Question

A compound is found to contain 31.42 % sulfur, 31.35 % oxygen, and 37.23% fluorine by weight. To answer the questions, enter the elements in the order presented above. 1. What is the empirical formula for this compound? 2. The molecular weight for this compound is 102.1 g/mol. What is the molecular formula for this compound?

Answer 1

Given, S = 31.42 %, O = 31.35 % , F = 37.23 %

So, if we suppose that the mass of the compound is 100g, then

Mass of S = 31.42 g

Mass of O = 31.35 g

Mass of F = 37.23 g

Number of moles = Mass of the substance/ Molar mass

So, Number of moles of S = 31.42 g / 32.065 g/mol = 0.979 mol

Number of moles of O = 31.35 g / 15.99 g mol = 1.96 mol

Number of moles of S = 37.23 g/ 18.99 g/mol = 1.96 mol

Divide with lowest value derived above to get the simplest molecular formula ,

S = 0.979 mol/0.979 mol = 1 mol

O = 1.96 mol/0.979 mol = 2 mol

F = 1.96 mol/0.979 mol = 2 mol

Hence, the empirical formula is SO2F2.

Molecular weight = 102.1 g/mol

Molar mass of the empirical formula is SO2F2.= 32.065 + (2 x 15.99) + (2 x 18.99) = 32.065 g/mol + 31.98 g/mol + 37.98 g/mol = 102.025 g/mol ≈ 102.1 g/mol

Since the molar mass of the compound and the molar mass of the derived empirical formula is equal so n = molecular molar mass/emprirical molar mass = 1. So the molecular same will also be SO2F2 .