1. What are chemical formulas? What can a chemical formula tell us? Number of atoms of a particular element present in a chemical compound – Ex. CO 2, NO 2 For ionic compounds, the ratio between the positive and negative ions within the chemical compound – Ex. NaCl, MgCl 2 Monoatomic ions— – Ions originating from ONE atom – Cation—named by element – Anion—named by element, drop ending and add –ide ending – Ex. O -2, N -3, Mg +2

2. Binary Compounds—Writing Chemical formulas 1) Cation, Anion. (write element names) 2) Write oxidation number/charge on each element. 3) Balance charges by the smallest common multiplier between the two elements. Charges must add to 0 for a neutral compound. 4) Add the subscripts for each element in the chemical compound.

3. 5) If the first element is a transition element, the oxidation number or charge of the element is given by a Roman numeral (I, II, III, IV, etc.) following the element name. Exceptions: Ag +1 and Zn +2 —they will always have these charges.

4. Binary Compounds—Writing Chemical formulas (Cont.) 6) If a binary compound has a prefix or prefixes in front of an element name, these prefixes are the subscripts for that element in the chemical formula. Prefixes represent the number of element atoms present in a molecule of the chemical compound.

5. Prefixes 1 mono6 hexa 2 di7 hepta 3 tri8 octa 4 tetra9 nona 5 penta10 deca

6. Example 1: Aluminum oxide

7. Example 2: Tin (IV) sulfide

8. Example 3: Iron (III) oxide

9. Example 4: Carbon monoxide

10. Polyatomic Chemical Compounds—Writing Chemical Formulas. 1) Cation, Anion—reference table, Element Symbols 2) Immediately put ( ) around polyatomic ion 3) Write the oxidation number/charge on each element and polyatomic ion. 4) Balance charges by the smallest common multiplier. Charges must add to 0 for a neutral compound. 5) Add the subscripts for each element/polyatomic ion and drop the ( ) around polyatomic if you can. *Polyatomic ions: place polyatomic ions in parentheses when adding subscript (ex. (NO 3 ) 2 )

11. Example 1: Silver nitrate

12. Example 2: Magnesium acetate

13. Binary Compounds: Chemical Nomenclature 1) Write an element’s oxidation number over the element symbol. 2) Cation, Anion—write the element name 3) Add the ending -IDE to the second element/anion

14. Example 1: LiCl

15. Example 2: Al 2 S 3

16. Transition Elements 4) Place a Roman numeral matching the charge/oxidation # on an element after the positive element’s name. Roman numeral is put in () (ex. (II)). Roman numeral indicates the oxidation number on an element can vary. -Go back to Step #3

17. Example 1: CuS

18. Example 2: CoBr 2

19. 5) If both elements are found on the RIGHT side of the periodic table, use the appropriate prefix for a given element. -Go back to Step #3

20. Example 1: CO 2

21. Example 2: CCl 4

22. Compounds with Polyatomic Ions: Chemical Nomenclature 1) Write oxidation # for each element/polyatomic ion in the formula 2) Write the element and/or polyatomic ion names. 3) If a transition element (an element where the oxidation # varies), use a Roman numeral to indicate an element’s oxidation # in the chemical name. *NO –IDE ENDING!!!!

23. Example 1: ZnCO 3

24. Example 2: CuSO 4

25. Chemical Equations ! !

26. Law of Conservation of Matter: Matter cannot be created or destroyed. AMOUNT OF REACTANTS = AMOUNT OF PRODUCTS ! ! ! !

27. Guidelines for Balancing Chemical Equations: 1) Do NOT change the subscripts ! 2) Every element MUST have the same amount of atoms on both sides of the equation. 3) To balance, coefficients (numbers) are added in front of the chemical formula until we have the same atom amount on both sides of the equation. 4) Once a coefficient is added, go back and check the rest of the equation 5) Balance every element EXCEPT Hs and Os. Then balance Hs, Os are balanced last. Balance Hs with an EVEN number 6) CHECK EQUATION TO BE SURE EVERYTHING BALANCES/CANCELS OUT ON BOTH SIDES ! ! !

28. Example 1: Mg + HCl  MgCl 2 + H 2

29. Example 2: KI (aq) + Pb(NO 3 ) 2  KNO 3 + PbI 2

30. Example 3: FeI 3 + K 2 SO 4  KI + Fe 2 (SO 4 ) 3

31. The Mole

32. The MOLE Main unit in chemistry we will use ! ! ! Describes the amount of atoms in a particular substance 1 mole = 6.02 x 10 23 atoms (KNOW THIS CONVERSION) 1 mole = formula mass (g) of a compound = atomic mass (g) of an element = 22.4 L of a solution under STP

33. Molar Mass = mass of 1 mole of a substance Units = grams/mole or g/mol

34. HOW many atoms are in 1 mole? 1 mole = 6.02 x 10 23 atoms “Avogadro’s number”

35. Example 1: Convert 10g CuSO 4 to moles

36. Example 2: Convert 2.5 moles of NaOH to molecules

37. Example 3: Convert 3g NaCl to molecules.

38. Stoichiometry

39. Example 1: How many grams of NH 3 will be formed from the reaction of 26 grams of N 2 with excess H 2 ? N 2 + 3H 2  2NH 3

40. Example 2: How many liters of CO 2 are produced when 25.0 grams of C 2 H 6 combusts? 2C 2 H 6 + 7O 2  4CO 2 + 6H 2 O

41. Example 3: How many grams of aluminum oxide will be produced when 12 grams of aluminum react with excess oxygen gas?

42. Example 4: For the following reaction: 2AgNO 3 + Ni  2Ag + Ni(NO 3 ) 2 112 grams of AgNO 3 reacts with 22.9 grams of Ni to produce calcium carbonate. Find the theoretical yield of nickel (II) nitrate and the limiting reactant. If the actual yield in a lab experiment was 57.3g of Ni(NO 3 ) 2, what is the percent yield for this lab?

43. Solution Chemistry

44. Solutions Uniform mixtures, composition is the same throughout Solvent—substance doing the dissolving, present in a larger amount. Solute—dissolved substance, present in a smaller amount, most of the time goes through phase change. Colloid—mixtures of particles existing between a suspension and a solution Suspension—mixture with larger particles settling to the bottom unless stirred.

45. Molarity (M) Main unit of concentration in chemistry Moles solute L solution First step—convert given values into moles and liters!! Get everything in the correct units!!

46. Example 1: What is the concentration/molarity of a 50.0 ml solution with 1.25g of CuSO 4 in water?

47. Example 2: Find the number of moles needed to make 250 ml of a 0.150 M solution of Na 2 SO 4.

48. Homework Review Packet #2