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Determine the empirical formula for a 100.00-g sample of a compound having the following percent composition.

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94.07% sulfur and 5.93% hydrogen
asked Aug 23, 2015 in Chemistry by frang (1,700 points)

1 Answer

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1 mol MnO x (1 mol Mn)/(1 mol MnO) x (59.4 g Mn)/(1 mol Mn) = 59.94 g Mn

1 mol MnO x (1 mol O)/(1 mol MnO) x (16.00 g O)/(1 mol O) = 16.00 g O

54.94 +16.00 = 70.94 g

The molar mass of manganese oxide is 70.94 g/mol

mass percent Mn = (54.94 g Mn)/(70.94 g MnO) x 100 = 77.45 % Mn

mass percent O = (16.00 g O)/(70.94 g MnO) x 100 = 22.55% O
answered Sep 13, 2015 by thb (2,000 points)
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