Acompound has a percent composition of 24.74% potassium (molar mass = 39.10 g/mol), 34.76% manganese (molar mass = 54.94 g/mol), and 40.50% oxygen (molar mass = 16.00 g/mol). assuming that the mass of the compound is 100 g, what is the compound’s empirical formula?

The empirical formua is KMnO4

Explanation:

Step 1: Data given

Mass of the compound = 100 grams

24.74%  = potassium

Molar mass of K = 39.10 g/mol)

34.76%  = manganese

Molar mass of Mn= 54.94 g/mol

40.50%  = oxygen

Molar mass = 16.00 g/mol

Step 2: Calculate moles of potassium

Moles K = mass K / molar mass K

Moles K = 24.74 grams / 39.10 g/mol

Moles K = 0.6327 moles

Step 3: Calculate moles of Mn

Moles Mn = 34.76 grams / 54.94 g/mol

Moles Mn = 0.6327 moles

Step 4: Calculate moles of O

Moles O = 40.50 grams / 16.00 g/mol

Moles O = 2.53

Step 5: Calculate the mol ratio

We divide by the smallest amount of mol

K: 0.6327 / 0.6327 = 1

Mn: 0.6327/0.6327 = 1

O: 2.53 / 0.6327 = 4

The empirical formua is KMnO4