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

Question

Owen Ambrose · Accepted Answer

The empirical formula of this compound is #"K"_2"MnO"_4"#.
This compound is potassium manganate.

We can assume that we have 100.0 g of the compound since the percentages add up to 100. This will translate the percentages to grams. By dividing each element's mass in the compound by its molar mass (atomic weight on the periodic table in grams/mole, or g/mol), you can find the moles of each element. I'll round to three significant figures at the end, saving some guard digits to prevent rounding errors. #39.6cancel"g K"xx(1"mol K")/(39.0983cancel"g K")="1.0128 mol K"# #27.9cancel"g Mn"xx(1"mol Mn")/(54.938049cancel"g Mn")="0.50784 mol Mn"# #32.5cancel"g O"xx(1"mol O")/(15.999cancel"g O")="2.0314 mol O"# To find the mole ratios for the formula, divide the moles of each element by the least number of moles. #"K":##(1.0128)/(0.50784)##=##"1.99##~~##2# #"Mn":##(0.50784)/(0.50784)="1.00"# #"O":##(2.0314)/(0.50784)="4.00"# The empirical formula of this compound is #"K"_2"MnO"_4"#, which is potassium manganate.

Cooper Anderson · Answer

undefined KMnO4 is the empirical formula for this compound.