Q: A white powder has %mass of Ba 69.6%, C 6.09% and O 24.3%. What is the empirical formula? (M(Ba) = 137.3 g mol-1, M(C) = 12.01 g mol-1, M(O) = 16.00 g mol-1)
(Worksheet 6)
In a 100 g sample, m(Ba) = 69.6 g, m(C) = 6.09g, and m(O) = 24.3 g
n(Ba) = m(Ba) ÷ M(Ba) = 69.6 ÷ 137.3 = 0.507 mol
n(C) = m(C) ÷ M(C) = 6.09 ÷ 12.01 = 0.507 mol
n(O) = m(O) ÷ M(O) = 24.3 ÷ 16.00 = 1.52 mol
Mole ratio is 0.507 : 0.507 : 1.52 = $\frac{0.507}{0.507} : \frac{0.507}{0.507} : \frac{1.52}{0.507}$ = 1 : 1 : 3
Therefore, the empirical formula is BaCO3
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