Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of co2 and 2.58 mg of h2o. what is the empirical formula of the compound

The empirical formula of 2.78 mg of ethyl butyrate which produces 6.32 mg of CO₂ and 2.58 mg of H₂O is C₃H₆O

Mass of CO₂ = 6.32 mg

Molar mass of CO₂ = 44 g/mol

Molar mass of C = 12 g/mol

Mass of C = 12/44 × 6.32

Mass of H₂O = 2.58 mg

Molar mass of H₂O  = 18 g/mol

Molar mass of H₂ = 1 × 2 = 2 g/mol

Mass of H = 2/18 × 2.58

Mass of C = 1.72 mg

Mass of H = 0.29 mg

Mass of compound = 2.78 mg

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

Mass of O = 2.78 – (1.72 + 0.29)

Mass of C = 1.72 mg

Mass of H = 0.29 mg

Mass of O = 0.77 mg

Divide by their molar mass

C = 1.72 / 12 = 0.143

H = 0.29 / 1 = 0.29

O = 0.77 / 16 = 0.048

Divide by the smallest

C = 0.143 / 0.048 = 3

H = 0.29 / 0.048 = 6

O = 0.048 / 0.048 = 1

Therefore, the empirical formula of the compound is C₃H₆O

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Explanation:

Let's call the ethyl butyrate X.

1. Calculate the mass of each element in 2.78 mg of X.

(a) Mass of C

(b) Mass of H

(d) Mass of O

Mass of O = 2.78 - 1.725 - 0.2886 = 0.7667 g

2. Calculate the moles of each element

3. Calculate the molar ratios

Divide all moles by the smallest number of moles.

4. Round the ratios to the nearest integer

C:H:O = 3:6:1

5. Write the empirical formula

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explanation: