Question

# When 3.795 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 13.03 grams...

When 3.795 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 13.03 grams of CO2 and 2.134 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.

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

= 13.03/44

= 0.2961

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

= 2.134/18

= 0.1186

Since 1 mol of CO2 has 1 mol of C

Number of moles of C in CO2= 0.2961

so, x = 0.2961

Since 1 mol of H2O has 2 mol of H

Number of moles of H = 2*0.1186 = 0.2371

Divide by smallest:

C: 0.2961/0.2371 = 1.25

H: 0.2371/0.2371 = 1

Multiply by 4 to get simplest whole number ratio:

C: 1.25*4 = 5

H: 1*4 = 4

So empirical formula is:C5H4

2)

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

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