1. Percent Composition, Empirical Formulas, & Molecular Formulas Section 10.4

2. 2 types of Chemists: Synthetic: Develops new compounds for pharmaceutical, industrial and home use Analytic: Analyzes the compound to provide experimental proof of its composition and its chemical formula

3. Percent Composition New compounds are first made on a small scale by a synthetic chemist. Then, an analytical chemist analyzes the compound to verify its structure and percent composition. Can be used to help determine the identity of an unknown compound Percent by mass of each element in a compound

4. Percent Composition Formula Percent by mass = mass of element x 100 mass of compound Percent Composition = the percent of an element in a compound Example: a 100 g sample of a new compound contains 55 g of element X and 45 g of element Y. What is the percent composition of the compound? 55 g element X x 100 = 55% element X 100 g compound 45 g element Y x 100 = 45% element Y 100 g compound

5. Percent Composition Practice 1.The % of N in KNO 2 2. The % of P in Al 2 (PO 4 ) 3 3. The % of Cl in LiCl 4. The % of Ca in CaSO 4 5. The % of H in NaOH

6. Empirical Formulas The next step in determining the identity of a compound is to determine its empirical formula Empirical formula = formula with the smallest whole number ratio of elements, a simplified formula Examples: C 4 H 9, H 2 O Molecular formula = formula with the actual number of atoms of each element Examples: C 8 H 18, H 6 O 3 The identities of compounds are given as percentages by the mass spectrometer and the percentages are then used to determine the empirical values. Use percent composition to calculate both empirical and molecular formulas

7. Which of the following is the empirical formula?  Problem 1  Problem 3  C 3 H 6 O 2  C 2 H 6  C 6 H 12 O 4  CH 3  C 4 H 12  Problem 2  Problem 4  C 22 H 24 O 12  N 4 O 12  C 11 H 12 O 6  N 2 O 6  NO 3

8. Determining Empirical Formulas Take the percentages determined by the spectrometer: 40.5 % S 59.05% 0 Because the % values equal 100% you can assume that you are working with a 100g sample which will give you the same values. So drop the % and replace it with grams. 40.5 g Sulfur 59.05 g 0xygen Now divide each value by its molar mass (from periodic table) 40.5 g S ÷ 32 g/mol = 1.25 mol Sulfur 59.05 g O ÷ 16 g/mol = 3.75 mol Oxygen

9. Determining Empirical Formulas As you can see, these values are not whole numbers, and cannot be used as subscripts in a formula. Therefore, we must convert them by taking the smallest number of moles and dividing the other mole values by this number. 1.25 moles ÷ 1.25 moles = 1 mole Sulfur 3.75 moles ÷ 1.25 moles = 3 moles Oxygen The simplest whole number ratio then becomes 1:3, and the empirical formula for this sample of sulfur and oxygen is: S0 3

10. Steps for Determining Empirical Formulas 1.Assume 100 grams of compound 2. Use the percent composition of each element in grams equal to 100 grams 3. Use the mass of elements to calculate moles of each element (divide each mass by its molar mass from periodic table) 4. If you do not get whole numbers, pick the smallest number of moles 5. Divide each set by the smallest number of moles 6. If you do not get whole numbers, then multiply to get whole numbers (start with 2x) 7. Write the empirical formula using whole numbers as subscripts

11. Example Empirical Formula Calculation: A blue solid is found to contain 36.43% nitrogen and 63.16% oxygen. What is the empirical formula of this compound?