Percent Yield & Limiting Reactant Chemistry

Percent Yield

Percent Yield describes how much product was actually made in the lab versus the amount that theoretically could be made. Actual Yield X 100 = Percent Yield Theoretical Yield Percent yield tells you how close you were to the 100% mark. – Reactions do not always work perfectly. Experimental error (spills, contamination) often means that the amount of product made in the lab does not match the ideal amount that could have been made.

Example: A chemist was supposed to produce 75.0 g of aspirin. However, he squandered some of the reactants for his own personal use (he was later fired), and so only actually made 50.0 g. What was his percent yield? 50.0 g 75.0 g X 100 = 66.7 %

More Examples! When I was a sophomore in college, we had a lab where we were supposed to isolate caffeine from tea leaves. Most of my caffeine was washed down the drain in a freak accident. Although I should have had 5.0 g of caffeine, I only ended up with g of caffeine and a bad grade on the lab. What was my percent yield? g 5.0 g X 100= 0.80 %

A very sloppy student did not wait until his NaCl was dry before he weighed it. As a result, his product weighed 2.25 g when it should have been 1.75 g. What was his percent yield? 2.25 g 1.75 g X 100 = 129%

Actual Yield Theoretical Yield Theoretical Yield = The maximum amount of product that could be formed from given amounts of reactants – found by doing a 3 step stoichiometry problem Actual Yield = The amount of product actually formed or recovered when the reaction is carried out in the laboratory – GIVEN in the problem x 100 = Percent Yield

Steps for Calculating % Yield Circle both givens (number, unit, substance) – Reactant – label start stoich – Product – label actual Perform stoichiometry to get theoretical (theo) Find % yield

What is the percent yield if 5.50 grams of hydrogen were used in a reaction that only produced 20.4 grams NH 3 ? H 2 + N 2  NH g H g H 2 1 mole H 2 3 moles H 2 2 moles NH g NH 3 1 mol NH 3 = grams NH 3 = Theoretical Yield!!!! Only 20.4 grams of ammonia is actually produced in the lab. What is the percent yield? 20.4 g g X 100 = % yield

Limiting Reactant

A limiting reactant in a chemical reaction is the substance that: Is used up first. Stops the reaction. Limits the amount of product that can form. 11

Reacting Amounts In a table setting, there is 1 plate, 1 fork, 1 knife, and 1 spoon. How many table settings are possible from: 5 plates, 6 forks,4 spoons, and 7 knives? What is the limiting item? 12 Copyright © 2008 by Pearson Education, Inc. Publishing as Benjamin Cummings

Reacting Amounts Four table settings can be made. Initially Use Left over plates forks 6 42 spoons knives The limiting item is the spoon. 13

Limiting Reactants When 4.00 mol H 2 is mixed with 2.00 mol Cl 2,how many moles of HCl can form? H 2 (g) + Cl(g)  2HCl (g) 4.00 mol 2.00 mol ??? mol Calculate the moles of product from each reactant, H 2 and Cl 2. The limiting reactant is the one that produces the smaller amount of product. 14

Limiting Reactants Using Moles HCl from H mol H 2 x 2 mol HCl = 8.00 mol HCl 1 mol H 2 (not possible) HCl from Cl mol Cl 2 x 2 mol HCl = 4.00 mol HCl 1 mol Cl 2 (smaller number) The limiting reactant is Cl 2 because it is used up first. Thus Cl 2 produces the smaller number of moles of HCl. 15

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Limiting Reactants Using Mass If 4.80 mol Ca mixed with 2.00 mol N 2, which is the limiting reactant? 3Ca(s) + N 2 (g)  Ca 3 N 2 (s) Moles of Ca 3 H 2 from Ca 4.80 mol Ca x 1 mol Ca 3 N 2 = 1.60 mol Ca 3 N 2 3 mol Ca (Ca used up) Moles of Ca 3 H 2 from N mol N 2 x 1 mol Ca 3 N 2 = 2.00 mol Ca 3 N 2 1 mol N 2 (not possible) All Ca is used up when 1.60 mol Ca 3 N 2 forms. Thus, Ca is the limiting reactant. 17

Limiting Reactants Using Mass Calculate the mass of water produced when 8.00 g H 2 and 24.0 g O 2 react? 2H 2 (g) + O 2 (g) 2H 2 O(l) 18

Limiting Reactants Using Mass Calculate the grams of H 2 for each reactant. H 2 : 8.00 g H 2 x 1 mol H 2 x 2 mol H 2 O x g H 2 O g H 2 2 mol H 2 1 mol H 2 O = 71.5 g H 2 O (not possible) O 2 : 24.0 g O 2 x 1 mol O 2 x 2 mol H 2 O x g H 2 O g O 2 1 mol O 2 1 mol H 2 O = 27.0 g H 2 O (smaller) O 2 is the limiting reactant. 19

Checking Calculations InitiallyH mol Cl mol 2HCl 0 mol Reacted/ Formed mol mol Left after reaction 2.00 mol Excess 0 mol Limiting 4.00 mol 20