# How to calculate the mass of hydrogen in a borane sample?

In a $$\pu{45.9 g}$$ sample of borane, a compound containing only boron and hydrogen, that's $$85.63\%$$ boron, what is the mass of hydrogen in this sample?

I was just going to take $$\pu{45.9 g}$$ and multiply by $$0.1437$$ to get the mass of hydrogen, but I don't think it works that way because they don't equally contribute to the mass of the compound. Then I thought, maybe I should consider the molar masses of both atoms, hydrogen being $$\pu{1.008 g mol-1}$$ and boron being $$\pu{10.811 g mol-1}$$. I am stuck at this point. How to solve the question.

## 1 Answer

The problem does not need any difficult approach, not even relative atomic masses, if you trust the one who made this exercise. Since borane (whatever type it is)* consists (by exercise's hypothesis) only of hydrogen and boron and since boron is $$85.63\%$$ by mass, hydrogen's mass percentage would be the remaining $$100\% - 85.63\% = 14.37\%$$. Therefore, we have $$\pu{6.60 g}$$ of hydrogen in this sample.

The compound borane, $$\ce{BH3}$$, is about $$78.1\%$$ boron per mass. However, there is a class of compounds called boranes which also only consist of boron and hydrogen. From this class pentaborane(9), $$\ce{B5H9}$$, with about $$85.7\%$$ boron by mass would fit the exercise.