Question

Combustion analysis of a hydrocarbon produced 33.01 g CO2 and 13.51gH2O. Calculate the empirical formula of the hydrocarbon.

Answer 1

we have to calculate no.of moles of C and H

for this given mass of CO2 = 33.01 and mass of H2O = 13.51 gm after combustion

so

C+ O2 ...........> CO2

from the above equation

44 gm(1 mole) of CO2 can be formed from 12 gm(1 mole) of C

33.01 gm of CO2 formed from ................................?

        = 33.01*12/44 = 9.0 mole

similarly

2H2 + O2 .............> 2H2O

36gm (2 moles) of H2O formed from 4 gm(1mole) of H2

13.51gm of H2O formed from ...............?

      = 13.51*4/36 = 1.5 moles

Now assign these no. of moles of C and H at the suffixes of C and H

C = C₉

H = H_1.5

Divide thes numbers with smallest one between them

C = C₉ = C_9/1.5 = C₆

H = H1.5 = H_1.5/1.5 = H₁

So emperical formula is C6H