Abinary 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?

Empirical formula: BH3

Molecular Formula: B2H6

Explanation:

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.

the answer is c

tell me if wrong