Answer to Question #168559 in General Chemistry for H7uju h

Question #168559

A compound has the composition 62.1% C, 10.3% H and 27.6% O. What is its empirical formula? (Ar: C = 12, H = 1, O = 16)


1
Expert's answer
2021-03-04T06:33:13-0500

Mass of C = 62.1 g

Moles of C = "\\dfrac{62.1}{12}=5.175\\space mol"

Mass of H = 10.3 g

Moles of H = 10.3 mol

Mass of O = 27.6 g

Moles of O = "\\dfrac{27.6}{16}=1.725\\space mol"

C = "\\dfrac{5.175}{10.3}=0.50"

H = "\\dfrac{10.3}{10.3}=1"

O = "\\dfrac{1.725}{10.3}=0.167"

Multiplying by suitable integer 6,

"C=0.50\\times 6=3\\\\H=1\\times6=6\\\\O=0.167\\times6=1.002\\approx1"

Empirical formula = "C_3H_6O"

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