Question

# 5- Elemental analysis is sometimes carried out by combustion of the sample. For a hydrocarbon, the...

5- Elemental analysis is sometimes carried out by combustion of the sample. For a hydrocarbon, the only products formed are CO2 and H2O.
If a 1.36-g sample of an unknown hydrocarbon is burned and 2.21 g of H2O are produced along with 4.07 g of CO2, what is the empirical formula of the hydrocarbon?

To solve this problem you can take the 4.07g CO2 and convert that to mol of C :
4.07g CO2 x 1mol CO2/44.01g CO2 x 1 mol C/1 mol CO2=0.0924 mol C

Do the same process with the 2.21g H20 to get 0.245 mol H. Moles of H: (2.21÷18.02) × 2= 0.245 mol

Convert the moles of C and H into grams( 0.0924×12=1.11g of C and 0.245×1.008= 0.25g of H). Subtract this mass from the original mass of the sample to get mass of O if present. But in this case that will be zero (1.11g+ 0.25= 1.36g). So no O is present in the hydrocarbon.

Now since you have the mol for each C and H, you can get the empirical formula of the hydrocarbon by taking the ratio of moles of Hydrogen and Carbon:

0.245 mol÷ 0.092 mol= 2.66 or 8/3 for H:C

Thus the empirical formula is C3H8.

#### Earn Coins

Coins can be redeemed for fabulous gifts.