Now that we know how to relate quantities of atoms, elements or molecules to real-world measurable quantities such as grams, it is time to combine our knowledge of quantities and chemical equations to plan and predict chemical reactions.

LEARNING OBJECTIVES:

• Stoichiometry is the study of mole and mass calculations from balanced chemical equations.
• We will use mass to mole calculations, mole to mole ratios, and mole to mass calculations to predict the quantity of a product, given a specified quantity of starting reactant.
• Limiting Reactants are substances which "limit" the extent of the reaction.
• Assume that you have 5 pieces of chicken and 7 helpings of potato and one full dinner consists of 1 piece of chicken and 1 helping of potato--which is the limiting food? In other words, what is the maximum number of full dinners that can be prepared given the ingredients?--only 5, because the chicken is the limiting food (or reactant), and you will have two helpings of potato left over.
• We will learn how to identify the limiting reactant in a reaction and,
• We will learn how to perform calculations with limiting reactants.
• Percentage Yield is the relative actual yield of product compared to the maximum amount of product that could possibly be produced. In other words, it shows the difference between the actual yield and the calculated yield, which may be very different. The larger the %yield, the closer the actual yield is to the calculated yield (maximum amount that can be produced).
• Before performing quantitative calculations in chemistry, you must be able to understand and recognize:
• correct chemical formulas
• correctly balanced chemical equations
• that the balanced chemical equation provides the mole ratio of all species that occur in the equation
• that Molar Mass converts between grams and moles of the same species
• that Avogadro’s Number converts between moles and atoms (or molecules or formula units) of the same species
• that the mole ratio in a chemical equation can convert between moles of different species

Given the following balanced equation:

• Calculate the number of moles of HCl required to react completely with 0.42 mol of Mg(OH)₂


0.42 mol Mg(OH)₂     x      2 mol HCl       =      0.84 mol HCl
                      1 mol Mg(OH)₂

The mole ratio comes from the balanced chemical equation: 2 mol of HCl will always react exactly with 1 mol of Mg(OH)₂ in this reaction.
 

Problems:

• Aluminum reacts with oxygen to form aluminum oxide. How many moles of aluminum oxide can be obtained from 5.22 mol of aluminum and the required (or excess) amount of oxygen?

                      4Al_(s) + 3O_2(g) ® 2Al₂O_3(s)

we are concerned with the relative amounts of Al and Al₂O₃ for which the mole ratio in the balanced chemical equation is:      4 mol Al_(s)     (this is our conversion factor) 
                                                     2 mol Al₂O_3(s)

In other words, we can only make half as much Al₂O₃ as the amount of Al that we begin with (or another way of putting it, it takes twice as much Al to make Al₂O₃). If we begin with 5.22 mol Al, the maximum amount of Al₂O₃ that we can make is 2.61 mol Al₂O₃.

     mole A                           mole ratio B:A                                 calc. mole B
5.22 mol Al      x       2 mol Al₂O₃      =           2.61 mol Al₂O₃
                                  4 mol Al

• When iron rusts, the reaction is described by:

                  4Fe_(s) + 3O_2(g) ® 2Fe₂O_3(s)

Calculate the mass of oxygen required to react with 1.00 g of iron.

Again, we know all mole ratios for all species in the reaction because we have the balanced chemical equation.

4 mol Fe          4 mol Fe              3 mol O₂
3 mol O₂       2 mol 2Fe₂O₃      2 mol 2Fe₂O₃
 

initial mass of A             MM of A                 mole ratio B:A                 MM of B                  calc. mass of B
1.00 g Fe     x     1 mol Fe     x       3 mol O₂     x      32.0 g O₂   =     0.429 g O₂
                          55.9 g Fe             4 mol Fe              1 mol O₂
 

• Copper(II) oxide will decompose into copper(I) oxide when it is heated strongly. The equation is:

                     4CuO_(s) ® 2Cu₂O_(s) + O_2(g)

How many grams of oxygen can be obtained from 2.64 g of CuO?

Conversion factors:
   4 molCuO            1 mol O₂            1 mol O₂
2 mol Cu₂O           4 mol CuO         2 mol Cu₂O

initial mass of A                     MM of A                      mole ratio B:A                     MM of B
2.64 g CuO      x      1 mol CuO       x       1 mol O₂      x       32.0 g O₂
                               79.5 g CuO              4 mol CuO               1 mol O₂
 

      calc. mass of B
= 0.266 g O₂

Try this problem:

You have a 9.00 g mixture of KClO₃ and rust. Determine the %KClO₃ in the mixture if the mixture loses 2.00 g upon heating.

                                                    _{D                                                (remember D means heat)}
                        2KClO_3(s) ® 2KCl_(s) + 3O_2(g)

Note:

• rust does not emit O₂ when heated but KClO₃ does
• if you can calculate the amount (in moles) of O₂ produced, you can determine, stoichiometrically, the amount (in moles) of KClO₃ that you started with
• remember all the conversion factors: MM of all substances, mole ratios of all substances
• % KClO₃     =       g KClO₃      x      100

                             g tot. mix.

