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.
A chemical equation uses chemical symbols to show what happens during a chemical reaction (g) (g) (l) reactants product 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.
The Law of Conservation of Mass states that matter is neither created nor destroyed Chang, R. 2002. Chemistry 7 th 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 7 th ed. Singapore: McGraw-Hill.
A quick note on balancing chemical equations Translate the statement Balance the atoms Adjust the coefficients Check the atom balance Specify states of matter 3-5
Calculating the amounts of reactant and product
Stoichiometry of 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) + O 2(g) 2CO 2(g) 2 mol CO = 1 mol O 2 2 mol CO 1 mol O 2 = 1 1 mol O 2 2 mol CO = 1 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2 nd 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) + O 2(g) 2CO 2(g) 2 mol CO = 2 mol CO 2 2 mol CO 2 mol CO 2 = 1 2 mol CO 2 2 mol CO = 1 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2 nd 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) + O 2(g) 2CO 2(g) 1 mol O 2 = 2 mol CO 2 1 mol O 2 2 mol CO 2 = 1 2 mol CO 2 1 mol O 2 = 1 Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2 nd 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. 2 nd ed. New York: McGraw-Hill.
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)
How many moles of H 2 will be formed by the complete reaction of 6.23 moles of Li with water? 2Li (s) + 2H 2 O (l) 2LiOH (aq) + H 2(g) n H2 = 6.23 mol Li 1 mol H 2 x 2 mol Li = 3.12 mol H 2
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) 80.57 g Li 1 mol Li m H2 = x x 1 mol H 2 6.941 g Li 2 mol Li x 2.016 g H 2 1 mol H2 = 11.70 g H 2
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)
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 = 10.0 mol Cu 2 S 3 mol O 2 x 2 mol Cu 2 S = 15.0 mol O 2
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) 10.0 mol Cu 2 S m SO2 = x 2 mol SO 2 x 64.07 g SO 2 2 mol Cu 2 S 1 mol SO 2 = 641 g SO 2
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) 2.86 kg Cu 2 O 1000 g Cu 2 O m O2 = x x 1 mol Cu 2 O 1 kg Cu 2 O 143.10 g Cu 2 O x 3 mol O 2 32.00 g O 2 x 2 mol Cu 2 O 1 mol O 2 = 960 g O 2
Within the cylinders of a car s engine, the hydrocarbon octane (C 8 H 18 ), one of many components of gasoline, mixes with oxygen from the air and burns to form carbon dioxide and water vapor. 2C 8 H 18 (l) + 25O 2 (g) 16CO 2 (g) + 18H 2 O (g) a. How much carbon dioxide ( in kg) is produced when 2.00 kg of octane is burned? b. How much oxygen(in kg) is required to burn the same amount of octane?
How many double cheeseburgers can you make from 8 bun slices + 2 cheese slices + 6 burger patties?
What is the limiting factor? 8 bun slices 2 cheese slices 6 burger patties # Cheeseburgers= 2 cheese slices [1 cheeseburger/ 2 cheese slices]
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? How much C and D are produced?
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?
Strategy Check if the equation is balanced. 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: Al + 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 1000⁰C: TiCl 4 + Mg 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?
3-27 Mass Percent from the Chemical Formula Mass % of element X = atoms of X in formula x atomic mass of X (amu) molecular (or formula) mass of compound (amu) x 100 Mass % of element X = moles of X in formula x molar mass of X (g/mol) mass (g) of 1 mol of compound x 100
3-28 Sample Problem 3.3 PLAN: SOLUTION: Per mole glucose there are 6 moles of (a) C, 12 moles of H, 6 moles of O Calculating Mass Percents and Masses of Elements in a Sample of a Compound PROBLEM: In mammals, lactose (milk sugar) is metabolized to glucose (C 6 H 12 O 6 ), the key nutrient for generating chemical potential energy. (a) What is the mass percent of each element in glucose? (b) How many grams of carbon are in 16.55 g of glucose? We have to find the total mass of glucose and the masses of the constituent elements in order to relate them.
3-29 Sample Problem 3.3 continued Calculating the Mass Percents and Masses of Elements in a Sample of a Compound 6 mol C x 12.01 g C mol C = 72.06 g C 12 mol H x 1.008 g H = 12.096 g H mol H 6 mol O x 16.00 g O mol O = 96.00 g O M = 180.16 g/mol (b) mass percent of C = 72.06 g C 180.16 g glucose = 0.4000 x 100 = 40.00 mass % C mass percent of H = 12.096 g H 180.16 g glucose = 0.06714 x 100 = 6.714 mass % H mass percent of O = 96.00 g O 180.16 g glucose = 0.5329 x 100 = 53.29 mass % O
3-30 Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis and gives rise to the smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists; it may be a multiple of the empirical formula.
