1. Chemical Formulas Chemistry 1

2. Oxidation Numbers  “oxidation states”  The number of electrons that must be added to or removed from an atom in a combined state to convert the atom into the elemental form.

3. Assigning Oxidation States Rules for Assigning Oxidation States The Oxidation State of…  An atom in an element is zero  Na(s), O 2 (g), Hg(l)  A monatomic ion is the same as its charge  Na +, Cl -  Fluorine is -1 in its compounds  HF, PF 3  Oxygen is usually -2 in its compounds  H 2 O, CO 2  Exception: Peroxides (containing O 2 2- ) in which oxygen is -1  Hydrogen is +1 in its covalent compounds  H 2 O, HCl, NH 3

4. Oxidation Numbers - Model Assign oxidation numbers to the atoms of each element in the following compounds.  PI 3  HF  CS 2  As 2 S 3  NO 2 -

5. Oxidation Numbers Practice Assign oxidation numbers to the atoms of each element in the following compounds.  CI 4  HCl  SO 3  IO 3  H 2 CO 3  SO 4 2-

6. Percent Composition  The percentage by mass of each element in a compound.

7. Steps For Calculations 1.Using the periodic table determine the molar mass of the compound. 2.Take each mass for each element (molar mass times number of atoms) and divide it by the total mass of the compound. 3.Multiply this value by 100 to obtain a percentage. 4.Percentages should add up to equal 100%

8. Percent Composition  Determine the percent composition in the following compounds:  PbBr 4  Ba(OH) 2

9. Percent Composition  Determine the percent composition in the following compounds:  NaOH  (NH 4 ) 3 PO 4  CCl 2 F 2  Pb(NO 3 ) 2

10. Bite the Bubble Lab  See handout

11. Review Practice

12. Empirical Formula  Empirical Formula : The formula with the smallest whole number ratio of the elements in the compound (simplest formula)  Note: May not always be the same as the molecular formula. 1.To calculate, convert masses of elements to moles.  If a question gives the percentages of each compound, assume you have 100 g of the sample and then each percentage becomes the mass of the element. 2.Next, divide each mole by the smallest one. 3.This gives the ratio of atoms, which you will use to write the formula.

13. Empirical Formulas  The percent composition of a compound was found to be 63.5% silver, 8.2% nitrogen, and 28.3% oxygen. Determine this compound’s empirical formula.  Name the compound.

14. Empirical Formulas  A 170.00g sample of a compound contains 29.84 g sodium, 67.49 g chromium and 72.67 g oxygen. What is the empirical formula?  Name the compound.

15. Empirical Formula  200.00 grams of an organic compound is known to contain 83.884 grams of carbon, 10.486 grams of hydrogen, 18.640 grams of oxygen and the rest is nitrogen. What is the empirical formula of the compound?

16. Molecular Formulas  Molecular Formulas : A formula that specifies the actual number of atoms of each element in one molecule of the substance. -Molecular formula is a multiple of the empirical formula. Steps for Solving:  Determine the Empirical Formula first!  Use the formula : N= The number we will multiply the subscripts in the empirical formula by to obtain the new molecular formula.

17. Empirical and Molecular Formulas  A compound of boron and hydrogen has a percent composition of 78.14% boron and 21.86% hydrogen. If the molar mass is 27.6, what is the empirical and molecular formula?

18. Empirical and Molecular Formulas  A compound with the formula C 2 H 5 O is found to have a molar mass of 90g. What is the molecular formula of the compound?

19. Individual Practice  Unit Packet Part 4 and 5

20. Chapter Review  Page 251  # 2, 6-8, 10, 11, 16-18, 21, 23-25, 32, 33, 36-41, 44, 47- 50  Test will cover…  New  Ionic, Molecular, and Acid Nomenclature  Oxidation Numbers and Percent Composition  Empirical and Molecular Formulas  Chemical Bonding  Periodic Table Trends