Moles, masses, reactions, oh my! Stoichiometry Moles, masses, reactions, oh my!

Stoikheion + metria From the Greek for “element” and “the process of measuring” Quantitative analysis of reactants and products Moles Molar masses Balanced equations

Grinch Sandwiches Ingredients per sandwich Three pieces of bread Sauerkraut (10 g) Toadstool (15 g) Arsenic sauce (5 g)

Balanced Equations 2H2 + O2  2H20 Coefficients = moles 2 moles of hydrogen react with 1 mole of oxygen to produce 2 moles of water What if you only had 1 mole of hydrogen? What if you had 10 moles of both hydrogen and oxygen? What if you needed to make 5 moles of water?

Law of Conservation of Mass Convert each of the moles in the previous equation to grams Add the reactants’ masses together. Compare to the product’s mass. Be in awe!

Mole Ratio s Equation must be balanced Relate reactants to each other, products to each other, or reactants to products

Molar Masses If mass is mentioned, find that substance’s molar mass

Making Stoichiometry Work Dimensional Analysis Number, unit, substance…always!

Problem #1 When 2.50 g of hydrogen gas are mixed with an excess of oxygen gas, how many grams of water should be produced?

Problem #1—step 1 Write a balanced equation. 2H2 + O2  2H20

Problem #1—step 2 Begin with your given information and make a plan of conversions. Grams of hydrogen moles of hydrogen moles of water grams of water Write it as dimensional analysis. g H2 x mol H2 x mol H2O x g H2O = g H2 mol H2 mol H20

Problem #1—step 3 Fill in your numbers. 2.5 g H2 x 1 mol H2 x 2 mol H2O x 2.02 g H2 2 mol H2 18.02 g H2O = 1 mol H20 Cross out what divides out. Calculate your answer.

Problem #1—step 4 2.5 g H2 x 1 mol H2 x 2 mol H2O x 2.02 g H2 2 mol H2 18.02g H2O = 22.3 g H2O 1 mol H20 How much oxygen gas would be required?

Problem #2 The thermite reaction: aluminum foil reacts with ferric oxide

Problem #2 Wow! Now let’s assume that 0.010 g of aluminum reacted with the excess ferric oxide on the ball bearing. What mass of aluminum oxide should be produced?

Problem #2—Step 1 Write a balanced equation. 2Al + Fe2O3  Al2O3 + 2Fe

Problem #2—Step 2 Write your plan. g Al x mol Al x mol Al2O3 x g Al2O3 = g Al mol Al mol Al2O3

Problem #2—Step 3 Place the proper numbers and calculate. 0.010 g Al x 1 mol Al x 1 mol Al2O3 x 26.98 g Al 2 mol Al 101.96 g Al2O3 = 1 mol Al2O3

Problem #2—Answer 0.0189 g of Al2O3 should be formed

Problem #3 What if only 0.0150 g of Al2O3 were actually produced in the previous problem? What is the percent yield? So, you can calculate the % yield % yield = actual amt x 100 theoretical amt

Problem #3 Thus, the % yield calculation 0.0150 g x 100 = 79.4 % yield

Limiting Reagent Ideally, the perfect amount of each reactant Often one reactant limits how much product can be formed—limiting reagent (LR) Often the other reactant(s) is(are) in excess (ER) You will be given at least 2 quantities of reactants

Grinch Sandwiches Ingredients per sandwich Three pieces of bread Sauerkraut (10 g) Toadstool (15 g) Arsenic sauce (5 g)

Problem #4 0.500 g of silver nitrate in solution reacts with 0.750 g of tin (IV) chloride in solution. If 0.400 g of solid silver chloride are retrieved, what is the percent yield?

Problem #4—Step 1 Write a balanced equation. 4AgNO3 + SnCl4  4AgCl + Sn(NO3)4 Write shorthand, to save time. You’ll write the real stuff in the answer. A B C D

Problem #4—Step 2 Ask yourself, when I have 0.500 g of AgNO3, how many grams of SnCl4 do I really need? This will determine the LR. Set up the plan g A x mol A x mol B x g B = g A mol A mol B

Problem #4—Step 3 Plug in the numbers. 0.5 g A x 1mol A x 1 mol B 169.88g A 4 mol A x 260.50 g B = 0.192 g SnCl4 1 mol A

Problem #4—Step 4 Analyze and determine the LR 0.750 g of SnCl4 are available, and only 0.192 g of it need to be used to completely react with the AgNO3 Which substance will be in excess? Which substance will you run out of first?

Problem #4—Step 5 Now, use the given amount of the LR to find how much product you should be able to make. Set up your plan. g A x mol A x mol C x g C = g A mol A mol C

Problem #4—Step 6 Plug in your numbers and calculate. 0.5 g A x 1 mol A x 4 mol C x 169.88 g A 4 mol A 143.32 g C = 0.422 g AgCl should be 1 mol C formed

Problem #4—Step 7 Look at the actual amount of product that was formed, and determine percent yield using the theoretical you just calculated. 0.400 g AgCl x 100 = 94.8% yield 0.422 g AgCl

Practice Sit down with some bread, sauerkraut, toadstools, and arsenic sauce, and work some stoichiometry problems just for fun.