3-31 Sample Problem 3.4 Determining an Empirical Formula from Masses of Elements PROBLEM: PLAN: Elemental analysis of a sample of an ionic compound showed 2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What is the empirical formula and name of the compound? Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula).
Sample Problem 3.4 Determining an Empirical Formula from Masses of Elements continued SOLUTION: 2.82 g Na x 4.35 g Cl x 7.83 g O x mol Na 22.99 g Na mol Cl 35.45 g Cl mol O 16.00 g O Na1.00 Cl1.00 O3.98 = 0.123 mol Na = 0.123 mol Cl = 0.489 mol O NaClO4 NaClO4 is sodium perchlorate. 3-32
3-33 Sample Problem 3.5 PROBLEM: Determining a Molecular Formula from Elemental Analysis and Molar Mass During physical activity, lactic acid (M = 90.08 g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that this compound contains 40.0 mass % C, 6.71 mass % H, and 53.3 mass % O. (a) Determine the empirical formula of lactic acid. (b) Determine the molecular formula. PLAN: a) Assume 100 g of lactic acid and find the mass of each element. Convert mass of each to moles, get a ratio and convert to integer subscripts. b) Divide molar mass by empirical mass to get the multiplier then write the molecular formula accordingly.
3-34 Sample Problem 3.5 continued Determining a Molecular Formula from Elemental Analysis and Molar Mass SOLUTION: In 100.0 g of lactic acid, there are: 40.0 g C 6.71 g H 53.3 g O We convert the grams to moles and get a ratio for the empirical formula. 40.0 g C x mol C 12.01 g C = 3.33 mol C 6.71 g H x mol H 1.008 g H = 6.66 mol H 53.3 g O x mol O 16.00 g O = 3.33 mol O
Sample Problem 3.5 Determining a Molecular Formula from Elemental Analysis and Molar Mass continued SOLUTION: 3.33 mol C 6.66 mol H 3.33 mol O We now divide by the smallest number and get a ratio for the empirical formula. C3.33 H6.66 O3.33 3.33 3.33 3.33 molar mass of lactate CH2O empirical formula 90.08 g 3 mass of CH2O 3-35 30.03 g C3H6O3 is the molecular formula
3-36 Figure 3.5 Combustion apparatus for determining formulas of organic compounds. m m C n H m + (n + 2 ) O 2 = n CO(g) + H 2 O(g) 2
3-37 Sample Problem 3.6 PROBLEM: Determining a Molecular Formula from Combustion Analysis Vitamin C (M = 176.12 g/mol) is a compound of C,H, and O found in many natural sources, especially citrus fruits. When a 1.000-g sample of vitamin C is placed in a combustion chamber and burned, the following data are obtained: mass of CO 2 absorber after combustion mass of CO 2 absorber before combustion mass of H 2 O absorber after combustion mass of H 2 O absorber before combustion What is the molecular formula of vitamin C? = 85.35 g = 83.85 g = 37.96 g = 37.55 g PLAN: The difference in absorber mass before and after combustion is the mass of oxidation product of the element. Find the mass of each element from its combustion product, convert each to moles and determine the formula. Using the molar mass and empirical mass, determine the molecular formula.
3-38 Sample Problem 3.6 continued Determining a Molecular Formula from Combustion Analysis SOLUTION: CO 2 85.35 g - 83.85 g = 1.50 g H 2 O 37.96 g - 37.55 g = 0.41 g There are 12.01 g C per mol CO 2. 1.50 g CO2 x 12.01 g C 44.01 g CO 2 = 0.409 g C There are 2.016 g H per mol H 2 O. 0.41 g H 2 O x 2.016 g H 18.02 g H 2 O = 0.046 g H O must be the difference: 1.000 g - (0.409 + 0.046) = 0.545 g O 0.409 g C 12.01 g C = 0.0341 mol C 0.046 g H = 0.0456 mol H 0.545 g O = 0.0341 mol O 1.008 g H 16.00 g O C 1.00 H 1.3 O 1.00 C 3 H 4 O 3 176.12 g/mol 88.06 g = 2.000 = 2 C 6 H 8 O 6
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