A sample of a compound is made up of 57.53 g C, 3.45 g H, and 39.01 g F. Determine the empirical formula of this compound.

C7H5F3

Explanation:

The following data were obtained from the question:

Mass of Carbon (C) = 57.53g

Mass of Hydrogen (H) = 3.45g

Mass of Fluorine (F) = 39.01g

The empirical formula of the compound can be obtained as follow:

C = 57.53g

H = 3.45g

F= 39.01g

Divide each by their molar mass

C = 57.53/12 = 4.79

H = 3.45/1 = 3.45

F = 39.01/19 = 2.05

Divide each by the smallest

C = 4.79/2.05 = 2.3

H = 3.45/2.05 = 1.7

F = 2.05/2.05 = 1

Multiply through by 3 to express in whole number

C = 2.3 x 3 = 7

H = 1.7 x 3 = 5

F = 1 x 3 = 3

Therefore, the empirical formula for the compound is C7H5F3

i don,t know so just answer mine