Question

a compound has an empirical formula of c₂oh₄ and a molar mass of 88 g/mol. what is the molecular formula of this compound? c₂oh₄ c₄o₂h₈ co₀.₅h₂ c₃h₃h₁₂

Explanation:

Step1: Calculate the empirical formula mass

The atomic masses are: \(C = 12\space g/mol\), \(O=16\space g/mol\), \(H = 1\space g/mol\).
For the empirical formula \(C_{2}OH_{4}\), the empirical formula mass \(M_{e}\) is \(M_{e}=(2\times12)+16+(4\times1)=24 + 16+4=44\space g/mol\)

Step2: Find the ratio \(n\)

The ratio \(n=\frac{\text{Molar mass}}{\text{Empirical formula mass}}\). Given molar mass \(M = 88\space g/mol\) and \(M_{e}=44\space g/mol\), then \(n=\frac{88}{44} = 2\)

Step3: Determine the molecular formula

Multiply each sub - script in the empirical formula by \(n\).
For \(C_{2}OH_{4}\), the molecular formula is \(C_{2\times2}O_{1\times2}H_{4\times2}=C_{4}O_{2}H_{8}\)

Answer:

\(C_{4}O_{2}H_{8}\)