Unit #5

1. Unit #5 Polyatomic Ions / Mole / Molarity / Electrochemistry

2. This unit builds upon previous topics: Electron Configuration Orbital Notation Electron Dot Notation Octet Rule

3. Remember: • Ion- • an atom or group of atoms that has either lost or gained electron(s) and as a result has an electric charge. Cation- is a positively charged ion. Anion- is a negatively charged ion.

4. Recall how we represented atoms with the electron dot notation.

5. We will represent ions the same way with two exceptions: • 1) If the atom has at least 5, but less than 8 valence electrons; we will draw as many additional electrons as needed to acquire a total of 8 valence electrons. (This will give the ion an Octet of valence electrons.)

6. 1st exception con’t: • After the "atomic symbol" of the ion, in the upper right corner of the symbol, we will write a negative sign (or digit, if more than one) for each additional electron we added to achieve the nearest noble gas core. An octet of valence electrons.

7. 1st exception con’t: Example #1:

8. 1st exception con’t: Example #2: Bromine

9. 1st exception con’t: Example #3: Oxygen

10. We will represent ions the same way with two exceptions (con’t.): • 2) If the atom has less than 4 valence electrons; we will remove as many electrons as needed to move back to the nearest noble gas core. A total of 8 valence electrons. (This will give the ion an Octet of valence electrons.)

11. 2nd exception con’t: • After the "atomic symbol" of the ion, in the upper right corner of the symbol, we will write a positive sign (or digit, if more than one) for each additional electron we subtracted to achieve the nearest noble gas core. An octet of valence electrons.

12. 2nd exception con’t: • Example #1: Sodium

13. 2nd exception con’t: • Example #2: Aluminum

14. 2nd exception con’t: • Example #3: Barium

15. Polyatomic ion- A type of ion made of more than 2 atoms. These will also be classified as either an anion or a cation.

16. 4 steps to drawing polyatomic ions: Step 1) Determine and write down the "4 key numbers". These are: p e c v

17. "4 key numbers” (con’t.) • “p” # of protons (add up the atomic number of all involved atoms). • “e” # of total electrons (add up the atomic numbers (which equals the number of total electrons in a neutral atom) of all atoms in the ion.

18. "4 key numbers” (con’t.) • “c” # of core electrons (atomic number minus the number of "s" and "p" electrons in the highest energy level). • “v” # of valence electrons (determined by subtracting the # of core electrons from the # of total electrons).

19. Draw the "skeleton" of the ion • Step 2) Step 3) • Give all atoms eight electrons (except Hydrogen, it only wants two), (remember each bond (represented by a dash) counts as 2 electrons).

20. Count the number of electrons drawn and compare this to the calculated number ("Key number #4” the “v” ). • Step 4) • Adjust as necessary by adding double or triple bonds to reduce the number of electrons while still ensuring all atoms still have an octet of valence electrons. • Draw brackets and valence charge to finish.

21. Examples: Carbonate Cyanide Ammonium

22. The Mole Concept.

23. Mole - The amount of substance that contains Avagadro's number of particles of that substance. Abbreviated: mol. NOT m, M, or M.

24. Avagadro's number - • This is equal to 6.02 * 1023 . • It is specifically the number of C12 atoms in 12 grams (the chart mass in grams) of this isotope. • Think of this word like the word "Dozen", which always means twelve. Well this number always means 6.02 * 1023 particles.

25. Atomic weight - The weighted average of the masses of the isotopes of an element. • Based on C12 , 1 atom = 12 AMUs (atomic mass units). AMU - arbitrary unit equal to 1/12 the mass of a C12 atom. (Atomic Mass Unit)

26. AMU continued: • Imagine if we had to determine the "weight" of everyone in the classroom, but, we didn't have a scale. We could select someone and then assign to that individual an arbitrary "weight" unit. We might say he/she weight 10 CRUs (Chem. Room Units). Everyone in the room will now be weighed relative to that student. • The atomic weight of all elements is relative to the "weight" of carbon 12.

