Combustion of a 1.031-g sample of a compound containing only carbon, hydrogen, and oxygen produced 2.265 g of CO2(g) and 1.236 g of H2O(g). What is the
empirical formula of the compound?
3 국
Molar masses in g/mol: CO2 = 44.01; H20 = 18.02; C = 12.01; H = 1.01
C3H50​

C₃H₈O

Explanation:

From the question given above, the following data were obtained:

Mass of compound = 1.031 g

Mass of CO₂ = 2.265 g

Mass of H₂O = 1.236 g

Empirical formula =?

Next, we shall determine the mass of carbon, hydrogen and oxygen in the compound. This can be obtained as follow:

For carbon, C:

Mass of CO₂ = 2.265 g

Molar mass of CO₂ = 44.01 g/mol

Molar mass of C = 12.01 g/mol

Mass of C =?

Mass of C = molar mass of C / molar mass of CO₂ × mass of CO₂

Mass of C = 12.01/44.01 × 2.265

Mass of C = 0.618 g

For hydrogen, H:

Mass of H₂O = 1.236 g

Molar mass of H₂O = 18.02 g/mol

Molar mass of H₂ = 2 × 1.01 = 2.02 g/mol

Mass of H =?

Mass of H = Molar mass of H₂ / Molar mass of H₂O × Mass of H₂O

Mass of H = 2.02/18.02 × 1.236

Mass of H = 0.139 g

For oxygen, O:

Mass of compound = 1.031 g

Mass of C = 0.618 g

Mass of H = 0.139 g

Mass of O =?

Mass of O = Mass of compound – (mass of C + mass of H)

Mass of O = 1.031 – ( 0.618 + 0.139)

Mass of O = 1.031 – 0.757

Mass of O = 0.274 g

Finally, we shall determine the empirical formula. This can be obtained as follow:

C = 0.618 g

H = 0.139 g

O = 0.274 g

Divide by their molar mass

C = 0.618 / 12.01 = 0.051

H = 0.139 / 1.01 = 0.138

O = 0.274 / 16 = 0.017

Divide by the smallest

C = 0.051 / 0.017 = 3

H = 0.138 / 0.017 = 8

O = 0.017 / 0.017 = 1

Therefore, the empirical formula of the compound is C₃H₈O

3

explanation:

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