A 0.2500 g sample of a compound known to contain carbon, hydrogen and oxygen undergoes complete combustion to produce 0.3664 grams carbon dioxide and 0.1500 grams water. What is the empirical formula of this compound

C₂H₅O₂

Explanation:

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

Mass of compound = 0.25 g

Mass of CO₂ = 0.3664 g

Mass of H₂O = 0.15 g

Empirical formula =?

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

For Carbon (C):

Mass of CO₂ = 0.3664 g

Molar mass of CO₂ = 12 + (2×16) = 44 g/mol

Mass of C = 12/44 × 0.3664

Mass of C = 0.1

For Hydrogen (H):

Mass of H₂O = 0.15 g

Molar mass of H₂O = (2×1) + 16 = 18 g/mol

Mass of H = 2/18 × 0.15

Mass of H = 0.02 g

For Oxygen (O):

Mass of C = 0.1 g

Mass of H = 0.02 g

Mass of compound = 0.25 g

Mass of O =?

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

Mass of O = 0.25 – (0.1 + 0.02)

Mass of O = 0.25 –0.12

Mass of O = 0.13 g

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

C = 0.1

H = 0.02

O = 0.13

Divide by their molar mass

C = 0.1 / 12 = 0.0083

H = 0.02 / 1 = 0.02

O = 0.13 / 16 = 0.0081

Divide by the smallest

C = 0.0083 / 0.0081 = 1

H = 0.02 / 0.0081 = 2.47

O = 0.008 / 0.008 = 1

Multiply by 2 to express in whole number

C = 1 × 2 = 2

H = 2.47 × 2 = 5

O = 1 × 2 = 2

Thus, the empirical formula for the compound is C₂H₅O₂

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