Question

What is the empirical formula of a compound containing 24.56% potassium, 34.81% manganese, and 40.50% oxygen?

Answer 1

Consider these percentages as actual masses out of a 100 g sample of the compound. That is, there is 24.56 g K, 34.81 g Mn, and 40.50 g O.

Next, convert these figures to moles. I'm assuming that you know how to do such conversions.

Once you have mols K, Mn, and O, find the ratio of each. Suppose you have \(X\) mols of K, \(Y\) mols of Mn, and \(Z\) mols of O. Find \(X:Y:Z\). If any of these values are not whole numbers, you can either round them or multiply them by a factor to make them a whole number. NOTE: If you multiply one of the values by a factor, you must do it to the rest.

With your finalized ratio, it will form the empirical formula for \(K_X Mn_Y O_Z\)

Answer 2

I'm just gonna guess it's KMnO4 considering that this compound has K, Mn, and O. I didn't calculate it btw, you can calculate it by yourself.