Question

1. When 1.389 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 4.771 grams of CO₂ and 0.7813 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 128.2 g/mol. Determine the empirical formula and the molecular 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
= 4.771/44
= 0.1084

Number of moles of H2O = mass of H2O / molar mass H2O
= 0.7813/18
= 4.341*10^-2

Since 1 mol of CO2 has 1 mol of C
Number of moles of C in CO2= 0.1084
so, x = 0.1084

Since 1 mol of H2O has 2 mol of H
Number of moles of H = 2*4.341*10^-2 = 8.681*10^-2

Divide by smallest:
C: 0.1084/8.681*10^-2 = 1.25
H: 8.681*10^-2/8.681*10^-2 = 1

Multiply by 4 to get simplest whole number ratio:
C: 1.25*4 = 5
H: 1*4 = 4
So empirical formula is:C5H4

Molar mass of C5H4,
MM = 5*MM(C) + 4*MM(H)
= 5*12.01 + 4*1.008
= 64.082 g/mol

Now we have:
Molar mass = 128.2 g/mol
Empirical formula mass = 64.082 g/mol
Multiplying factor = molar mass / empirical formula mass
= 128.2/64.082
= 2

So molecular formula is:C10H8
Answer: C10H8