Question

A 10.81 g sample of a compound contains 3.45 g of potassium, K , 3.13 g of chlorine, Cl , and oxygen, O . Calculate the empirical formula. Insert subscripts as needed. Empirical formula:

Answer 1

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