Question

a compound with the molar mass 70.13 g/mol has an empirical formula of ch₂. what is the molecular formula of the compound? c₅h₁₀ c₂h c₆h₁₂ ch₂

Explanation:

Step1: Calculate the empirical formula mass

The empirical formula is \(CH_2\). The atomic mass of \(C = 12.01\ g/mol\) and \(H= 1.008\ g/mol\).
The empirical formula mass \(M_{empirical}=(1\times12.01)+(2\times1.008)=12.01 + 2.016=14.026\ g/mol\)

Step2: Find the ratio \(n\)

We know that \(n=\frac{M_{molecular}}{M_{empirical}}\). Given \(M_{molecular} = 70.13\ g/mol\) and \(M_{empirical}=14.026\ g/mol\)
\(n=\frac{70.13}{14.026}\approx5\)

Step3: Determine the molecular formula

If the empirical formula is \(CH_2\) and \(n = 5\), then the molecular formula is \((CH_2)_n\). Substituting \(n = 5\), we get \(C_5H_{10}\)

Answer:

\(C_5H_{10}\)