Unit 9: Stoichiometry Chapter 12 Exam: February 11, 2009

Reminders… The mole is a counting unit 1 mole = 6.02 x particles (atoms, molecules, formula units) Molar mass is the mass of 1 mole Measured in grams Represented by the atomic mass shown on the periodic table Example: 1 mole of water (H 2 O) is about 18 grams. It contains 2 moles of hydrogen atoms and 1 mole of oxygen atoms.

What is Stoichiometry? Using the methods of stoichiometry, we can measure the amounts of substances involved in chemical reactions and relate them to one another. Pronounced stoy-kee-AW-muh-tree From the Greek words stoikheion meaning element and metria meaning measure

Stoichiometric Calculations There are three basic stoichiometric calculations: mole-to-mole conversions mole-to-mass / mass-to-mole conversions mass-to-mass conversions All stoichiometric calculations are based on a balanced equation and mole ratios.

Mole Ratios Indicates the molar relationship between two chemicals in an equation Use the coefficients (not subscripts) to make ratios Reduce to lowest whole-number ratio Example: Water Formation 2H 2 + O 2  2H 2 O What is the ratio of hydrogen to oxygen? What is the ratio of oxygen to water? What is the ratio of water to hydrogen? 2:1 1:2 1:1

Mole-to-Mole Conversions The following reaction shows table salt production. How many moles of sodium chloride are produced from 0.02 moles of chlorine? Solve through dimensional analysis: ? mol NaCl = 0.02 mol Cl 2 x 2 mol NaCl = 0.04 mol NaCl 1 mol Cl 2 Solve through ratios: ? mol NaCl = 2 mol NaCl 0.02 mol Cl 2 1 mol Cl 2 (cross multiply and solve) = 0.04 mol NaCl

Mole-to-Mass Conversions The following reaction shows photosynthesis. How many grams of glucose are produced when 24 moles of carbon dioxide reacts with water? Solve through dimensional analysis: ? g C 6 H 12 O 6 = 24 mol CO 2 x 1 mol C 6 H 12 O 6 x 180 g C 6 H 12 O 6 = 720 g 6 mol CO 2 1 mol C 6 H 12 O 6 C 6 H 12 O 6 Solve through ratios: ? mol C 6 H 12 O 6 = 1 mol C 6 H 12 O 6 24 mol CO 2 6 mol CO 2 (cross multiply and solve) = 4 mol C 6 H 12 O 6 ? g C 6 H 12 O 6 = 180 g C 6 H 12 O 6 4 mol C 6 H 12 O 6 1 mol C 6 H 12 O 6 (cross multiply and solve) = 720 g C 6 H 12 O 6

Mass-to-Mole Conversions The following reaction shows photosynthesis. How many moles of glucose are produced when 132 grams of carbon dioxide reacts with water? Solve through dimensional analysis: ? mol C 6 H 12 O 6 = 132 g CO 2 x 1 mol CO 2 x 1 mol C 6 H 12 O 6 = 0.5 mol 44 g CO 2 6 mol CO 2 C 6 H 12 O 6 Solve through ratios: ? mol CO 2 = 1 mol CO g CO 2 44 g CO 2 (cross multiply and solve) = 3 mol CO 2 ? mol C 6 H 12 O 6 = 1 mol C 6 H 12 O 6 3 mol CO 2 6 mol CO 2 (cross multiply and solve) = 0.5 mol C 6 H 12 O 6

Mass-to-Mass Conversions The following reaction shows the production of ammonia. How many grams of nitrogen are required to produce 85 grams of ammonia? Solve through dimensional analysis: ? g N 2 = 85 g NH 3 x 1 mol NH 3 x 1 mol N 2 x 28 g N 2 = 70 g N 2 17 g NH 3 2 mol NH 3 1 mol N 2 Solve through ratios: ? mol NH 3 = 1 mol NH 3 85 g NH 3 17 g NH 3 (cross multiply and solve) = 5 mol NH 3 ? mol N 2 = 2 mol NH 3 5 mol NH 3 1 mol N 2 (cross multiply and solve) = 10 mol N 2 ? g N 2 = 28 g N 2 10 mol N 2 1 mol N 2 (cross multiply and solve) = 70 g N 2

Percent Yield Percent yield compares the amount of product collected in an experiment (actual) to the amount anticipated according to calculations (theoretical). Actual mass will always be less than theoretical mass due to human error actual mass of product theoretical mass of product x 100 % yield =

Percent Yield The following reaction shows the production of methanol (CH 3 OH), which was used as one of several embalming fluids in ancient Egypt. In an experiment, you used 25 grams of CO 2 with ample hydrogen gas to produce 15 grams of methanol. What is the percent yield? Calculate theoretical yield ? g CH 3 OH = 25 g CO 2 x 1 mol CO 2 x 1 mol CH 3 OH x 32 g CH 3 OH = g 44 g CO 2 1 mole CO 2 1 mol CH 3 OH CH 3 OH Calculate percent yield 15 g CH 3 OH g CH 3 OH x 100 = 82.5% yield

Limiting Reactant In the real world, reactants are not present in the exact mole ratio described by the balanced equation. This will cause the reaction to stop as soon as one of its reactants is spent. The limiting reactant is used up first and restricts the reaction The excess reactant(s) remain after the reaction stops Example: If you placed a single drop of water on an Alka-Seltzer tablet… What would be the limiting reactant? Excess reactant? How could you switch the roles of these two reactants?

Limiting Reactant Experiment: Equal amounts of copper (II) chloride are combined with varying amounts of aluminum to form copper and aluminum chloride. Which pairing will react completely? 2 Al(s) + 3 CuCl 2 (aq)  2 AlCl 3 (aq) + 3 Cu(s) 1:1 mass 1:1 mole 2:3 mass 2:3 mole A B C D Al:CuCl 2 ratio

Limiting Reactant Results: In reactions A, B, & C, aluminum is still visible - the excess reactant. Copper (II) chloride is the limiting reactant. Reaction D, which follows the correct ratio in the balanced equation, reacted completely. 2 Al(s) + 3 CuCl 2 (aq)  2 AlCl 3 (aq) + 3 Cu(s) A B C D 1:1 mass 1:1 mole 2:3 mass 2:3 mole Al:CuCl 2 ratio

Limiting Reactant Practice Diagram A represents reactant particles, and Diagram B represents product particles in a fictitious chemical reaction. C Z Z C Z Z A A A A A C Z Z C Z Z A A A A A A A Diagram A Diagram B A A Write the balanced equation for this reaction. What is the limiting reactant? Excess?

ABC Limiting Reactant Practice The three particle diagrams to the right represent the burning of methane. In which diagram is the limiting reactant oxygen gas? CH 4 + 2O 2  CO H 2 O

Limiting Reactant Calculations In a reaction, 80 grams of sodium hydroxide is combined with 60 grams of sulfuric acid. What is the limiting reactant? Calculate the moles for each reactant: ? mol NaOH = 80 g NaOH x 1 mol NaOH = 2 mol NaOH 40 g NaOH ? mol H 2 SO 4 = 60 g H 2 SO 4 x 1 mol H 2 SO 4 = mol H 2 SO 4 98 g H 2 SO 4 Compare the actual molar amounts to the mole ratio The correct ratio of NaOH to H 2 SO 4 is 2:1 There is less H 2 SO 4 than required to react with 2 moles of NaOH Sulfuric acid is the limiting reactant