Answer:
The empirical formula is KClO3
Explanation:
Step 1: Data given
Mass of the sample = 10.81 grams
Mass of potassium = 3.45 grams
Mass of chlorine = 3.13 grams
molar mass O = 16.0 g/mol
Molar mass K = 39.1 g/mol
Molar mass of chlorine = 35.5 g/mol
Step 2: Calculate mass of oxygen
Mass of oxygen = mass of sample - mass of potassium - mass of chlorine
Mass oxygen = 10.81 grams - 3.45 grams - 3.13 grams
Mass oxygen = 4.23 grams
Step 3: Calculate moles
Moles = mass / molar mass
Moles K = 3.45 grams / 39.1 g/mol
Moles K = 0.0882 moles
Moles Cl = 3.13 grams / 35.5 g/mol
Moles Cl = 0.0882 moles
Moles O = 4.23 grams / 16.0 g/mol
Moles O = 0.264 moles
Step 4: Calculate mol ratio
We divide by the smallest amount of moles
K: 0.0882 / 0.0882 = 1
Cl: 0.0882 / 0.0882 = 1
O: 0.264/0.0882 = 3
for 1 mol K we have 1 mol Cl and 3 moles O
The empirical formula is KClO3
