If 1.3860 g of Mg ribbon combusts to form 2.1252 g of Magnesium oxide product, calculate the experimental empirical formula from this data.

Answer 1

mass Mg = 1.3860 g

mass O = 2.1252 g - 1.3860 g = 0.7392 g O

mols Mg = 1.3860 g x 1 mol / 24.31 g = 0.0570 mols Mg

mols O = 0.7392 g x 1 mol / 16 g = 0.0462 mols O

Assuming these values are the same within experimental error, we have 1 mol Mg and 1 mol O

Empirical formula would be MgO