Question

The combustion of 2.95 grams of a compound that contains only C, H and S yields 5.48
grams of CO2 and 1.13 grams of H20. What is the empirical formula of the compound?

Answer 1

The empirical formula of the compound is C₃H₃S.

Explanation:

molecular weight of CO₂ = 44 g/mole

molecular weight of H₂O = 18 g/mole

Knowing the molecular weights of the compounds we devise the following reasoning:

if in           44 g of CO₂ there are 12 g of C

then in  5.48 g of CO₂ there are X g of C

X = (5.48 × 12) / 44 = 1.49 g C

if in           18 g of H₂O there are 2 g of H

then in    1.13 g of H₂O there are Y g of H

Y= (1.13 × 2) / 18 = 0.126 g H

mass of compound = mass of C + mass of H + mass of S

mass of S = mass of compound - mass of C - mass of H

mass of S = 2.95 - 1.49 - 0.126 = 1.33 g S

Now to find the empirical formula of the compound we use the fallowing algorithm:

we devise each mass by the molar weight of the element:

for C   1.49 / 12 = 0.124

for H   0.126 / 1 = 0.126

for S   1.33 / 32 = 0.0416

now we divide the result by the lowest number which is 0.0416

for C    0.124 / 0.0416 ≈ 3

for H    0.126 / 0.0416 ≈ 3

for S    0.0416 / 0.0416 = 1

The empirical formula of the compound is C₃H₃S.

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