Percent Yield

Stoichiometric calculations determine the quantity of product that will form in a reaction ; this represents the maximum possible yield or what is referred to as the theoretical yield
when we carry out a reaction in the laboratory we often obtain less than the theoretical yield; this is the actual yield. The percent yield indicates how much product we produced as a percentage. This is calculated by the expression:
                        percent yield =   actual yield      x 100%
                                               theoretical yield
Sample calculation:
Copper (II) oxide may be reduced to copper by heating it in a stream of hydrogen gas:
                        CuO(s)  + H2(g)  →  Cu(s)  +  H2O(g)
In an experiment 13.65 g of copper(II) oxide produces 10.75 g of copper metal. What is the percent yield?
determine the number of moles of copper(II) oxide
moles of CuO = 13.65 g CuO ÷ 79.55 g/mol CuO
                       = 0.1716 mol CuO
determine the number of moles of Cu
from the equation we see that 1 mol CuO → 1 mol Cu
                                therefore 0.1716 mol CuO → 0.1716 mol Cu
determine the possible yield of Cu (theoretical yield)
mass of Cu = 0.1716 mol Cu x 63.55 g/mol
                  = 10.91 g Cu
determine the percent yield
percent yield = 10.75 g Cu x 100%  = 98.53%
                         10.91 g Cu
1.      “Red lead” is an oxide of lead that has the formula Pb3O4. When boiled with dilute nitric acid, the red lead reacts to produce lead(II) nitrate and lead(IV) oxide:
                                    Pb3O4(s)  +  4HNO3(aq)→ 2Pb(NO3)2(aq)  + PbO2(s)  +  2H2O(l)
a)      How many moles of lead(II) nitrate would be produced from the complete reaction of 0.46 mol of red lead with an excess of nitric acid?
b)      What volume of 2.50 mol/L nitric acid would be required to react with 15.85 g of red lead.
c)      What mass of red lead would be required to react to producer 14.75 g of lead ((IV) oxide?
d)      What is the percent yield if 11.25 g of red lead reacts with an excess of nitric acid to produce 3.85 g of lead(IV) oxide?
2.      Bromic acid will oxidize sulfur dioxide to sulfuric acid
2HBrO3(aq)  +  5SO2(g)  + 4H2O(l)  →  Br2(aq)  +  5H2SO4(aq)
In an experiment 0.100 L of 0.200 mol/L bromic acid is reacted with 0.100L of a solution of sulfur dioxide.
a)      What volume of sulfur dioxide measured at 23 oC and 98.7 kPa would neede to be dissolved in the aqueous solution to react will all the bromic acid?
b)      What is the maximum number of moles of sulfuric acid that could be produced?
c)      Calculate the percent yield of sulfuric acid if the concentration of the solution produced was found to be 0.230 mol/L?
3.      Aluminum metal can displace silver from a solution of silver nitrate.  In an experiment 0.270 g of aluminum is added to 40.0 mL of 1.00 mol/L silver nitrate solution.
a)      Write a balanced equation for the reaction.
b)      Determine which of the reactants is the limiting reagent.
c)      What is the theoretical yield of silver?
d)      The mass of silver collected in the experiment was 2.98 g. Calculate the percent yield.