Question

A sample contains 18.0 g N2 (MW = 28.02 g/mol), 50.5 g He (MW = 4.00 g/mol), and 34.6 g CO2 (MW = 44.01 g/mol). Calculate the mole fraction of carbon dioxide in the sample.

Answer 1

Answer:

The mole fraction of carbon dioxide is 0.0559

Explanation:

• Mole fraction of a gas in a mixture of gases its given by dividing the number of moles of the gas by the total number of moles of the mixture.

• $Mole fraction =\frac{number of moles of a gas}{Total number of moles in a mixture}$

In this case, we are given a mixture of N₂, CO₂ and  He

To find the mole fraction of CO₂, first we need to calculate the total number of moles in the mixture.

Individual number of moles for each gas are;

Moles = mass ÷ Molar mass

Therefore;

Nitrogen gas (N₂)

Mass =18.0 g

Molar mass = 28.02 g/mol

$Moles=\frac{18.0g}{28.02g/mol}$

                 = 0.642 moles N₂

Helium gas (He)

Mass = 50.5 g

Molar mass = 4.0 g/mol

$Moles = \frac{50.5g}{4.00g/mol}$

                = 12.625 moles He

Carbon dioxide (CO₂)

Mass = 34.6 g

Molar mass = 44.01 g/mol

$Moles=\frac{34.6g}{44.01g/mol}$

                = 0.786 moles CO₂

Therefore;

The total number of moles in the mixture;

= 0.642 moles + 12.625 moles + 0.786 moles

= 14.053 moles

Thus;

The mole fraction of CO₂= Moles of CO₂ ÷ Total number of moles

                                      = 0.786 moles ÷ 14.053 moles

                                      = 0.0559

Hence the mole fraction of carbon dioxide is 0.0559