Answer to Question #168559 in General Chemistry for H7uju h

Question

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)

Answer 1

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 = $C_3H_6O$