Question

When 4.748 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 16.05 grams of CO₂ and 3.286grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

empirical formula =

?

molecular formula =

?

Answer 1

Empirical formula is the least coefficient formula, that is

CxHy

we must find x,y via gravimetry.

Typically we do this relating to moles of C,H

mol of CO2 = mass of CO2/MW of CO2 = (16.05)/(44) = 0.3647 mol of CO2

1 mol of CO2 = 1 mol of C ---> 0.3647 mol of CO2 = 0.3647 mol of C

mol of H2O = mass of H2O/MW of H2O= (3.286)/(18) = 0.1825 mol of H2O

1 mol of H2O= 2 mol of H ---> 0.1825 mol of H2O= 0.1825*2 = 0.365 mol of H

Ratios

H:C =0.365 / 0.3647 = 1:1

so

CH is the empirical formula

if MW 26.04

MW of empirical = 12 + 1 = 13

then

ratio = 2x

C2H2 is the sample