Question

Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 4.82 g H2O. Calculate the empirical formula of the hydrocarbon.​

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

= 33.01/44

= 0.7502

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

= 4.82/18

= 0.2678

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.7502

so, x = 0.7502

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.2678 = 0.5356

Divide by smallest:

C: 0.7502/0.5356 = 1.4

H: 0.5356/0.5356 = 1

Multiply by 5 to get simplest whole number ratio:

C: 1.4*5 = 7

H: 1*5 = 5

So empirical formula is: C7H5

Answer: C7H5