LIMITING REACTANTS:

Note that in the aluminum problem above, we specified a given amount of one reactant, reactant A, and indicated an unspecified amount of the other reactant. The "required amount", or the stoichiometric amount, is that amount which would react exactly with reactant A as specified by the mole ratio in the balanced chemical equation. An "excess amount" is just that, more than what is exactly needed for the reaction.

In some problems you will be given non-stoichiometric amounts of both reactants and you will have to determine which is the limiting reactant or the reactant which is present in the smaller stoichiometric amount.

for example:

You have 10 bicycle frames and 16 bicycle wheels and you need to put them together to produce as many bicycles as possible, how many bicycles can be produced, what is your limiting "reactant", and how much excess "reactant" do you have left over?

Balanced equation:

                       one frame + two wheels ® one bicycles
                              10 frames           16 wheels                8 bicycles

the only way to produce one bicycle is with a ratio of:      two wheels
                                                                                            one frame

Is:         16 wheels the same ratio, larger ratio, or smaller ratio than 2 wheels?
             10 frames                                                                               1 frame

The ratio is smaller so the wheels are the limiting "reactant" (the wheels are stoichiometrically fewer in number than the frames) and the frames are the excess reagent. So how many bicycles can be produced? We have enough wheels for 8 bicycles but enough frames for 10 bicycles. It is obvious that we can only make as many bicycles as we have wheels for, or 8 bicycles.

Always use the limiting reactant for stoichiometric calculations:

16 wheels     x     1 bicycle     =       8 bicycles
                            2 wheels

8 bicycles    x       1 frame      =       8 frames
                            1 bicycle

10 frames - 8 frames    =    2 frames are left over

The stoichiometric quantities for this problem are 8 frames and all 16 wheels produce 8 bicycles – the other two frames are useless because there are no wheels available to produce a bicycle.

Problems:

• When copper(I) oxide is heated in oxygen, copper(II) oxide is formed. The chemical equation is

                              2Cu₂O_(s) + O_2(g) ® 4CuO_(s)
       given amounts          3.00 mol       3.00 mol

If 3.00 mol of each of the reactants are heated together, how many moles of the reactants and product will be present when the reaction is complete?

First, you must determine what the limiting reactant is because all stoichiometric calculations are based only upon the limiting reactant. We can determine the limiting reactant several ways (it does not matter which method you use):

• First: Just by inspection of the balanced equation, we can see that we need twice as much Cu₂O as O₂ (the ratio is 2 mol Cu₂O  : 1 mol O₂).  If we have 3.00 mol of each reactant, we do not have enough Cu₂O to react with all the O₂ (remember, we would need twice as much, or 6.00 mol of Cu₂O). Therefore, our limiting reactant is Cu₂O.
  
  
• Second: If the ratios are more complex, and you cannot determine the limiting reactant by simple inspection, compare the mole ratio in the balanced chemical equation to the mole ratio given in the problem:


mole ratio from balanced chemical equation:       2 mol Cu₂O 
                                                                          1 mol O₂

mole ratio from available given amounts:     3 mole Cu₂O or 1 mol
                                                                  3 mol O₂            1 mol

The ratio of actual quantities of Cu₂O to O₂ is smaller than in the balanced chemical equation ratio, therefore, Cu₂O (the numerator) is the limiting reactant. What if the ratio were larger? What would the limiting reactant be then?



• Third: If neither of the above methods works for you, simply perform two separate stoichiometric calculations to determine the amount of product formed given each initial amount of reactant:


3.00 mol Cu₂O     x        4 mol CuO      =      6.00 mol CuO 
                                      2 mol Cu₂O

3.00 mol O₂        x         4 mol CuO      =       12.00 mol CuO
                                        1 mol O₂

Whichever calculation produces the smallest amount of product, used the limiting reactant in the calculation. Therefore, Cu₂O is the limiting reactant

• The complete combustion of carbon disulfide occurs by the reaction

                           CS_2(g) + 3O_2(g) ® CO_2(g) + 2SO_2(g)

If 254 g of each reactant are used, which reactant, if any, is not completely consumed at the end of the reaction (which is the excess reactant)? How many grams of this reactant will be left, and how many grams of carbon dioxide will be formed?

Note: You will only have limiting and excess reactants if the reactants are not in exact stoichiometric ratio to one another.

The mole ratio of reactants in the balanced chemical equation is:     3 mol O₂
                                                                                                           1 mol CS₂

Convert the mass of each reactant to moles:

254 g CS₂    x     1 mol CS₂      =        3.33 mol CS₂
                           76.2 g CS₂

254 g O₂    x        1 mol O₂       =       7.94 mol O₂
                            32.0 g O₂

Take the mole ratio of the amounts of O₂ : CS₂:    7.94 mol O₂     =     2.38
                                                                 3.33 mol CS₂            1.0

Compare the mole ratio from the balanced chemical equation and the mole ratio from the given amounts. The mole ratio of O₂ : CS₂ is less than 3 : 1, therefore O₂ is the limiting reactant and CS₂ is the excess reactant. In other words, we need 3 times as much O₂ as CS₂ but we don't have that much.

