1. Percent Composition and Empirical Formulas What is 73% of 150? 110 The relative amounts of each element in a compound are expressed as the percent composition or the percent by mass of each element in a compound.

2. The percent composition of a compound has as many percent values as there are different elements in the compound From the graph above, you can see the percent composition of K 2 CrO 4 is K= 40.3%, Cr= 26.8%, and O= 32.9% These percents must total 100

3. Equation for Percent Mass: % mass of element A = grams of element A x 100% grams of compound

4. An 8.20 g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound? Knowns: mass of magnesium= 8.20 g Mg  mass of oxygen= 5.40 g O  mass of compound= 8.20 + 5.40 =13.60g Unknowns: percent Mg = ? % Mg  percent O = ? % O Calculate:

5. To calculate the percent composition of a known compound, use the chemical formula to calculate the molar mass. This gives the mass of one mole of the compound % mass = grams of element in 1 mol compound X 100% (composition) molar mass of compound

6. Calculate the percent composition of propane (C 2 H 6 )

7. Empirical Formulas Determining the percent composition of a compound can be used to calculate the empirical formula of a compound. empirical formula: gives the lowest whole- number ratio of the atoms of the elements in a compound. The empirical formula is the simplest formula for a compound. A molecular formula is the same as or a multiple of the empirical formula, and is based on the actual number of atoms of each type in the compound.

8. For example, if the empirical formula of a compound is C 3 H 8, its molecular formula may be C 3 H 8, C 6 H 16, etc. An empirical formula is often calculated from elemental composition data. The weight percentage of each of the elements present in the compound is given by this elemental composition.

9. Let's determine the empirical formula for a compound with the following elemental composition: 40.00% C, 6.72% H, 53.29% O. The first step will be to assume exactly 100 g of this substance.

10. This means in 100 g of this compound, 40.00 g will be due to carbon, 6.72 g will be due to hydrogen, and 53.29 g will be due to oxygen. We will need to compare these elements to each other stoichiometrically. In order to compare these quantities, they must be expressed in terms of moles.

11. So the next task will be to convert each of these masses to moles, using their respective atomic weights: (Take notice that since the composition data was given to four significant figures, the atomic weights used in the calculation were to at least four significant figures. Using fewer significant figures may actually lead to an erroneous formula. See, sig figs do make a difference!!!!)

12. Now that the moles of each element are known, a stoichiometric comparison between the elements can be made to determine the empirical formula. This is achieved by dividing through each of the mole quantities by which ever mole quantity is the smallest number of moles. In this example, the smallest mole quantity is either the moles of carbon or moles of oxygen (3.331 mol):

13. The ratio of C:H:O has been found to be 1:2:1, thus the empirical formula is: CH 2 O. Again, as a reminder, this is the simplest formula for the compound, and not necessarily the molecular formula.

14. Suppose we know that the molecular weight of this compound is 180 g/mol. With this information, the molecular formula may be determined. The formula weight of the empirical formula is 30 g/mol. Divide the molecular weight by the empirical formula weight to find a multiple: The molecular formula is a multiple of 6 times the empirical formula: C( 1 x 6 ) H( 2 x 6 ) O( 1 x 6 ) which becomes C 6 H 12 O 6

15. Another Example (a) Given the following mass percent composition, determine the empirical formula. 49.5%C, 5.2%H, 28.8%N, 16.5%O Assume 100g sample [RULE 1 49.5g C, 5.2g H, 28.8g N, 16.5g O

16. Divide each mass number by the molar mass [RULE 2]: 49.5 g C --------------- = 4.121 mol C 12.011 g/mol C 5.2 g H --------------- = 5.159 mol H 1.008 g/mol H 28.8 g N --------------- = 2.056 mol N 14.007 g/mol N 16.5 g O --------------- = 1.031 mol O 15.999 g/mol O

17. Divide each molar amount by smallest amount [RULE 3]: 4.121 mol C 4 mol C ----------- = --------- 1.031 mol O mol O 5.159 mol H 5 mol H ----------- = --------- 1.031 mol O mol O 2.056 mol N 2 mol N ----------- = --------- 1.031 mol O mol O 1.031 mol O 1 mol O ----------- = --------- 1.031 mol O mol O Empirical Formula is then C 4 H 5 N 2 O

18. Molecular Formulas Like you calculated in the first sample problem, the molecular formula of a compound can be determined if you know the empirical formula and the compound’s molar mass. Known molar mass of compound = Empirical formula mass multiplier to convert the empirical formula to the molecular formula

19. Example: Calculate the molecular formula of the compound whose molar mass is 60.0 g and empirical formula is CH 4 N