Question

1. the molecule borazine has an empirical formula of bh₃n. what is the molecular formula if the molecular weight is 80.5 g/mol?

Explanation:

Step1: Calculate the empirical formula weight

The atomic weights are: \(B = 10.81\space g/mol\), \(H= 1.008\space g/mol\), \(N = 14.01\space g/mol\).
The empirical formula \(BH_3N\) weight \(M_{empirical}=(10.81 + 3\times1.008+14.01)\space g/mol=(10.81 + 3.024 + 14.01)\space g/mol = 27.844\space g/mol\)

Step2: Find the ratio \(n\)

\(n=\frac{M_{molecular}}{M_{empirical}}\), where \(M_{molecular} = 80.5\space g/mol\) and \(M_{empirical}=27.844\space g/mol\)
\(n=\frac{80.5}{27.844}\approx 2.9\approx3\)

Step3: Determine the molecular formula

Multiply each sub - script in the empirical formula by \(n\).
The empirical formula is \(BH_3N\), so the molecular formula is \((BH_3N)\times3=B_3H_9N_3\)

Answer:

\(B_3H_9N_3\)