1. Topic 1 Quantitative Chemistry

2. Describe and Apply Mole [2-6] 1 mole = 6.02 x 10 23 – Avogadro’s constant 1 mole is the number of particles contained in exactly 12 g of Carbon-12 Applied to all particles: atoms, ions, molecules, formula units

3. “Representative Particles”

4. Mole problems 1)How many atoms are in 5.59 moles of S? 2)What is the mass of one atom of gold? 3)How many atoms are contained in 0.10 mole of C 2 H 4 ? 4)How many atoms are present in 36 molecules of C 6 H 2 O 6 ? 5)How many moles of hydrogen atoms are in 0.50 mole water? 6)How many chloride ions are in 0.30 moles of gold(III) chloride? 7)Exercises 1-4 page 6

5. Distinguish between atomic, molecular, and formula mass Atomic mass → 1 mole element Molecular mass → 1 mole molecules – Covalent compound Formula mass → 1 mole formula units – Ionic compound

6. Define empirical & molecular formula [7-9] Percent composition = proportion by mass of each element in compound converted to percent Empirical formula = simplest whole number ratio of atoms in a compound Molecular formula = actual number of atoms in a molecule, often multiple of empirical formula

7. Determine empirical & molecular formulas [7-9] Steps for Determining an Empirical Formula Step 1 If you have masses go onto step 2. If you have %. Assume the mass to be 100g, so the % becomes grams. Step 2 Determine the moles of each element. Step 3 Determine the mole ratio by dividing each elements number of moles by the smallest value from step 2. Step 4 Double, triple … to get an integer if they are not all whole numbers

8. Determine empirical & molecular formulas [7-9] Additional Steps for Determining Molecular Formula The question should have included a molecular mass. Step 5 Determine the mass of your empirical formula Step 6 Divide the given molecular mass by your E.F. mass in step 5 Step 7 Multiply the atoms in the empirical formula by this number

9. Determine empirical and molecular formulas [7-9] Exercises 7-9 page 9

10. Determine empirical & molecular formulas Steps Determining Formula with Combustion Analysis Step 1 Determine grams carbon in CO 2 and grams hydrogen in H 2 O. C: __g CO 2 x (12.011 g / 44.0098 g) = __g C H: __g H 2 O x (2.0158 g / 18.0152 g) = __g H Step 2 Subtract the mass of C and H from the sample mass to find the mass of O Step 3 Convert grams of C, H and O to their respective amount of moles. You can take it from here.

11. Determine empirical & molecular formulas Combustion analysis example 6.00 g of an organic compound(contains C, H, and O)form 8.80 g of CO 2 and 3.60 g of H 2 O when it undergoes complete combustion. What is its empirical formula? If the molecular mass is 90 g/mole, what is the molecular formula?

12. chemical equations [10-11] 1)Exercises 11 and 12 page 11

13. Apply State Symbols [10-11] (s) = solid (l) = liquid (aq)= aqueous solution (g) = gas Note-not many chemicals are pure liquids, water, bromine, mercury

14. Stoichiometry: Basic [11-12] Stoichiometry steps 1)Balance the equation 2)Convert units of given to moles 3)Find moles of unknown using mole ratio 4)Convert moles of unknown to desired units

15. Stoichiometry: Basic [11-12] Remember these conversions: Mass to mole Gas volume to mole Particles to mole

16. Stoichiometry: Basic [11-12] Exercise 13 page 13 Exercises 21 and 22 page 24

17. Stoichiometry: Theoretical Yield [13] Stoichiometry calculate the most product that can be made from the amount of reactant under perfect and complete conditions. – Therefore the theoretical yield Exercise 14 page 13

18. Stoichiometry: Limiting reactant [14] and percent yield [15] Exercises 15 and 16 page 16

19. Stoichiometry: excess reactant [14] Calculate the amount of excess reactant in exercises 15 and 16 page 16

20. Stoichiometry: Solutions [28-33] You must be able to calculate concentration of solutions Molarity (M) = moles solute dm 3 solution Remember if you know moles of a substance, you can calculate mass and visa versa dm 3 = L cm 3 = mL

21. Stoichiometry: Solutions [28-33] Exercises 28-30 page 30

22. Stoichiometry: Solutions [28-33] 1)500 mL of 0.500 M NaCl is added to 500 mL of 1.00 M Na 2 CO 3. Calculate the final concentration of Na + ions. 2)1.86 g of lead (II) carbonate is added to 50.0 mL (an excess) of nitric acid. What is the concentration of lead (II) nitrate in the resulting solution?

23. Stoichiometry: Solutions [28-33] Exercises 31-33 page 33

24. States of Matter [17]

25. Kelvin temperature scale [17] Temperature in Kelvin is directly proportional to kinetic energy – Higher K, higher KE K = °C + 273.15 Absolute zero = 0 K

26. Changes of state [17-18]

28. Gas Laws [24-28] Avogadro’s Law: Equal number of moles of any gas at the same temperature and pressure will occupy equal volume Dalton’s Law: The sum of partial pressures of gases in a mixture will equal the total pressure Graham’s Law: The smaller the molar mass of a gas, the faster it will travel

29. Gas Laws [24-28] CombinedBoyle’s Charles’Gay-Lussac

30. Gas Laws [24-28] Ideal gas law: PV = nRT The ideal gas law can be rearranged to solve for molar mass of a gas and density of a gas

31. Gas Laws [24-28] Exercises 23-24 page 27

32. Gas Laws [24-28] Exercises 25-27 page 28