Question

An unknown compound contains only C , H , and O . Combustion of 5.40 g of this compound produced 12.7 g CO2 and 3.47 g H2O . What is the empirical formula of the unknown compound? Insert subscripts as needed.

Answer 1

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

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

= 12.7/44

= 0.2886

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

= 3.47/18

= 0.1928

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2886

so, x = 0.2886

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.1928 = 0.3856

Molar mass of O = 16 g/mol

mass O = total mass - mass of C and H

= 5.4 - 0.2886*12 - 0.3856*1

= 1.551

number of mol of O = mass of O / molar mass of O

= 1.551/16.0

= 9.693*10^-2

so, z = 9.693*10^-2

Divide by smallest to get simplest whole number ratio:

C: 0.2886/9.693*10^-2 = 3

H: 0.3856/9.693*10^-2 = 4

O: 9.693*10^-2/9.693*10^-2 = 1

So empirical formula is:C3H4O

Answer: C3H4O