Question

A compound is found to contain 15.94 % boron and 84.06 % fluorine by mass. To answer the question, enter the elements in the order presented above. Question 1: the empirical formula for this compound is . Question 2: the molar mass for this compound is 67.81 g/mol. The molecular formula for this compound is .

Answer 1

given:

% B = 15.94 % by mass

% F = 84.06 % by mass

This means that for 100 g of the compound:

Mass of B = 15.94 g

Mass of F = 84.06 g

Now, atomic mass of B = 10.811 g/mol

Atomic mass of F = 18.998 g/mol

# moles of B = Mass of B/Atomic mass of B = 15.94 g/10.811 gmole-1 = 1.474 moles

# moles of F = Mass of F/ Atomic mass of F = 84.06/18.998 = 4.425 moles

The molar ratio of B and F in the given compound would be:

B = 1.474/1.474 = 1

F = 4.425/1.474 = 3

Therefore, the empirical formula is BF3

The corresponding empirical formula mass = 1* 10.811 + 3*18.998 = 67.805 g/mol i.e, 67.81 g/mol

It is given that the molar mass of the compound is 67.81 g/mol

The ratio of the two masses is:

n = empirical mass/molar mass = 67.81/67.81 = 1

Molecular formula = n*Empirical formula = 1*(BF3) = BF3

Hence, the molecular formula of the compound is BF3