What is the empirical formula of stannous fluoride, the first fluoride compound added to toothpaste to protect teeth against decay? its mass percent composition is 24.25% f, 75.75% sn. express your answer as a chemical formula.

The answer to your question is:    SnF₂  

Explanation:

Empirical formula of Stannous fluoride

Mass percent composition:

F = 24.25%

Sn = 75.75%

Express percents as grams

F = 24.25 g               AW F = 19 g/mol

Sn = 75.75g              AW Sn = 119 g(mol

Rule of three

                        19 g of F  1 mol

                        24.25 g    x

                        x = (24.25 x 1) / 19 = 1.28 mol

                        119 g of Sn  1 mol

                       75.75 g of Sn   x

                        x = (75.75 x 1) / 119 = 0.64 mol

Divide both results by the lowest number of moles

F = 1.28 / 0.64 = 2

Sn = 0.64 / 0.64 = 1

Then, the empirical formula is:     Sn₁ F₂   =  Sn F₂