A certain manganese oxide contains 72% manganese. What is the empirical formula?

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A certain manganese oxide contains 72% manganese. What is the empirical formula?

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Answer 1 · 2017-05-01T12:39:02

Since $%$ $%$ litterally means "per hundred" let's take a 100g sample.

Explanation:

We convert the mass ratio into mol ratio:

Then 72g of this is Manganese.
The molar mass of $M n = 55$ $Mn=55$ , so $72 g \leftrightarrow \frac{72}{55} = 1.31 m o l$ $72gharr72/55=1.31mol$ of $M n$ $Mn$

The 28g of Oxygen (molar mass $O = 16$ $O= 16$ ) breaks down to:
$28 g \leftrightarrow \frac{28}{16} = 1.75 m o l$ $28gharr28/16=1.75mol$ of $O$ $O$

The mol ratio $M n \div O = 1.31 \div 1.75 = 0.75 \div 1 = 3 \div 4$ $MndivO=1.31div1.75=0.75div1=3div4$

Empirical formula: $M n_{3} O_{4}$ $Mn_3O_4$

Note:
This works if you take any other sample size, but then there are more calculations to do.

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A certain manganese oxide contains 72% manganese. What is the empirical formula?

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Explanation:

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