1. Stoichiometry Study of mass and chemical amounts. 1. Compound stoichiometry 2. Reaction stoichiometry

2. Connecting the Macro and Atomic Scales: The Mole Chemical macroscopic counting unit: 1 mole contains 6.022 x 10 23 particles 6.022 x 10 23 is Avogadro’s number, named after Amedeo Avogadro. Avogadro’s number is not special, like pi, but is invented by us for our convenience.

3. Why it works. The mass of 1 mol of an element is equal to its atomic mass in grams Molar mass is in g/mol Molar mass of Al = 26.98 g/mol Unit sign: mole = mol sometimes, molar mass = M

4. How it works Two new conversion factors: Avogadro’s number: Molar mass:

5. How many atoms are in 2.60 mol of Ne?

6. How many atoms are in 3.84 mol of B? 1.3.84 2.2.31 x 10 25 3.2.31 x 10 24 4.6.38 x 10 -24

7. How many moles are in 4.00 x 10 22 atoms of Ne?

8. How many moles are in 6.84 x 10 21 atoms of B? 1.6.84 x 10 21 mol 2.0.0114 mol 3.4.12 x 10 45 mol 4.6.84 x 10 -21 mol

9. How many moles are in 45.0 g of Ne?

10. How many moles are in 45.0 g of Al? 1.1.67 mol 2.1214 mol 3.7.47 x 10 -23 mol 4.2.71 x 10 25 mol

11. What is the mass of 2.60 mol of C?

12. What is the mass of 2.60 mol of O? 1.8.90 g 2.0.163 g 3.2.60 g 4.41.6 g

13. How many moles are represented by 1.00 g of C?

14. How many atoms are in 1.00 g of O?

15. What is the mass of one Au atom?

16. How many C atoms are in a 1.00 caret diamond?

17. Compounds and Moles 1 mole of a compound contains 6.022 x 10 23 molecules or “units” Molar mass = sum of atomic molar masses –H 2 O –CO 2 –Fe(NO 3 ) 3

18. Molecules, Atoms and Moles Consider UF 6 –0.50 mol UF 6 contains …

19. Language for nonmolecular compounds … 1 mol CO 2 contains 6.022 x 10 23 molecules 1 mol NaCl contains 6.022 x 10 23 formula units

20. Percent Composition What fraction of a compound is made up of one of its elements? Consider FeO …

22. Determining Formulas from Percent Composition The key: the ratio of atoms in the formula is the same as the ratio of moles of those elements

23. Determining Formulas from Percent Composition The key: the ratio of atoms in the formula is the same as the ratio of moles of those elements If you determine the ratio of moles, you know the formula.

24. Determining Formulas from Percent Composition The key: the ratio of atoms in the formula is the same as the ratio of moles of those elements If you determine the ratio of moles, you know the formula. But! You get the empirical formula.

25. Example: A compound has Ca, S, and O. Ca:29.45% S:23.55% O:47.00% What is the empirical formula?

28. Writing Lab Reports

29. What time is it? Is it quiz time?

30. Last Name: A – M 1.What is the charge on a magnesium ion? 2.What is the charge on a sulfide ion? 3.What is the formula of sulfate ion? 4.What is the name of this ion: MnO 4 - ? 5.What element is below nitrogen on the periodic table? 6.What element is to the left of Be on the periodic table? Last Name: N – Z 1.What is the charge on a nitride ion? 2.What is the charge on an aluminum ion? 3.What is the formula of nitrate ion? 4.What is the name of this ion: SO 3 2- ? 5.What element is below oxygen on the periodic table? 6.What element is to the left of C on the periodic table?

31. Empirical vs. Molecular Formulas Example: Ethene is C 2 H 4 Percent composition tells us mol H/mol C = 2 The empirical formula is CH 2. The molecular formula is C 2 H 4.

32. Example: A hydrocarbon has 82.65% C and 17.34% H Molar mass is 58.12 g/mol What are the empirical and molecular formulas?

34. Hydrated Compounds Solids in which molecules of water are trapped and become part of the compound. Ex: Gypsum: CaSO 4 2H 2 O is hydrated calcium sulfate CaSO 4 is anhydrous calcium sulfate

36. Determining the number of waters of hydration 1.023 g CuSO 4 x H 2 O is heated to drive off the water. The resulting anhydrous CuSO 4 has a mass of 0.654 g. What is the value of x?

38. Next Week’s Lab You have a mixture of CuSO 4 5H 2 O and CuSO 4. The total mass is 4.300 g. After heating, the new mass is 3.658 g. What are the percentages of each compound in the mixture? CuSO 4 : 159.61 g/mol CuSO 4 5H 2 O: 249.69 g/mol

39. Initial mass: 4.300 g. After heating: 3.658 g. CuSO 4 : 159.61 g/mol CuSO 4 5H 2 O: 249.69 g/mol

40. Example: You react 0.125 g Ni with CO and form 0.364 g Ni(CO) x. What is the value of x?