Question

A 10.0 g sample of an unknown liquid is vaporized at 120.0°C and 5.0 atm. The volume of the vapour is found to be 568.0 mL. The liquid is determined to be made up of 84.2% carbon and 15.8% hydrogen. What is the molecular formula for the liquid?

Answer 1

Answer:

molecular formula of liquid = C₈H₁₈

Explanation:

First we determine the empirical formula of the liquid:

Number of moles of each element present in the liquid = % mass / molar mass

For Carbon, (molar mass = 12.01 g/mol) : 84.2/12.01 =7.011 moles

For Hydrogen (molar mass = 1.01 g/mol) : 15.8/1.01 = 15.643

Simplest mole ratio of the elements, C : H  is given by:

C = 7.011/7.011 = 1.0

H = 15.643/7.011 = 2.23

Multiplying through with 5, C:H = 5:11

Therefore, empirical formula is C₅H₁₁

The molecular mass of the liquid is next determined:

Using PV = nRT to find the number of moles of the liquid present

P = 5.0 atm; V = 568.0 mL = 0.568 L; R = 0.082 L*atmmol⁻¹ K⁻¹; T = 273 + 120 = 393 K

n = PV/RT = (5*0.568)/0.082*393

n = 0.088 moles

Molar mass of liquid = mass/no of moles = 10.0 g/ 0.088 moles = 113.63 gmol⁻¹

Molecular formula = n(empirical formula)

Molar mass of empirical formula, C₅H₁₁ = 71 gmol⁻¹

n = molecular mass/empirical mass = 113.63/71 = 1.6

Therefore, molecular formula =  1.6*(C₅H₁₁) = C₈H₁₈