Answer:
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
