A compound of B and H is found to have a mass percent composition of 81.1% Boron. Its empirical formula is B2H5. A mass spectrometry experiment tells us that the molecular mass is 53.3 g/mol. What is its molecular formula?

Question

A compound of B and H is found to have a mass percent composition of 81.1% Boron. Its empirical formula is B2H5. A mass spectrometry experiment tells us that the molecular mass is 53.3 g/mol. What is its molecular formula?

So I feel like this is a trick question, because I thought molecular formula was when it is simplified to its smallest like if it was originally B4H10 then its molecular formula would be B2H5 so I feel like the answer is B2H5 because it cannot be reduced. Am I completely off base on this?

Answer 1

Almost. You have the right idea but you've reversed the facts. The empirical formula is the simplest; the molecular formula is how many of the empirical units you have together.

So the empirical mass is 2*B + 5*H or 2*10.81 + 5*1 = 26.62
The molecular mass is 53.3 from the problem. so 53.3/25.52 = 2.00 so the molecular formula is
(B2H5)2 or B4H10.

Answer 2

Actually, you're on the right track! The empirical formula represents the simplest whole number ratio of atoms in a compound. In this case, the empirical formula B2H5 indicates that the ratio of boron to hydrogen is 2:5. However, the question asks for the molecular formula, which represents the true number of atoms of each element in a compound.

To determine the molecular formula, we need to find the ratio of empirical formula mass to the actual molecular mass of the compound. This can be done using the following equation:

(n × empirical formula mass) = molecular mass

Where "n" is the integer representing the number of empirical formula units in the molecular formula.

To calculate "n", we need to find the ratio of the molecular mass to the empirical formula mass:

n = (molecular mass) / (empirical formula mass)

Let's calculate the empirical formula mass first:

B2H5:
(2 × atomic mass of boron) + (5 × atomic mass of hydrogen)

Using the atomic masses from the periodic table:
(2 × 10.81 g/mol) + (5 × 1.01 g/mol) = 21.62 g/mol + 5.05 g/mol = 26.67 g/mol

Now we can determine "n":

n = (53.3 g/mol) / (26.67 g/mol) = 2

Therefore, the molecular formula is n times the empirical formula:

2 × B2H5 = B4H10

So the molecular formula for the compound with a mass percent composition of 81.1% Boron and a molecular mass of 53.3 g/mol is B4H10.