1. We measure ordinary objects either by counting or weighing them, depending on which method is more convenient Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. http://weyume.com/wp-content/uploads/2011/04/rice.jpg http://farm1.static.flickr.com/21/90994367_5613e69fd9.jpg

2. Dozen = 12 Pair = 2 Certain nouns can be used to define a collection of objects

3. The mole

4. The mole (n or mol) is the amount of matter that contains as many entities (atoms, molecules, ions, or other particles) as there are atoms in exactly 12 g of the carbon-12 isotope ( 12 C) The actual number of atoms in 12 g of carbon-12 was determined experimentally Avogadros number (N A ) N A = 6.02 x 10 23 Brown,, E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8 th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill.

5. Just as 1 dozen of oranges contains 12 oranges, 1 mole of matter contains 6.02 x 10 23 entities Brown,, E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8 th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill. 1 mole 12 C atoms = 6.02 x 10 23 12 C atoms 1 mole H 2 O molecules = 6.02 x 10 23 H 2 O molecules 1 mole NO 3 - ions = 6.02 x 10 23 NO 3 - ions

6. Each of these contains one mole of the substance Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill. copperiron carbonsulfur mercury

7. One mole (or an Avogadros number) is an extremely big number One mole of softdrink cans would cover the surface of the earth to a depth of over 300 kilometers If we were able to count the number of atoms at a rate of 10 million per second, it would take about 2 billion years to count a mole of atoms

8. Molar mass

9. The molar mass (M) of a substance is the mass of one mole of its entities (atoms, molecules, ions, or other particles) in units of g/mol M H2O = 18.0 g/mol (one mole of H 2 O molecule weighs 18.0 g) M NO3- = 62.0 g/mol (one mole of NO 3 - ion weighs 62.0 g) M C = 12.01 g/mol (one mole of C atom weighs 12.01 g) Brown, T., E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

10. The periodic table is indispensable for calculating the molar mass of a substance Elements –M is the numerical value from the periodic table Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. M H = 1.008 g/mol M O = 16.00 g/mol

11. The periodic table is indispensable for calculating the molar mass of a substance Compounds –M is the sum of the molar masses of the atoms of the elements in the formula Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. M SO2 = M S + (2 x M O ) = 32.07 g/mol + (2 x 16.00 g/mol) = 64.07 g/mol

12. The periodic table is indispensable for calculating the molar mass of a substance Compounds –M is the sum of the molar masses of the atoms of the elements in the formula Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. M K2S = (2 x M K ) + M S = (2 x 39.10 g/mol) + 32.07 g/mol = 110.27 g/mol

13. Interconverting moles, mass, and chemical entities (atoms, molecules, ions, or other particles)

14. The factor-label method is used to convert from one unit to another 1 peso = 4 25-centavos 1 peso 4 25-centavos = 1 1 peso 4 25-centavos = 1 unit factor Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill.

15. Alexa bought 18 fresh chickens eggs. How many dozens of egg did she buy? dozens of egg = x 18 eggs 1 dozen egg 12 eggs = 1.5 dozens of egg unit factor

16. In order to convert between moles, mass, and chemical entities, the factor label method is used Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

17. Methane (CH 4 ) is the principal component of natural gas. How many moles of methane are present in 6.07 g of CH 4 ? M CH4 = M C + (4 x M H ) = 12.01 g/mol + (4 x 1.01 g/mol) = 16.05 g/mol Report final answer with the correct number of significant figures! n CH4 = x 6.07 g CH 4 1 mol CH 4 16.05 g CH 4 = 0.378 mol CH 4

18. How many molecules of methane are present 6.07 g of CH 4 ? molecules CH 4 = x 6.07 g CH 4 1 mol CH 4 16.05 g CH 4 = 2.28 x 10 23 molecules CH 4 x 6.02 x 10 23 molecules CH 4 mol CH 4

19. Glucose (C 6 H 12 O 6 ), also known as blood sugar, is used by the body as energy source. How many moles of glucose are present in 1.75 x 10 22 molecules of glucose? n C6H12O6 = x 1.75 x 10 22 molecules C 6 H 12 O 6 1 mol C 6 H 12 O 6 6.02 x 10 23 molecules C 6 H 12 O 6 = 0.0291 mol C 6 H 12 O 6

20. How many grams of glucose are present in 1.75 x 10 22 molecules of glucose? M C6H12O6 = (6 x M C ) + (12 x M H ) + (6 x M O ) = (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol) = 180.18 g/mol

21. How many grams of glucose are present in 1.75 x 10 22 molecules of glucose? n C6H12O6 = x 1.75 x 10 22 molecules C 6 H 12 O 6 1 mol C 6 H 12 O 6 6.02 x 10 23 molecules C 6 H 12 O 6 180.18 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 = 5.24 g C 6 H 12 O 6

22. 1.Urea [(NH 2 ) 2 CO] is used in animal feeds, as fertilizer and in the manufacture of polymers. a.Draw the Lewis structure of area. C is surrounded by O and the Ns. (Where are the Hs connected?) b.Calculate its molar mass. c.Consider 25.6 g of urea. How many moles of urea present? d.How many moles of N are present? e.How many moles of C are present? f.How many molecules of urea are present? g.How many atoms of N are present?

23. 2. Vitamin C, ascorbic acid, is often sold as sodium ascorbate. a.Calculate its molar mass. b.Consider a 500.-mg tablet. How many moles of sodium ascorbate are present? c.How many moles of C are present? d.How many moles of Na are present? e.How many formula units of sodium ascorbate are present? f.How many atoms of Na are present?

24. Chemical reactions and chemical equations

25. A chemical reaction shows the process in which a substance (or substances) is changed into one or more new substances Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill.

