A mixture contains methane (CH4), ethane (C2H6), and propane (C3H8). The mixture contains 25.9% methane, 31.4% ethane, and the balance propane by mass. Converting the mass percents to mole fractions for all three components gives:

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Chemistry

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23 June, 08:50

A mixture contains methane (CH4), ethane (C2H6), and propane (C3H8). The mixture contains 25.9% methane, 31.4% ethane, and the balance propane by mass. Converting the mass percents to mole fractions for all three components gives:

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Answers (1)

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Aron Mccormick · Answer 1

Mol fraction methane = 0.4453

Mol fraction ethane = 0.2878

Mol fraction propane = 0.2669

Explanation:

Step 1: Data given

Suppose the mass of the mixture = 100 grams

The mixture contains:

25.9 % methane = 25.9 grams

31.4 % ethane = 31.4 grams

propane = 100 - 25.9 - 31.4 = 42.7 grams

Molar mass of methane = 16.04 g/mol

Molar mass of ethane = 30.07 g/mol

Molar mass of propane = 44.1 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles methane = 25.9 grams / 16.04 g/mol

Moles methane = 1.615 moles

Moles ethane = 31.4 grams / 30.07 g/mol

Moles ethane = 1.044 moles

Moles propane = 42.7 grams / 44.1 g/mol

Moles propane = 0.968 moles

Step 3: Calculate total moles

Total moles = moles methane + moles ethane + moles propane

Total moles = 1.615 + 1.044 + 0.968

Total moles = 3.627 moles

Step 4: Calculate mo lfraction

Mol fraction = mol component / total moles

Mol fraction methane = 1.615 moles / 3.627 moles

Mol fraction methane = 0.4453

Mol fraction ethane = 1.044 moles / 3.627 moles

Mol fraction ethane = 0.2878

Mol fraction propane = 0.968 moles / 3.627 moles

Mol fraction propane = 0.2669