Question

A 0.5242-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.9740 g of CO2 and 0.1994 g of H2O. What is the empirical formula of the compound?

Answer 1

C₃H₃O₂

Step-by-step explanation:

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

Mass of compound = 0.5242 g

Mass of CO₂ = 0.9740 g

Mass of H₂O = 0.1994 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 = 0.9740 g

Molar mass of CO₂ = 12 + (16×2)

= 12 + 32 = 44 g/mol

Mass of C = 12/44 × 0.9740

Mass of C = 0.2656 g

For hydrogen, H:

Mass of H₂O = 0.1994 g

Molar mass of H₂O = (2×1) + 16

= 2 + 16 = 18 g/mol

Mass of H = 2/18 × 0.1994

Mass of H = 0.0222 g

For oxygen, O:

Mass of compound = 0.5242 g

Mass of C = 0.2656 g

Mass of H = 0.0222 g

Mass of O =?

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

Mass of O = 0.5242 – (0.2656 + 0.0222)

Mass of O = 0.5242 – 0.2878

Mass of O = 0.2364 g

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

Mass of C = 0.2656 g

Mass of H = 0.0222 g

Mass of O = 0.2364 g

Divide by their molar mass

C = 0.2656 /12 = 0.0221

H = 0.0222 /1 = 0.0222

O = 0.2364 /16 = 0.0148

Divide by the smallest

C = 0.0221 / 0.0148 = 1.5

H = 0.0222 / 0.0148 = 1.5

O = 0.0148 / 0.0148 = 1

Multiply through by 2 to express in whole number.

C = 1.5 × 2 = 3

H = 1.5 × 2 = 3

O = 1 × 2 = 2

Empirical formula => C₃H₃O₂