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Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

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

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Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

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

  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 (12C) • The actual number of atoms in 12 g of carbon-12 was determined experimentally • Avogadro’s number (NA) NA = 6.02 x 1023 Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

  5. Just as 1 dozen of oranges contains 12 oranges, 1 mole of matter contains 6.02 x 1023 entities 1 mole 12C atoms = 6.02 x 1023 12C atoms 1 mole H2O molecules = 6.02 x 1023 H2O molecules 1 mole NO3- ions = 6.02 x 1023 NO3- ions Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

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

  7. One mole (or an Avogadro’s 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 MC = 12.01 g/mol (one mole of C atom weighs 12.01 g) MH2O = 18.0 g/mol (one mole of H2O molecule weighs 18.0 g) MNO3- = 62.0 g/mol (one mole of NO3- ion weighs 62.0 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. 2nd 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 MH= 1.008 g/mol MO= 16.00 g/mol Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

  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 MSO2= MS + (2 x MO) = 32.07 g/mol + (2 x 16.00 g/mol) = 64.07 g/mol Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

  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 MK2S= (2 x MK) + MS = (2 x 39.10 g/mol) + 32.07 g/mol = 110.27 g/mol Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

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

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

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

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

  17. 6.07 g CH4 1 mol CH4 nCH4 = x 16.05 g CH4 = 0.378 mol CH4 Methane (CH4) is the principal component of natural gas. How many moles of methane are present in 6.07 g of CH4? MCH4= MC + (4 x MH) = 12.01 g/mol + (4 x 1.01 g/mol) = 16.05 g/mol Report final answer with the correct number of significant figures!

  18. 6.07 g CH4 1 mol CH4 molecules CH4 = x 16.05 g CH4 6.02 x 1023 molecules CH4 x mol CH4 = 2.28 x 1023 molecules CH4 How many molecules of methane are present 6.07 g of CH4?

  19. 1.75 x 1022 molecules C6H12O6 1 mol C6H12O6 nC6H12O6 = x 6.02 x 1023 molecules C6H12O6 = 0.0291 mol C6H12O6 Glucose (C6H12O6), also known as blood sugar, is used by the body as energy source. How many moles of glucose are present in 1.75 x 1022 molecules of glucose?

  20. How many grams of glucose are present in 1.75 x 1022 molecules of glucose? MC6H12O6= (6 x MC) + (12 x MH) + (6 x MO) = (6 x 12.01 g/mol) + (12 x 1.01 g/mol) + (6 x 16.00 g/mol) = 180.18 g/mol

  21. 1.75 x 1022 molecules C6H12O6 1 mol C6H12O6 nC6H12O6 = x 6.02 x 1023 molecules C6H12O6 180.18 g C6H12O6 x 1 mol C6H12O6 = 5.24 g C6H12O6 How many grams of glucose are present in 1.75 x 1022 molecules of glucose?

  22. Chemical reactions and chemical equations

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

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

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

  26. 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 7th ed. Singapore: McGraw-Hill.

  27. Calculating the amounts of reactant and product

  28. Figure on slices of bread + ham to make sandwich. Use this to relate to stoich

  29. 2 mol CO 1 mol O2 = 1 = 1 1 mol O2 2 mol CO 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) + O2(g)  2CO2(g) 2 mol CO = 1 mol O2 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

  30. 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) + O2(g)  2CO2(g) 2 mol CO = 2 mol CO2 2 mol CO 2 mol CO2 = 1 = 1 2 mol CO2 2 mol CO Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

  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) + O2(g)  2CO2(g) 1 mol O2 = 2 mol CO2 1 mol O2 2 mol CO2 = 1 = 1 2 mol CO2 1 mol O2 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

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

  33. 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide

  34. 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) 6.23 mol Li 1 mol H2 nH2 = x 2 mol Li = 3.12 mol H2 How many moles of H2 will be formed by the complete reaction of 6.23 moles of Li with water?

  35. 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) 80.57 g Li 1 mol Li 1 mol H2 mH2 = x x 6.941 g Li 2 mol Li 2.016 g H2 x 1 mol H2 = 11.70 g H2 How many grams of H2 will be formed by the complete reaction of 80.57 g of Li with water?

  36. 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 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 Cu2S) by a multistep process. After an initial grinding, the first step is to “roast” the ore (heat it strongly with O2) to form Cu2O and SO2

  37. 10.0 mol Cu2S 3 mol O2 nO2 = x 2 mol Cu2S 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) = 15.0 mol O2 How many moles of oxygen are required to roast 10.0 mol of Cu2S?

  38. 10.0 mol Cu2S 2 mol SO2 64.07 g SO2 mSO2 = x x 2 mol Cu2S 1 mol SO2 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) = 641 g SO2 How many grams of SO2 are formed when 10.0 mol of Cu2S is roasted?

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

  40. 2.86 kg Cu2O 1000 g Cu2O 1 mol Cu2O mO2 = x x 1 kg Cu2O 143.10 g Cu2O 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) 3 mol O2 32.00 g O2 x x 2 mol Cu2O 1 mol O2 = 960 g O2 How many grams of O2 are required to form 2.86 kg of Cu2O?

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