2. An empirical formula consists of the symbols for the elements combined in a compound, with the subscripts showing the smallest whole-number mole ratio of the different atoms in the compound.

3. 1. Assume a 100 g sample. 2. Convert from g to mol and round to 3 sig figs. 3. Divide molar amounts by the smallest molar amount. 4. If necessary, multiply result by a small, whole number (2,3,4) to eliminate fractions/decimals. Example- What is the empirical formula of a compound that is 25.9% nitrogen and 74.1% oxygen?

4. 1. Convert from g to mol and round to 3 sig figs. 2. Divide molar amounts by smallest molar amount. 3. If necessary, multiply the result by a small, whole number to eliminate fractions/decimals. Example- Calculate the emperical formula of a compound that contains 1.67 g of cerium and 4.54 g of iodine.

5. Example Problem 1 Quantitative analysis shows that a compound contains 32.83% sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical formula for this compound.

6. Example Problem 2 Analysis of a 10.15 g sample of a compound known to contain only phosphorus and oxygen indicates a phosphorus content of 4.43 g. What is the empirical formula of this compound?

7. Example Problem 3 A compound is found to contain 63.52% iron and 36.48% sulfur. Find the empirical formula.

8. The molecular formula is the actual formula of a molecular compound. It is either the emperical or a multiple of it.

9. The relationship between a compound’s empirical formula and its molecular formula can be written as follows. X (Empirical Formula) = Molecular Formula X = whole-number multiple indicating the factor by which the subscripts in the empirical formula must be multiplied to obtain the molecular formula.