Empirical Formulas Chapter 7-4, p. 229 - 231

Empirical Formula Defined as: the symbols for the elements combined in a chemical compound, with subscripts showing the smallest whole number mole ratio of the different atoms in the compound.

Example C2H6 = dicarbon hexahydride (ethane) has an empirical formula of CH3 (notice the common multiple is 2, which you use to divide to get the smallest whole number ratio) Ethane’s molecular formula is C2H6

Determining Empirical Formula from Percent Composition Assume you have 100.00 g of the compound. Use the number in percent of each element and change it to grams. For example, the percent oxygen in NaOH is 32.6%. If you have a 100.00 g sample of NaOH, 32.6 g of it is oxygen. Divide by the molar mass of each element from the periodic table. Compare the mole ratios by dividing all values by the smallest mole value.

Example A compound is 32.38% Na, 22.65% S and 44.99% O. Determine the empirical formula. 32.38 g Na 1 mol Na = 1.408 mol Na 22.99 g Na 22.65 g S 1 mol S = 0.7063 mol S 32.07 g S 44.99 g O 1 mol O = 2.812 mol O 16.00 g O Which molar value is the smallest?

Example The empirical formula is Na2SO4 A compound is 32.38% Na, 22.65% S and 44.99% O. Determine the empirical formula. 1.408 mol Na = 1.993 mol Na = 2 Na 0.7063 0.7063 mol S = 1.000 mol S = 1 S 2.812 mol O = 3.981 mol O = 4 O The empirical formula is Na2SO4

You Try It! A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula.

You Try It! Solution FeS A compound is found to contain 63.52% iron and 36.48% sulfur. Find its empirical formula. 63.52 g Fe 1 mol Fe = 1.137 mol Fe 55.85 g Fe 36.48 g S 1 mol S = 1.138 mol S 32.07 g S Which molar value is the smallest? 1.137 mol Fe = 1 Fe 1.138 mol S = 1 S 1.137 1.137

Determining Empirical Formula from Mass Convert the mass of each element to moles by dividing by its molar mass from the periodic table. Compare the mole ratios by dividing all values by the smallest mole value.

Example A sample of a compound containing only phosphorus and oxygen has a mass of 10.150 g. The phosphorus content is 4.433 g. Find the empirical formula. 10.150 g Total – 4.433 g P = 5.717 g O 4.433 g P 1 mol P = 0.1431 mol P 30.97 g P 5.717 g O 1 mol O = 0.3573 mol O 16.00 g O Which molar value is the smallest?

Example Uh Oh… can you have 2.5 atoms? A sample of a compound containing only phosphorus and oxygen has a mass of 10.150 g. The phosphorus content is 4.433 g. Find the empirical formula. 0.1431 mol P = 1.000 mol P = 1 P 0.1431 0.3573 mol O = 2.497 mol O = 2.5 O Uh Oh… can you have 2.5 atoms?

No Worries! The empirical formula is P2O5 Multiply both molar values by 2… you will get a whole number… 2 (1 P) = 2 P 2 (2.5 O) = 5 O The empirical formula is P2O5

You Try It! Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of Ca are present. What is the empirical formula?

You Try It! Solution Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of Ca are present. What is the empirical formula? 20.00 g Total – 4.00 g Ca = 16.00 g Br 4.00 g Ca 1 mol Ca = 0.0998 mol Ca 40.08 g Ca 16.00 g Br 1 mol Br = 0.2003 mol Br 79.90 g Br Which molar value is the smallest?

You Try It! Solution Analysis of 20.0 g of a compound containing only calcium and bromine indicates that 4.00 g of Ca are present. What is the empirical formula? 0.0998 mol Ca = 1.00 mol Ca = 1 Ca 0.0998 0.2003 mol Br = 2.01 mol Br = 2 Br The empirical formula is CaBr2