What is the empirical formula for a compound which contains 67.1 zinc and the rest is oxygen

The empirical formula is ZnO2

Explanation:

What is the empirical formula for a compound which contains 67.1% zinc and the rest is oxygen?

Step 1: Data given

Suppose the compound has a mass of 100.0 grams

A compound contains:

67.1 % Zinc = 67.1 grams

100 - 67.1 = 32.9 % oxygen = 32.9 grams

Molar mass of Zinc = 65.38 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles of Zinc

Suppose the compound is 100 grams

Moles Zn = 67. 10 grams / 65.38 g/mol

Moles Zn = 1.026 moles

Step 3: Calculate moles of O

Moles O = 32.90 grams / 16.00 g/mol

Moles O = 2.056 moles

Step 4: Calculate mol ratio

We divide by the smallest amount of moles

Zn: 1.026/1.026 = 1

O: 2.056/1.026 = 2

The empirical formula is ZnO2

To control this we can calculate the % Zinc for 1 mol

65.38 / (65.38+2*16) = 0.67.1 = 67.2 %