26. A chemical equation uses chemical symbols to show what happens during a chemical reaction reactantsproduct (g) (l) Two molecules of hydrogen react with one molecule of oxygen to yield two moles of water Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill.

27. The Law of Conservation of Mass states that matter is neither created nor destroyed Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill.

28. To conform with the Law of Conservation of Mass, there must be the same number of each type of atom on both sides of the arrow. Hence, we balance the equation by adding coefficients before each chemical symbol Chang, R. 2002. Chemistry 7 th ed. Singapore: McGraw-Hill.

29. Calculating the amounts of reactant and product

30. Stoichiometry a double cheeseburger 2 bun slices + 2 cheese slices + 2 burger patties =

31. In a balanced equation, the number of moles of one substance is equivalent to the number of moles of any of the other substances 2CO (g) + O 2(g) 2CO 2(g) 2 mol CO = 1 mol O 2 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. 2 mol CO 1 mol O 2 = 1 1 mol O 2 2 mol CO = 1

32. In a balanced equation, the number of moles of one substance is equivalent to the number of moles of any of the other substances 2CO (g) + O 2(g) 2CO 2(g) 2 mol CO = 2 mol CO 2 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. 2 mol CO 2 mol CO 2 = 1 2 mol CO 2 2 mol CO = 1

33. In a balanced equation, the number of moles of one substance is equivalent to the number of moles of any of the other substances 2CO (g) + O 2(g) 2CO 2(g) 1 mol O 2 = 2 mol CO 2 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. 1 mol O 2 2 mol CO 2 = 1 2 mol CO 2 1 mol O 2 = 1

34. The amount of one substance in a reaction is related to that of any other Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

35. All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide 2Li (s) + 2H 2 O (l) 2LiOH (aq) + H 2(g)

36. How many moles of H 2 will be formed by the complete reaction of 6.23 moles of Li with water? n H2 = x 6.23 mol Li 1 mol H 2 2 mol Li = 3.12 mol H 2 2Li (s) + 2H 2 O (l) 2LiOH (aq) + H 2(g)

37. How many grams of H 2 will be formed by the complete reaction of 80.57 g of Li with water? 2Li (s) + 2H 2 O (l) 2LiOH (aq) + H 2(g) m H2 = x x 80.57 g Li 1 mol Li 1 mol H 2 6.941 g Li 2 mol Li 2.016 g H 2 x 1 mol H 2 = 11.70 g H 2

38. In a lifetime, the average American uses about 794 kg of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores (such as Cu 2 S) by a multistep process. After an initial grinding, the first step is to roast the ore (heat it strongly with O 2 ) to form Cu 2 O and SO 2 2Cu 2 S (s) + 3O 2(g) 2Cu 2 O (s) + 2SO 2(g)

39. How many moles of oxygen are required to roast 10.0 mol of Cu 2 S? 2Cu 2 S (s) + 3O 2(g) 2Cu 2 O (s) + 2SO 2(g) n O2 = x 10.0 mol Cu 2 S 3 mol O 2 2 mol Cu 2 S = 15.0 mol O 2

40. How many grams of SO 2 are formed when 10.0 mol of Cu 2 S is roasted? 2Cu 2 S (s) + 3O 2(g) 2Cu 2 O (s) + 2SO 2(g) m SO2 = x x 10.0 mol Cu 2 S 2 mol SO 2 64.07 g SO 2 2 mol Cu 2 S 1 mol SO 2 = 641 g SO 2

41. Ch 2 F No meeting this Friday Lab discussion moved to March 2 1:30-3:30 pm SOM 201

42. How many grams of O 2 are required to form 2.86 kg of Cu 2 O? 2Cu 2 S (s) + 3O 2(g) 2Cu 2 O (s) + 2SO 2(g) m O2 = x x 2.86 kg Cu 2 O 1000 g Cu 2 O 1 mol Cu 2 O 1 kg Cu 2 O 143.10 g Cu 2 O 3 mol O 2 32.00 g O 2 x 2 mol Cu 2 O 1 mol O 2 = 960 g O 2

43. Limiting Reactants The reactant used up first in a chemical reaction is called the limiting reactant. Excess reactants are present in quantities greater than necessary to react with the quantity of the limiting reactant. A + B --- C + D Given the amounts of A and B, which is the limiting reactant?

44. Urea is prepared by reacting ammonia with carbon dioxide: 2NH 3 (g) + CO 2 (g) -- (NH 2 ) 2 CO(aq) + H 2 O(l) In one process, 637.2 g of NH 3 are allowed to react with 1142 g of CO 2. (a) Which is the limiting reactant? (b) How much urea (in grams) is produced? (c) How much of the excess reactant (in grams) is left at the end of the reaction?

45. Strategy Convert mass of each reactant to moles Calculate the amount of product formed from the each of the reactants. The reactant the produces the less amount is the limiting reactant.

46. 1. The reaction between aluminum and iron (III) oxide can generate temperatures around 3000 C and is used in welding metals: 2Al + Fe 2 O 3 -- Al 2 O 3 + 2Fe In one process, 124 g of Al are reacted with 601 g of ferric oxide. (a)Which is the limiting reactant? (b)How much Al 2 O 3 (in grams) is produced? (c)How much of the excess reactant (in grams) is left at the end of the reaction? 2. Titanium is a strong & light metal used in rockets & aircrafts. It is prepared by the reaction between titanium (IV) chloride with molten magnesium at around 1000 C:TiCl 4 + 2Mg -- Ti + 2MgCl 2 In a certain industrial operation, 3.54 x 10 7 g of TiCl 4 are reacted with 1.13 x 10 7 g of magnesium. (a)Which is the limiting reactant? (b)How much Ti (in grams) is produced? (c)How much of the excess reactant (in grams) is left at the end of the reaction?