A compound consists of 50.04% carbon, 5.59% hydrogen, and 44.37% oxygen. What is the empirical formula of this compound?

The empirical formula of the compound is C₃H₄O₂

From the question given above, the following data were obtained:

Carbon = 50.04%

Hydrogen = 5.59%

Oxygen = 44.37%

Divide by their molar mass

C = 50.04 / 12 = 4.17

H = 5.59 / 1 = 5.59

O = 44.37 / 16 = 2.773

Divide by the smallest

C = 4.17 / 2.773 = 1.5

H = 5.59 / 2.773 = 2

O = 2.773 / 2.773 = 1

Multiply by 2 to express in whole number

C = 1.5 × 2 = 3

H = 2 × 2 = 4

O = 1 × 2 = 2

Therefore, the empirical formula of the compound is C₃H₄O₂

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increased stability in the completion of fluorine's outer shell.