An analysis of a 4.2-gram sample of a compound is found to contain 54.76 percent sodium (na) and 45.24 percent fluorine (f). determine the empirical formula of the compound

NaF

Explanation:

If the 54.76% of the sample is sodium and 45.24% is fluorine. Mass of each atom is:

Na: 4.2g × 0.5476 = 2.3g

F: 4.2g × 0.4524 = 1.9g

In moles:

Na: 2.3g × (1mol / 22.99g) = 0.1 moles

F: 1.9g × (1mol / 19) = 0.1 moles

As empirical formula is defined as the simplest whole-number ratio of atoms in a compound, the ratio of Na:F is:

0.1 / 0.1 = 1. Thus, empirical formula is:

NaF

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