Answer:
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