27. Gram Atomic Weight - • The mass in grams of 1 mole of a substance. Example 1: Hydrogen. 1 atom weighs ? 1.0079 amu 1 mole of hydrogen atoms weighs ? 1.0079 g. However what is special about Hydrogen ? Its diatomic

28. H2 • What is the formula for a molecule of Hydrogen? 1 molecule of Hydrogen weighs ? 2.0158 amu 1 mole of Hydrogen gas weighs ? 2.0158 g.

29. Example #2: CO2 1 molecule of Carbon dioxide weighs ? 44.0098 amu 1 mole of Carbon dioxide weighs ? 44.0098 g.

30. Make sure you are aware of what you are working with: Atoms or Molecules AMUs or Grams

31. Molarity - Is a concentration unit in: #moles / liter. (The liter is usually water) Note that the numerator is # of moles. We will learn how to convert this to grams in the next topic. • Abbreviated: MNOT mol, M, m, or mol.

32. DIMO - An acronym that means: Divide In Multiply Out. It is a drawing that you have to be able to reproduce and use.

33. # of grams # of particles Avagadro’s # Chart mass # of moles 22.4 # of liters of a gas

34. 4 Steps to use the DIMO chart. 1 ) Determine the chart mass of the substance you're working with. • 2) Deal with the concentration of that substance. • 3) Deal with the volume of the solution you're working with. • 4) Use "DIMO" to solve.

35. Example 1: • How many grams of NaOH are in 1 liter of a 1M solution of the NaOH? STEP 1: Determine the chart mass

36. Example 1 (con’t.): STEP 1: 40 g/mol STEP 2: Deal with the concentration

37. Example 1 (con’t.): STEP 1: 40 g/mol STEP 2: 1 mol/L STEP 3: Deal with the volume

38. Example 1 (con’t.): STEP 1: 40 g/mol STEP 2: 1 mol/L STEP 3: 1 mol of NaOH in sample STEP 4: Use "DIMO" to solve

39. Solution to Example 1: • How many grams of NaOH are in 1 liter of a 1M solution of the NaOH? There are 40 g of NaOH in 1 liter of a 1M solution of NaOH.

40. Example 2: • 0.75 liters of a 0.5M solution contains how many grams of CaCl2? STEP 1: Determine the chart mass

41. Example 2 (con’t.): STEP 1: 111 g/mol STEP 2: Deal with the concentration

42. Example 2 (con’t.): STEP 1: 111 g/mol STEP 2: 0.5 mol/L STEP 3: Deal with the volume

43. Example 2 (con’t.): STEP 1: 111 g/mol STEP 2: 0.5 mol/L STEP 3: 0.375 mol of CaCl2 in sample STEP 4: Use "DIMO" to solve

44. Solution to Example 2: • How many grams of CaCl2 are in 0.75 liter of a 0.5M solution of the CaCl2? There are 41.625 g of CaCl2 in 0.75 liter of a 0.5M solution of CaCl2.

45. Notice in the previous examples we calculated the number of grams of a substance. • In the next example, we will determine how many moles of a substance are in a given volume of a solution.

46. Example 3: • What is the Molarity of a solution created by dissolving 150 grams of NaI into 250 mL of distilled water? STEP 1: Determine the chart mass

47. Example 3 (con’t.): STEP 1: 150 g/mol STEP 2: Deal with the concentration

48. Example 3 (con’t.): STEP 1: 150 g/mol STEP 2: 150 g / 0.25 L STEP 3: Deal with the volume

49. Example 3 (con’t.): STEP 1: 150 g/mol STEP 2: 150 g / 0.25 L STEP 3: 600 g / 1 L STEP 4: Use "DIMO" to solve

50. Solution to Example 3: • What is the Molarity of a solution created by dissolving 150 grams of NaI into 250 mL of distilled water? 600 g of NaI is 4 moles There are 4 moles of NaI in 1 liter of thesolution of NaI. Therefore the molarity is ____ 4M