1. % composition I can determine the % by mass of any element in a compound.

2. % composition The percentage composition is the percentage by mass of each element in a compound

3. Percentage Composition Mass of element in sample of compound / mass of sample of compound X 100 = % element in compound. Why do we do this. Well if you need Oxygen but all you have is KCLO 3, you will need to find out how much oxygen you have in your compound so that you know how much can be used.

4. % Composition Cont.. The mass percentage of an element in a compound is the same regardless of the sample’s size. Therefore, a simpler way to calculate the percentage of an element in a compound is to determine how many grams of the element are present in one mole of the compound. Then divide this value by the molar mass of the compound and multiply by 100.

5. Practice Determine the % comp of each of the following –NaCl –AgNO 3 –Mg(OH) 2

6. Empirical Formula Calculations I can determine the empirical formula for any compound given the % composition or mass of the components.

7. Empirical Formulas Empirical formula consist of the symbol for the elements combined in a compound, with subscripts showing the smallest whole number ratio of the different atoms in the compound. Ionic compounds the formula unit is usually the compound’s empirical formula. Molecular compounds the formula is not always the empirical formula. Example: BH 3 is the empirical formula for B 2 H 6.

8. Calculating the Empirical Formula You can determine the empirical formula from the % composition. 1. You need to turn the % into grams or find the grams of each substance percent. 2. change the mass to moles. (so divide by the molar mass) 3. this gives you the mole ratio. (however you do not have the smallest mole ratio so divide all answers by the smallest number in the existing ratio. 4. If not whole numbers, round to the nearest.

9. Example: A compound is composed of 32.38 % sodium, 22.65% sulfur, and 44.99% oxygen what is the empirical formula? 1. 32.38 g, 22.65 g, 44.99g, 2. change to moles –32.38 g Na X 1mol/22.99g Na = 1.408 mol Na –22.65 g S X 1 mol S/32.07 g s = 0.7063 mol S –44.99 g O X 1 mol / 16 g O = 2.812 mol O 3. smallest mole ratio –1.408 / 0.7063 =1.993 mol –0.7063/ 0.7063 = 1 mol –2.812 mol O/ 0.7063 = 3.981 mol –Ratio is 2:1:4 –Empirical Formula is Na 2 SO 4

10. Empirical Formulas when you are given grams In a 10.150 g sample of a compound that contains 4.433 g of phosphorus and some oxygen, what is the empirical formula. 1. If you have 4.433 g P and the total is 10.150 g if we subtract we will get the amount of O. 10.150 g – 4.433 g P = 5.717g O. 2. change to moles: –4.433 g P X 1 molP/30.97 g P = 0.1431 mol P –5.717 g O X 1 mol O/16 g O = 0.3573 mol O 3. smallest mole ratio –0.1431 mol P / 0.1431 = 1 mol P –0.3573 mol O / 0.1431 = 2.497 mol O –Empircial formula is 2:5, THIS IS NOT WHAT YOU WOULD THINK! BECAUSE 2.497 IS NOT COLEST TO 3 OR 2 WE ADD A STEP. SO WE MULTIPLE EACH NUMBER IN THE RATIO BY 2. –1X2 = 2 2.497 X 2 = 5

11. PRACTICE: Chemical analysis of citric acid shows that it contains 37.51 % C, 4.20% H, and 58.295% O. What is it’s empirical formula?

12. Molecular Formulas I can determine the molecular formula of a molecule if the appropriate mass is given.

13. Molecular Formulas The molecular formula is not always the empirical formula. Look at the following: B 2 H 6 B 3 H 9 B 4 H 12 BH 3 All of these are different molecules, however they all have the same empirical formula of BH 3

14. Molecular formulas Cont. The relationship between the empirical formula and the molecular formula can be written as follows: X (empirical formula) = molecular formula X (empirical formula mass )= molecular formula mass. If the empirical formula mass can be found and the molecular formula mass is given you can divide to find X. then you multiple the empirical formula by X to get the molecular formula.

15. Example: The empirical formula is P 2 O 5 Experimentation shows that the molar mass of this compound is 283.89 g/mol. What is the compound’s molecular formula? 1. x = molecular mass / empirical mass –283.89 g/mol (given) divided by the empirical formula mass (2 X 30.97 amu + 5 x 16.00amu) 141.94 amu. –X = 2.001 2. X (empirical formula) = 2(P 2 O 5 ) 3. molecular formula is P 4 O 10

16. Practice: IF 4.04 g of N combined with 1.46 g O to produce a compound with a formula mass of 108.0 amu, what is the molecular formula of this compound? (hint: you have to find the empirical formula first)