Determine the empirical formula of a compound having the following percent composition by mass: K: 24.75%, Mn: 34.77%, and O: 40.51%

To determine the empirical formula of a compound, we need to find the simplest whole number ratio of the elements present in the compound.

Step 1: Assume we have 100g of the compound.

Step 2: Convert the percent composition to grams:
- K: 24.75% of 100g = 24.75g
- Mn: 34.77% of 100g = 34.77g
- O: 40.51% of 100g = 40.51g

Step 3: Convert the mass of each element to moles using their atomic masses:
- Moles of K = 24.75g / atomic mass of K
- Moles of Mn = 34.77g / atomic mass of Mn
- Moles of O = 40.51g / atomic mass of O

Step 4: Divide each of the amounts of moles by the smallest number of moles obtained in step 3 to obtain the empirical formula.

Let's calculate the moles of each element:
- The atomic mass of K is 39.10 g/mol. So the moles of K are: 24.75g / 39.10 g/mol = 0.6330 mol
- The atomic mass of Mn is 54.94 g/mol. So the moles of Mn are: 34.77g / 54.94 g/mol = 0.6330 mol
- The atomic mass of O is 16.00 g/mol. So the moles of O are: 40.51g / 16.00 g/mol = 2.5319 mol

Step 5: Divide each of the amounts of moles by the smallest number of moles obtained in step 4 to obtain the empirical formula.

Since the smallest number of moles obtained is approximately 0.6330 mol, we divide the number of moles of each element by this value:
- Moles of K / 0.6330 ≈ 0.6330 mol / 0.6330 mol = 1
- Moles of Mn / 0.6330 ≈ 0.6330 mol / 0.6330 mol = 1
- Moles of O / 0.6330 ≈ 2.5319 mol / 0.6330 mol = 4

Therefore, the empirical formula of the compound is K1Mn1O4.

Answer: The empirical formula of the compound is KMnO_4