Question

What is the empirical formula for ethyl fluoride if the compound contains 49.97% carbon, 10.51% hydrogen, and 39.52% fluorine by mass?

A) C₂₅F₂
B) C₄H₁₀F₄
C) C₂H₅F
D) C₄H₁₀F₂

Answer 1

The empirical the moles of each element, we divide the mass formula for ethyl fluoride is C2H5F.

Step-by-step explanation:

The empirical formula is C2H5F. To find the empirical formula, we need to determine the ratio of the elements present in the compound. We can assume a 100g sample of the compound, which will give us 49.97g of carbon, 10.51g of hydrogen, and 39.52g of fluorine. To find the moles of each element, we divide the mass by the molar mass. The molar mass of carbon is 12.01 g/mol, hydrogen is 1.01 g/mol, and fluorine is 19.00 g/mol.

Carbon: 49.97g / 12.01 g/mol = 4.16 mol

Hydrogen: 10.51g / 1.01 g/mol = 10.41 mol

Fluorine: 39.52g / 19.00 g/mol = 2.08 mol

Next, we divide each mole value by the lowest mole value to get the simplest whole-number ratio.

Carbon: 4.16 mol / 2.08 mol = 2

Hydrogen: 10.41 mol / 2.08 mol = 5

Fluorine: 2.08 mol / 2.08 mol = 1

Therefore, the empirical formula for ethyl fluoride is C2H5F.