What is the empirical formula of a compound that contains 46.3% lithium and 53.7% oxygen?

Step 1: Assume you have 100g of the compound. This means you have 46.3g of lithium (Li) and 53.7g of oxygen (O).

Step 2: Convert the mass of each element to moles. The atomic mass of Li is approximately 6.94g/mol and the atomic mass of O is approximately 16.00g/mol.

For Li: 46.3g ÷ 6.94g/mol = 6.67 mol For O: 53.7g ÷ 16.00g/mol = 3.36 mol

Step 3: Write the ratio of the moles. This gives us Li6.67O3.36.

Step 4: Divide each of these by the smallest number of moles calculated (in this case, 3.36) to get the simplest whole number ratio.

For Li: 6.67 ÷ 3.36 = 1.99 ≈ 2 For O: 3.36 ÷ 3.36 = 1

Step 5: Write the empirical formula. The empirical formula of the compound is Li2O.