_2\text{SO}_4}}{m_\text{solution}}\\ &=\frac{196.16}{1190}\\ &=0.16

Practice Problems

Find the mass fraction in the following problems.

Problem 1: Isopropyl Alcohol

An isopropyl alcohol solution contains 40 g of isopropyl alcohol and 20 g of water. The molar mass of isopropyl alcohol  and water is 60 g mol−1 and 18 g mol−1.

Problem 2: Sodium Hydroxide

The molarity of a sodium hydroxide solution is 0.51 M. The molar mass of sodium hydroxide is 40 g mol−1. The density of the solution is 1.02 g cm−3.

Problem 3: Exhaust from Chimney

Exhaust from a chimney contains 10 mol of oxygen (O2), 53 mol of nitrogen (N2), and 37 mol of carbon dioxide (CO2). The molar mass of oxygen, nitrogen, and carbon dioxide is 32 g mol−1, 28 g mol−1, and 44 g mol−1.

  1. 0.67 of isopropyl alcohol and 0.33 of water.
  2. 0.020 of sodium hydroxide.
  3. 0.09 of oxygen, 0.43 of nitrogen, and 0.48 of carbon dioxide.


  • The mass fraction is the ratio of the mass of a component in the mixture to the total mass of the mixture.
  • w_i =\frac{m_i}{m_\text{T}} =\frac{m_i}{\sum m_i}
  • For a solution, it is the ratio of the mass of a solute to the mass of the solution.
  • w_i=\frac{m_\text{solute}}{m_\text{solution}}
  • The sum of the mass fraction of each component is always equal to one.
  • \sum{w_i} =1
  • It is an intensive property. Thus, it would not change with the quantity of a sample. Also, it is independent of temperature, pressure, and location.

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Dionice Joshua
07th Dec 2020
thank you so very much for your clear and understandable clarifications
23rd Nov 2020
Please for the examples you gave for finding the mass percentage, the question didn't ask us to find the mass percentage because you did so I wanted to know how to determine if you're supposed to find the mass percentage.

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