Use the following information to determine the empirical formula if a compound is found to have 18.7% Li, 16.3% C, and 65.5% O Answer questions 4-6 with this data

The empirical formula is Li2CO3

Explanation:

Step 1: Data given

Suppose the mass of the compound = 100 grams

The compound contains:

18.7% Li = 18.7 grams of Li

16.3 % C = 16.3 grams of C

65.0% O = 65.0 grams of O

Total = 100%

Molar mass of Li = 6.94 g/mol

Molar mass of C = 12.01 g/mol

Molar mass of O = 16 g/mol

Step 2: Calculate moles

Moles = Mass / molar mass

Moles Li = 18.7 grams / 6.94 g/mol = 2.65 moles

Moles C = 16.3 grams / 12.01 = 1.36 moles

Moles O = 65.0 grams / 16.0 g/mol = 4.06 moles

Step 3: Calculate the mol ratio

We divide by the smallest number of moles

Li: 2.65/1.36 ≈ 2

C: 1.36/1.36 = 1

O: 4.06/1.36 ≈ 3

The empirical formula is Li2CO3