At this point, all stoichiometric calculations will be based upon the initial amount of the limiting reactant or O₂.

To calculate how much CS₂ is left over, first calculate how much is used in the reaction. We know that all of the O₂ will be used, all 7.94 mol O₂, so how much CS₂ will be used, stoichiometrically relative to 7.94 mol O₂?

7.94 mol O₂       x         1 mol CS₂      x       76.2 g CS₂      =       202 g CS₂
                                     3 mol O₂                 1 mol CS₂                     used

254 g CS₂ given amount - 202 g CS₂ used in reaction      =     52 g CS₂ excess

To calculate the amount of CO₂ produced, again, use only the limiting reactant amount in the calculation:

7.94 mol O₂       x        1 mol CO₂        x       44.0 g CO₂     =      116 g CO₂
                                    3 mol O₂                    1 mol CO₂              produced

PERCENT YIELD AND THEORETICAL YIELD:

The Theoretical Yield is the maximum amount of product that can be obtained from a reaction given the specified amounts of the reactants. In other words, the Theoretical Yield is just the calculated yield of product based upon the limiting reactant.

Few reactions will actually produce the maximum amount of product that can be made due to experimental error, incomplete reaction, poor technique in isolating the product, etc. Therefore we calculate the %Yield to give us some indication as to how complete the reaction is and how good our experimental technique is.

The Percent Yield       =         actual yield      x       100
                                           theoretical yield
 
 

Problems:

• What is the theoretical yield of NaCl in the following reaction if you start with 1.00 g each of NaOH and HCl?

                      NaOH + HCl ® NaCl + H₂O
given amounts        1.00 g        1.00 g

The mole ratio in the balanced chemical equation of NaOH : HCl is  1
Calculate the mole ratio of the given amounts:                                    1

1.00 g NaOH       x      1 mol NaOH      =      0.0250 mol NaOH
                                   40.0 g NaOH

1.00 g HCl         x         1 mol HCl        =       0.0274 mol HCl
                                    36.5 g HCl

mole ratio NaOH : HCl    =       0.0250    =      0.91
                                                   0.0274              1

Since the ratio is less than 1 : 1 the limiting reactant is NaOH (there is not enough NaOH for all of the HCl present, therefore the NaOH will be completely used and there will be some HCl left over).  Any stoichiometric calculations will be based upon the original amount of the NaOH, only.

The theoretical yield of NaCl is then:

      initial mol of A                                 mole ratio B:A                     MM of B                        Th.Y. of B
0.0250 mol NaOH       x         1 mol NaCl      x      58.5 g NaCl    =     1.46 g
                                              1 mol NaOH             1 mol NaCl             NaCl



• Nitrobenzene, C₆H₅NO_2(l), is the substance that gives shoe polish its odor. It is made from the reaction of benzene (C₆H₆) and nitric acid (HNO₃) according to the reaction

                     C₆H_6(l) + HNO_3(aq) ® C₆H₆NO_2(l) + H₂O_(l)

If the actual yield of nitrobenzene is 28.7 g when 20.3 g of benzene reacts with excess nitric acid, calculate the theoretical yield and the percentage yield.

We do not have to determine limiting reactant here because we have only one given amount (the other reactant is present in excess amount). We just need to calculate the theoretical yield of nitrobenzene, based upon the given starting amount of 20.3 g of benzene.

20.3 g C₆H₆       x       1 mol C₆H₆    x      1 mol C₆H₆NO₂    x      124 g C₆H₆NO₂
                                  78.0 g C₆H₆            1 mol C₆H₆                   1 mol C₆H₆NO₂

Theoretical Yield = 32.3 g C₆H₆NO₂

Calculate the % Yield:

actual yield       x       100       =        28.7 g C₆H₆NO₂     x     100 =
theo. Yield                                        32.3 g C₆H₆NO₂

% Yield = 88.9%

Tips for Stoichiometry Problems:

• Always start with a balanced chemical equation.
• Determine if the problem is a limiting reactant problem (do you have two or more specific reactant amounts given?).
• If the given amounts of the substances are given in grams, change amounts to moles.

                          2H₂ + O₂ ® 2H₂O

The mole ratio is:         2 mole H₂          in conversion factor form
                                     1 mole O₂

                                     2 mole H₂O
                                     1 mole O₂   


                                     2 mole H₂O
                                     2 mole H₂

The mole ratio converts between moles of "different substances" in the balanced chemical equation

The pattern for solving stoichiometry problems is:

    _{start           convert using                           convert using                        convert using            end
    ^{with           molar mass A                          mole ratio B:A                         molar mass            with}}
grams A             ®            moles A          ®            moles B         ®           grams B

If the problem is a limiting reactant problem, you must first determine which reactant is the limiting reactant, in other words, which reactant is present in the smallest stoichiometric amount or will be used up completely in the reaction.

All stoichiometric calculations will be based only on the initial limiting reactant amount, in other words, start all calculation procedures with the limiting reactant amount (whether it be in grams or moles). In other words, A (in the above scheme) is always the limiting reactant.