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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 Certain nouns can be used to define a collection of objects Dozen = 12 Pair = 2 The mole 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. 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. Each of these contains one mole of the substance carbon sulfur mercury copper iron Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. 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 Molar mass 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. 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. 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. 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. Interconverting moles, mass, and chemical entities (atoms, molecules, ions, or other particles) The factor-label method is used to convert from one unit to another 1 peso = 4 25-centavos 1 peso 4 25-centavos unit factor 4 25-centavos 1 peso =1 =1 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Alexa bought 18 fresh chicken’s eggs. How many dozens of egg did she buy? unit factor dozens of egg = 18 eggs x 1 dozen egg 12 eggs = 1.5 dozens of egg 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. 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 nCH4 = 6.07 g CH4 x 1 mol CH4 16.05 g CH4 = 0.378 mol CH4 Report final answer with the correct number of significant figures! How many molecules of methane are present 6.07 g of CH4? molecules CH4 = 6.07 g CH4 x 1 mol CH4 16.05 g CH4 x 6.02 x 1023 molecules CH4 mol CH4 = 2.28 x 1023 molecules CH4 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? nC6H12O6 = 1.75 x 1022 molecules C6H12O6 x 1 mol C6H12O6 6.02 x 1023 molecules C6H12O6 = 0.0291 mol C6H12O6 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 How many grams of glucose are present in 1.75 x 1022 molecules of glucose? nC6H12O6 = 1.75 x 1022 molecules C6H12O6 x 1 mol C6H12O6 6.02 x 1023 molecules C6H12O6 x 180.18 g C6H12O6 1 mol C6H12O6 = 5.24 g C6H12O6 1. Urea [(NH2)2CO] 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 N’s. (Where are the H’s 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? 2. Vitamin C, ascorbic acid, is often sold as sodium ascorbate. OH HO O O a. Calculate its molar mass. Na+ O OH 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? Chemical reactions and chemical equations 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. A chemical equation uses chemical symbols to show what happens during a chemical reaction (g) (g) reactants (l) product “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. The Law of Conservation of Mass states that matter is neither created nor destroyed Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. 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. Calculating the amounts of reactant and product Stoichiometry a double cheeseburger 2 bun slices + 2 cheese slices + 2 burger patties = 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 2 mol CO 1 mol O2 =1 1 mol O2 =1 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. 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 2 mol CO2 =1 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. 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 2 mol CO2 =1 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. 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. All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) How many moles of H2 will be formed by the complete reaction of 6.23 moles of Li with water? 2Li(s) + 2H2O(l) 6.23 mol Li nH2 = x 2LiOH(aq) + H2(g) 1 mol H2 2 mol Li = 3.12 mol H2 How many grams of H2 will be formed by the complete reaction of 80.57 g of Li with water? 2Li(s) + 2H2O(l) 80.57 g Li mH2 = 2LiOH(aq) + H2(g) 1 mol Li x x 6.941 g Li x 2.016 g H2 1 mol H2 = 11.70 g H2 1 mol H2 2 mol Li 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 2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g) How many moles of oxygen are required to roast 10.0 mol of Cu2S? 2Cu2S(s) + 3O2(g) nO2 = 10.0 mol Cu2S 2Cu2O(s) + 2SO2(g) x 3 mol O2 2 mol Cu2S = 15.0 mol O2 How many grams of SO2 are formed when 10.0 mol of Cu2S is roasted? 2Cu2S(s) + 3O2(g) mSO2 = 10.0 mol Cu2S 2Cu2O(s) + 2SO2(g) x 2 mol SO2 2 mol Cu2S = 641 g SO2 x 64.07 g SO2 1 mol SO2 Ch 2 F • • • • No meeting this Friday Lab discussion moved to March 2 1:30-3:30 pm SOM 201 How many grams of O2 are required to form 2.86 kg of Cu2O? 2Cu2S(s) + 3O2(g) mO2 = 2.86 kg Cu2O x 2Cu2O(s) + 2SO2(g) 1000 g Cu2O x 1 kg Cu2O x 3 mol O2 2 mol Cu2O = 960 g O2 x 32.00 g O2 1 mol O2 1 mol Cu2O 143.10 g Cu2O 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? Urea is prepared by reacting ammonia with carbon dioxide: 2NH3(g) + CO2(g) -- (NH2)2CO(aq) + H2O(l) In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2. (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? 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. 1. The reaction between aluminum and iron (III) oxide can generate temperatures around 3000⁰C and is used in welding metals: 2Al + Fe2O3 -- Al2O3 + 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 Al2O3 (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 1000⁰C: TiCl4 + 2Mg -- Ti + 2MgCl2 In a certain industrial operation, 3.54 x 107g of TiCl4 are reacted with 1.13 x 107 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?