# A compound was found to contain "39% K", "27.9% Mn", and "32.5% O". What is the empirical formula for this compound?

Nov 6, 2015

The empirical formula of this compound is $\text{K"_2"MnO"_4}$.
This compound is potassium manganate.

#### Explanation:

Since the percentages add up to 100, we can assume that we have 100.0 g of the compound, which will convert the percentages to grams.

Determine the moles of each element by dividing their masses in the compound by their molar masses (atomic weight on the periodic table in grams/mole, or g/mol). I will be saving some guard digits to reduce rounding errors. I will round to three significant figures at the end.

$39.6 \cancel{\text{g K"xx(1"mol K")/(39.0983cancel"g K")="1.0128 mol K}}$

$27.9 \cancel{\text{g Mn"xx(1"mol Mn")/(54.938049cancel"g Mn")="0.50784 mol Mn}}$

$32.5 \cancel{\text{g O"xx(1"mol O")/(15.999cancel"g O")="2.0314 mol O}}$

Next divide the moles of each element by the least number of moles. This will give you the mole ratios for the formula.

$\text{K} :$$\frac{1.0128}{0.50784}$$=$"1.99$\approx$$2$

$\text{Mn} :$$\frac{0.50784}{0.50784} = \text{1.00}$

$\text{O} :$$\frac{2.0314}{0.50784} = \text{4.00}$

The empirical formula of this compound is $\text{K"_2"MnO"_4}$, which is potassium manganate.