Isobutyl propionate is the substance that provides the flavour for rum extract. Combustion of a 1.152 g sample of C-H-O compound yields 2.726 g CO2 and 1.116 g H2O. What is the E. F of the compound?

C₇H₁₄O₂

Explanation:

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

Mass of compound = 1.152 g

Mass of CO₂ = 2.726 g

Mass of H₂O = 1.116 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 = 2.726 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Mass of C = 12/44 × 2.726

Mass of C = 0.743 g

For hydrogen (H):

Mass of H₂O = 1.116 g

Molar mass of H₂O = (2×1) + 16

= 2 + 16

= 18 g/mol

Mass of H = 2/18 × 1.116

Mass of H = 0.124 g

For oxygen (O):

Mass of compound = 1.152 g

Mass of C = 0.743 g

Mass of H = 0.124 g

Mass of O =?

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

Mass of O = 1.152 – (0.743 + 0.124)

Mass of O = 1.152 – 0.867

Mass of O = 0.285 g

Finally, we shall determine the empirical formula for the compound as follow:

C = 0.743 g

H = 0.124 g

O = 0.285 g

Divide by their molar mass

C = 0.743 / 12 = 0.062

H = 0.124 / 1 = 0.124

O = 0.285 / 16 = 0.018

Divide by the smallest

C = 0.062 / 0.018 = 3.44

H = 0.124 / 0.018 = 7

O = 0.018 / 0.018 = 1

Multiply by 2 to express in whole number.

C = 3.44 × 2 = 7

H = 7 × 2 = 14

O = 1 × 2 = 2

Empirical formula => C₇H₁₄O₂