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Molar fraction: how it is calculated, examples, exercises

A molar fraction is a way of expressing the concentration of elements present in a compound or the concentration of a compound in a mixture.

The molar fraction of the elements of a compound is defined as a quotient between the number of moles of each of the different elements present in the compound and the total number of moles of them.

For example, if a compound has an element A and an element B, the molar fraction of A is the number of moles of A divided by the number of moles of A plus the number of moles of B. Likewise, for the molar fraction of B the same operation is performed, but placing the moles of B in the numerator.

This operation is represented in the image above. The sum of the molar fractions is equal to 1 (one). The molar fraction is a dimensionless (dimensionless) number. Many laws can be expressed in terms of them, such as Dalton’s Law.

Features and Symbol

The molar fraction value is independent of temperature and, in an ideal gas mixture, can be used to calculate the partial pressures of each of the gases present in the gas mixture; as stated by Dalton’s law.

The molar fraction is usually represented or symbolized with a capital X (X) on the right, as a subscript, element symbol or compound formula is placed if you have a mixture of compounds.

How is it calculated

If the number of moles of each of the elements that make up a given compound is known, adding the moles of the elements can get the total number of moles that exist in the compound.

So, to obtain the molar fraction of each element, the number of its springs is divided by the total number of springs present in the compound. The sum of the molar fraction values ​​of the different elements is equal to unity (1).

Examples

The following are examples of uses of the molar fraction.

Example 1

The molality of a solution, expressed in moles of solute per kilogram of water, can be transformed into the molar fraction of the solute. To do this, convert the 1,000 g of water into moles of water by simply dividing the mass of 1,000 g of water by the molecular weight of the water (18 g/mol).

Then, dividing the number of moles of the solute by the number of moles of water plus those of the solute, the molar fraction of the solute will be obtained.

For example, substance A has a molality of 0.03 m. That means you have 0.3 moles of A dissolved in one kilogram of water. One kilogram of water corresponds to 55.55 moles of water (1,000 g ± 18 g / mol). Thus, the molar fraction of A becomes:

X (A) or X A = 0.03 ÷ (55.55 + 0.03)

= 0.0005398 or 5.39810 -4

Example 2

Calculation of partial pressures of gases based on their molar fractions. The Law of Partial Pressures was enunciated by Dalton and points out that in a mixture of gases each gas exerts its pressure as if it occupied the entire volume of the mixture of gases.

The total pressure of the gas mixture is the sum of the pressures exerted, separately, by each one of the gases that are part of the gas mixture.

The atmosphere is mainly composed of a mixture of four gases: nitrogen, oxygen, carbon dioxide and water vapor, each exerting the following partial pressures separately:

Nitrogen: 596 mmHg

Oxygen: 158 mmHg

Carbon dioxide: 0.3 mmHg

Water vapor: 5.7 mmHg.

This produces an atmospheric pressure value of 760 mmHg. Using these gas pressures, the following values ​​of their molar fractions can be calculated:

Nitrogen

N2 = 596 mmHg / 760 mmHg

= 0.7842

Oxygen

O2 = 158 mmHg / 760 mmHg

  = 0.2079

Carbon dioxide

CO2 = 0.3 mmHg / 760 mmHg

= 0.00039

Steam

H2O = 5.7 mmHg / 760 mmHg

= 0.075

Conversely, the partial pressure of each of the gases present in a mixture can be calculated by multiplying the value of its molar fraction by the total pressure exerted by the gas mixture.

Exercises

Exercise 1

What is the molar fraction of a mixture of methanol (CH 3 OH) and water (H 2 O) solution containing 145 g of CH 3 OH and 120 g of H 2 O? Molecular weights: CH 3 OH = 32 g / mol and water = 18 g / mol.

First we calculate the moles of methanol and water:

Moles of CH 3 OH = 145 g · 1 mol CH 3 OH ÷ 32 g of CH 3 OH

= 4.53 mole of CH 3 OH

Moles of H 2 O = 120 g · 1 mole of H 2 O ÷ 18 g of H 2 O

= 6.67 mol of H2O

So we calculate the total moles:

Total moles of CH 3 OH and H 2 O = 4.53 + 6.67

= 11.2 moles

And so we determined the molar fractions of methanol and water:

X (CH 3 OH) = 4.53 moles / 11.2 moles

= 0.404

X (H 2 O) = 6.67 moles / 11.2 moles

= 0.596

Exercise 2

A mixture of 1.56 moles of nitrogen (N 2 ) and 1.2 moles of oxygen (O 2 ) exerts a pressure of 0.8 atmospheres (atm). Calculate the partial pressure exerted by each of the gases.

The first step in solving the problem is the calculation of the molar fractions of gases. In a second step, the partial pressures exerted by the gases are obtained, multiplying their molar fraction by the total pressure exerted by the gas mixture.

Nitrogen molar fraction:

N2 = 1.56 moles / (1.56 moles + 1.2 moles)

= 0.565

Related:   Benzaldehyde: properties, synthesis, structure and uses

Oxygen molar fraction:

O2 = 1.2 moles / (1.56 moles + 1.2 moles)

= 0.435

And finally we calculate the partial pressures of each gas:

N2 = X N2  · P T

= 0.565 · 0.8 atm

= 0.452 atm

O2 = X O2 · P t

= 0.435 · 0.8 atm

= 0.348 atm

Exercise 3

What is the molar fraction of formaldehyde (CH 2 O) if 23 g of the compound are dissolved in 4 moles of carbon tetrachloride (CCl 4 )? Molecular weight of CH 2 O = 30.03 g/mol.

We first calculate the moles of formaldehyde:

Moles CH 2 O = 23 g CH 2 O · 1 mol CH 2 O ÷ 30.03 g CH 2 S

= 0.766 moles

And for the second we calculate the molar fraction:

CH2OH = 0.766 moles CH 2 OH / (0.766 moles CH 2 OH + 4 moles CCl 4 )

= 0.161

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