Question

1. what is the molecular formula of a compound with a molecular weight of 84 g/mol and an empirical formula of ch₂?

Explanation:

Step1: Calculate the empirical formula weight

The empirical formula is \(CH_2\). The atomic weight of \(C\) is \(12\space g/mol\) and of \(H\) is \(1\space g/mol\).
The empirical formula weight \(M_{empirical}=(1\times12)+(2\times1)=14\space g/mol\)

Step2: Find the ratio \(n\)

Let \(n = \frac{M_{molecular}}{M_{empirical}}\)
Given \(M_{molecular} = 84\space g/mol\) and \(M_{empirical}=14\space g/mol\)
\(n=\frac{84}{14} = 6\)

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by \(n\).
The empirical formula \(CH_2\), when multiplied by \(n = 6\) gives \((CH_2)_6=C_6H_{12}\)

Answer:

\(C_6H_{12}\)