Question

Analysis of gas produced during combustion determined a compound contained 43.74% C, 3.14% H, 16.61% O A) What is the empirical formula? B) If the molar mass of the compound was experimentally determined to be 244.26 g/mol. What is the molecular formula of this compound?

Answer 1

Answer:

A) C₇H₆O₂

B) C₁₄H₁₂O₄

Explanation:

To determine the empirical and molecular formula of the compound, we need to follow a series of steps.

Step 1: Divide each percentage by the atomic mass of the element

C: 43.74/12.01 = 3.64

H: 3.14/1.01 = 3.11

O: 16.61/16.00 = 1.04

Step 2: Divide all the numbers by the smallest one, i.e. 1.04

C: 3.64/1.04 = 3.5

H: 3.11/1.04 = 3

O: 1.04/1.04 = 1

Step 3: Multiply all the numbers by 2, so that all of them are integers

C: 3.5 × 2 = 7

H: 3 × 2 = 6

O: 1 × 2 = 2

The empirical formula of the compound is C₇H₆O₂.

Step 4: Determine the molar mass of the empirical formula

M(C₇H₆O₂) = 7 × M(C) + 6 × M(H) + 2 × M(O)

M(C₇H₆O₂) = 7 × 12.01 g/mol + 6 × 1.01 g/mol + 2 × 16.00 g/mol = 122.13

Step 5: Calculate "n"

We will use the following expression

n = M(compound) / M(empirical formula)

n = (244.26 g/mol) / (122.13 g/mol) = 2

Step 6: Determine the molecular formula of the compound

We determine the molecular formula by multiplying the empirical formula by "n".

C₇H₆O₂ × 2 = C₁₄H₁₂O₄