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04-10-2018
Chemistry

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What is the empirical formula of a compound that is 14 g calcium, 11g oxygen, and 0.7 g hydrogen

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12-10-2018

CaO2H2

convert to moles, divide all by smallest mole mass, multiply til whole

14 g Ca ÷ 40.08 = .35 mol
11 g O ÷ 16 = .69 mol
.7 g H ÷ 1.008 = .69 mol

.35 ÷ .35 = 1
.69 ÷ .35 = 2
.69 ÷ .35 = 2

CaO2H2

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