Question

An unknown compound contains only carbon and hydrogen (CxHy). Combustion of this compound produced 3.67 g of carbon dioxide and 1.50 g of water. Determine the molecular formula of this compound if the molecular mass is between 40 and 45 amu.

Answer 1

let in compound number of moles of C and H be x and y respectively

Number of moles of CO2 = mass of CO2 / molar mass CO2

= 3.67/44

= 0.0834

Number of moles of H2O = mass of H2O / molar mass H2O

= 1.5/18

= 0.0833

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.0834

so, x = 0.0834

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.0833 = 0.1667

Divide by smallest to get simplest whole number ratio:

C: 0.0834/0.0834 = 1

H: 0.1667/0.0834 = 2

So empirical formula is:CH2

CH2

Molar mass of CH2,

MM = 1*MM(C) + 2*MM(H)

= 1*12.01 + 2*1.008

= 14.026 g/mol

Now we have:

Molar mass = 45.0 g/mol

Empirical formula mass = 14.026 g/mol

Multiplying factor = molar mass / empirical formula mass

= 45.0/14.026

= 3

Hence the molecular formula is : C3H6