a substance is analyzed to have a percent composition of 74.186% sodium and 25.814% fluorine. calculate the empirical formula

Na₂₆F₁₁

Explanation:

We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:

74.186 g Sodium | 1 mol Sodium/23 g              =>       3.2255 mol Na    

25.814 g Fluorine | 1 mol Fluorine/19 g             =>       1.3586 mol F

Divide each by smallest number of moles:

3.2255/1.3586 = 2.37

1.3586/1.3586 = 1

Multiply by common number to get a smallest whole number:

2.37*11 = 26,

1*11 = 11

The empirical formula is Na₂₆F₁₁

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