Question

An artificial compound with the following elemental composition was synthesized: 8.0268%C, 1.3378%H, 19.0635%F,
23.4114%N, 37.4582%O, and 10.7023%S. What is the empirical and molecular formula of the artificial compound if the molar
mass of the compound is 598,000 mg/mol?

Answer 1

Assuming 100 g of sample in a given compound then

8.0268 %C = 8.0268 g of C

1.3378 %H = 1.3378 g of H

19.0635 % F = 19.0635 g of F

23.4114 % N = 23.4114 g of N

37.4582 % O = 37.4582 g of O

10.7023 % S = 10.7023 g of S

Atomic weight of C = 12 g/mol

Atomic weight of H = 1 g/mol

Atomic weight of F = 19 g/mol

Atomic weight of N = 14 g/mol

Atomic weight of O = 16 g/mol

Atomic weight of S = 32 g/mol

Number of moles = given mass in gram / atomic weight

Number of moles of C = 8.0268 g / 12 g/mol = 0.6689 mol

Number of moles of H = 1.3378 g / 1 g/mol = 1.3378 mol

Number of moles of F = 19.0635 g / 19 g/mol = 1.00 mol

Number of moles of N = 23.4114 g / 14 g/mol = 1.6722 mol

Number of moles of O = 37.4582 g / 16 g/mol = 2.3411 mol

Number of moles of S = 10.7023 g / 32 g/mol = 0.3344 mol

Divide by lowest number of moles in each element

C = 0.6689 mol / 0.3344 mol = 2

H = 1.3378 mol / 0.3344 mol = 4

F = 1.00 mol / 0.3344 mol = 3

N = 1.6722 mol / 0.3344 mol = 5

O = 2.3411 mol / 0.3344 mol = 7

S = 0.3344 mol / 0.3344 mol = 1

Empirical formula (EF) = C₂H₄F₃N₅O₇S

Empirical formula weight = (2 * 12 + 4 * 1 + 3 * 19 + 5 * 14 + 7 * 16 + 1 * 32) g/mol = (24 + 4 + 57 + 70 + 112 + 32) g/mol = 299 g/mol

We know 1 mg = 10^-3 g

Molar mass of compound = 598000 mg / mol = 598000 * 10^-3 g/mol = 598 g/mol

Divide Molar mass of compound by empirical formula weight

598 g/mol / 299 g/mol = 2

Molecular formula = 2 * empirical formula = 2 * C₂H₄F₃N₅O₇S

Molecular formula = C₄H₈F₆N₁₀O₁₄S₂