Question

29 find the molecular formula of ethylene glycol, which is used in antifreeze. the molar mass is 62.0 g/mol, and the empirical formula is ch₃o. need a hint? 30 what is the molecular formula of a compound with the empirical formula ccln and a molar mass of 184.5 g/mol?

Explanation:

Step1: Calculate empirical formula mass (CH₃O)

Find sum of atomic masses: $M_{\text{empirical}} = 12.01 + 3(1.008) + 16.00 = 31.034\ \text{g/mol}$

Step2: Find ratio n of molar to empirical mass

$n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{62.0}{31.034} \approx 2$

Step3: Multiply empirical formula by n

Molecular formula: $2 \times \text{CH}_3\text{O} = \text{C}_2\text{H}_6\text{O}_2$

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Step1: Calculate empirical formula mass (CCIN)

Find sum of atomic masses: $M_{\text{empirical}} = 12.01 + 12.01 + 126.90 + 14.01 = 164.93\ \text{g/mol}$

Step2: Find ratio n of molar to empirical mass

$n = \frac{\text{Molar Mass}}{\text{Empirical Formula Mass}} = \frac{184.5}{164.93} \approx 1.12$ (round to 1, as n must be integer)

Step3: Multiply empirical formula by n

Molecular formula: $1 \times \text{CCIN} = \text{C}_2\text{ClN}$

Answer:

1. $\text{C}_2\text{H}_6\text{O}_2$
2. $\text{C}_2\text{ClN}$