Question

When 1.661 grams of a hydrocarbon, C_xH_y, were burned in a combustion analysis apparatus, 5.211grams of CO₂ and 2.134 grams of H₂O were produced.

In a separate experiment, the molar mass of the compound was found to be 42.08 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

Enter the elements in the order presented in the question.

Answer 1

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

= 5.211/44

= 0.118432

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

= 2.134/18

= 0.118556

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.118432

so, x = 0.118432

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.118556 = 0.237111

Divide by smallest to get simplest whole number ratio:

C: 0.118432/0.118432 = 1

H: 0.237111/0.118432 = 2

So empirical formula is:CH2

2)

Molar mass of CH2,

MM = 1*MM(C) + 2*MM(H)

= 1*12.01 + 2*1.008

= 14.026 g/mol

Now we have:

Molar mass = 42.08 g/mol

Empirical formula mass = 14.026 g/mol

Multiplying factor = molar mass / empirical formula mass

= 42.08/14.026

= 3

Hence the molecular formula is : C3H6