A binary compound of boron and hydrogen has the following percentage composition: 78.14% boron, 21.86% hydrogen. If the molar mass of the compound is determined by experiment to be between 27 and 28 g, what are the empirical and molecular formulas of the compound?

To solve the exercise, we need to know how many boron atoms and how many hydrogen atoms the compound has. We know that of the total weight of the compound, 78.14% correspond to boron and 21.86% to hydrogen. As the weight of the compound is between 27 g and 28 g, using the above percentages we can solve that the compound has between 21.1 g and 21.8 g of boron, and between 5.9 g and 6.1 g of hydrogen:

100% _____ 27 g

78.14% _____ x = 78.14% * 27g / 100% = 21.1 g boron

100% ______27 g

21.86% ______ x = 21.86% * 27g / 100% = 5.9 g hydrogen

100% _____ 28 g

78.14% _____ x = 78.14% * 28g / 100% = 21.8 g boron

100% _____ 28g

21.86% _____ x = 21.86% * 28g / 100% = 6.1 g hydrogen

So, if the atomic weight of boron is 10.8 g, there must be two boron atoms in the compound that sum 21.6 g. The weight of hydrogen is 1 g, so the compound must have six hydrogen atoms.

The molecular formula represents the real amount of atoms that form a compound. Therefore, the molecular formula of the compound is B2H6.

The empirical formula is the minimum expression that represents the proportion of atoms in a compound. For example, ethane has 2 carbon atoms and 6 hydrogen atoms, so its molecular formula is C2H6, however, its empirical formula is CH3. Therefore, the empirical formula of the boron compound is BH3.