Answer
The empirical formula is $K_{2}MnO_{4}$ .
Work Step by Step
Strategy: 1- Use percentage composition to find the mass of each element in the compound. For this assume you have 100 g compound. If for example you are given 21% as the percentage composition of an element X in the compound, this is $\frac{21}{100}\times100g$ = 21 g of element X since you assumed a mass 100 g for the compound. Thus the mass of the element equals the numerical value of percentage composition. 2- Convert the masses to moles, using conversion factor 3- Divide each mole number by the smallest one in order to find the smallest integers. 4- If at the end of step 3 you do not get integers, than find a whole number to multiply the results of step 3 to get integers.
The compound contains these percentages by mass of each element: 39.6% K, 27.9% Mn,, and 32.5% O. Assuming 100 g compound we get the masses of each element: 39.6 g K, , 27.9 g Mn, 32.5 g O.
Now we convert masses to moles:
Moles of K: 39.6 g$\times\frac{1 mol(K)}{39.0 g (K)}$ = 1.02 mol K
Moles of Mn: 27.9 g$\times\frac{1 mol(Mn)}{54.9 g (Mn)}$ = 0.51 mol Mn
Moles of oxygen: 32.5g$\times\frac{1 mol (O)}{16.0 g (O) }$ = 2.03 mol O
Divide the mole number by the smallest one.
For K: $\frac{1.02mol}{0.51mol}$ = 2.00
For Mn: $\frac{0.51mol}{0.51mol}$ = 1.00
For O: $\frac{2.03mol}{0.51mol}$ = 4.00
We found that all the numbers are integers, and the ratio is 2:1:4, so the empirical formula is $K_{2}MnO_{4}$ .
