Question

A 6.52 g sample of a hydrocarbon compound containing only C and hydrogen produced 20.5 g of CO2.   A 9.29 ml sample weighing 21.6 mg was collected over water at a pressure of 796 torr at 25 °C (The vapor pressure of water is 23.8 torr at 25°C) What is the molecular formula of the compound?

Answer 1

moles of CO2 = 20.5 / 44 = 0.4659

mass of C = 0.4659 x 12 = 5.591 g

mass of hydrogen = 6.52 - 5.591 = 0.929 g

moles of hydrogen = 1 x 0.929 = 0.929

moles ration of hydorgen and carbon = 0.929 : 0.4659

= 2 : 1

empirical formula = CH2

pressure of gas = 796 -23.8 = 772.2 torr = 1.016 atm

volume = 9.29 mL = 9.29 x 10^-3 L

mass = m = 21.6 mg = 21.6 x 10^-3 g

T = 25 + 273 = 298 K

P V = n R T

P V = w R T / M

1.016 x 9.29 x 10^-3 = 21.6 x 10^-3 x 0.0821 x 298 / M

M = 56 g / mol

molar mass = 56 g / mol

empirical formula mass = CH2 = 12 + 2 = 14 g /mol

n = molar mass / empirical formula mass

= 56 / 14

= 4

molecular formula = n x empirical formula

= 4 x CH2

= C4H8

molecular formula of the compound = C4H8