Question

question 1
1 pts
a compound has an empirical formula bh₃ and a molar mass of 27.6 g/mol. what is the molecular formula of the compound?

question 2
1 pts
a compound is 31.1 % s, the rest is chlorine. the compound has a molecular mass of 205.94 amu. what is its molecular formula?

Explanation:

Step1: Calculate the empirical formula mass

The atomic mass of \(B = 10.81\space g/mol\) and \(H= 1.008\space g/mol\).
The empirical formula \(BH_3\) has a mass \(M_{empirical}=(10.81 + 3\times1.008)\space g/mol=(10.81+3.024)\space g/mol = 13.834\space g/mol\)

Step2: Find the ratio \(n\)

We know that \(n=\frac{M_{molecular}}{M_{empirical}}\)
Given \(M_{molecular} = 27.6\space g/mol\) and \(M_{empirical}=13.834\space g/mol\)
\(n=\frac{27.6}{13.834}\approx2\)

Step3: Determine the molecular formula

Multiply the sub - scripts in the empirical formula by \(n\).
The empirical formula is \(BH_3\), so the molecular formula is \(B_2H_6\)

Answer:

\(B_2H_6\)