AP Chemistry
 What is Stoichiometry?

Stoichiometry is the quantitative description of the proportions by moles of the substances in a chemical reaction. 
When doing experiments, it is unscientific and dangerous to mix chemicals in a haphazard manner. Before any practical laboratory work is done, chemists almost always start with a balanced reaction equation. This is true for a reaction that we know works, or for one that we predict works. A balanced equation helps tell the reaction's stoichiometry. 
The stoichiomentry of a reaction is the description of the relative quantitites by moles of the reactants and products as they appear by the coefficients of the balanced equation. Stoichiometry is the molar bookkeeping of chemistry, and in nature the books must balance. Quantitative information is only available through its stoichiometry. This applies to the pure study of chemistry, but also to the uses of chemistry in a wide variety of fields such as agriculture, clinical analysis, pharmaceuticals, food chemistry, inhalation therapy, nutrition, forensic science, geochemistry and the list can go on and on. The work begins with the writing and balancing of an equation, because this is the work that gives the coefficients  the numbers that disclose the proportions by moles. 
Stoichiometric Calculations: Mole to Mole Calculations 
When we balance an equation it is important to think if it in terms of atoms of each element. For example, in a simple reaction between hydrogen and oxygen to make water, the equation we get is 
2 H_{2} + O_{2} > 2 H_{2}O 
which can mean
2 molecules of H_{2}
+ 1 molecules of O_{2}
> 2 molecules of H_{2}O

However, when we use a balanced equaiton to plan how much of each reactant to use in an actual experiment, we have to shift our thinking to huge collections of molecules  to moles. The shift from molecules to moles is done by taking advantage of a simple rule from mathematics. Multiplying a set of numbers, such as the coefficients, by any constant number does not alter the ratios among the members of the set. If we select Avogadro's number as the multiplier then we get labsized units of each chemical. 
2 X (6.02 X 10^{23}
molecules) of H_{2} + 1 X (6.02 X 10^{23} molecules) of O_{2}
> 2 X (6.02 X 10^{23} molecules) of H_{2}O 
The essential 2:1:2 ratio has not been changed by this
multiplication. But the scale of the reaction has shifted to the mole
level.
2 moles of H_{2} + 1 moles of O_{2} > 2 moles of H_{2}O 
The ratio of moles of molecules is identical to the ratio of molecules  it has to be, since equal numbers of moles have equal numbers of molecules. 
The ratio of the coefficients for any given chemical reaction is set by nature. You cannot change this ratio. It is set when you write the formulae correctly and then balance the equation properly. Once this is done the coefficient numbers can be used as the basis for chemical calculations. The decision that is left for us is the scale of the reaction  how much do we want to use or make? The number of options is infinite. We could have 
0.02 moles of H_{2}
+ 0.01 moles of O_{2}
> 0.02 moles of H_{2}O
or 1.36 moles of H_{2} + 0.68 moles of O_{2} > 1.36 moles of H_{2}O or 88 moles of H_{2} + 44 moles of O_{2} > 88 moles of H_{2}O 
In every case, the relative mole quantities of H_{2} to O_{2} to H_{2}O are 2:1:2. We could say that 2 moles of H_{2}, 1 mole of O_{2}, and 2 moles of H_{2}O are equivalent to each other in this reaction. This does not mean that one chemical can actually substitute for any other chemical. It does mean that a specific mole quantity of one substance requires the presence of a specific mole quantity of each of the other substance in accordance with the ratio of coefficients. 
Below shows five different scales for the reaction of iron with sulphur to make iron sulphide, FeS. Notice that the mole ratios are the same regardless of the scale. 
1 atom of Fe + 1 atom of S
> 1 molecule of FeS
10 atoms of Fe + 10 atoms of S > 10 molecules of FeS 55.8 mg of Fe + 32.1 mg S > 87.9 mg FeS 5.58 g of Fe + 3.21 g of S > 8.79 g of FeS 55.8 g of Fe + 32.1 g of S > 87.9 g of FeS

This is an example of how to do mole to mole type problems:
Two atoms of sulphur react with three molecules of oxygen to form two molecules of sulphur trioxide, which is an air pollutant. 2 S + 3 O_{2} > 2 SO_{3} 
How many moles of sulphur react in this way with 9 moles
of O_{2}? Solution: From the balanced equation you can see that 2 S react with 3 O_{2} Set up your ratio like this: 2 S = 3 O_{2} x 9 moles Cross multiply to get 2 * 9 moles = 3 * x
x = (2 * 9 moles) / 3 = 6 moles 
Therefore if 9 moles of oxygen are reacted then 6 moles of
S must also be present. Note that the unit "moles" was carried through the calculation. 
Mole to Mole Calculations Worksheet 
The following file is recommended reading. 
Bridging the Stoichiometry Gap 
Stoichiometric Calculations: Gram to Gram Calculations 
The type of problems involved here can be solved easily with 23 simple steps. Use the Bridge Method for solving only after the equation is balanced. 
Never, ever assume that an equation is balanced. 
Example Solution #1 
During its combustion, ethane C_{2}H_{6}, combines with oxygen O_{2} to give carbon dioxide and water. A sample of ethane was burned completely and the water that formed has a mass of 1.61 grams. How much ethane, in moles and in grams, was in the sample? 
Solution: 
1. Set up the equation based on the words in the
problem. Then balance it correctly.
2 C_{2}H_{6} + 7 O_{2}
> 4 CO_{2} + 6 H_{2}O

2. Compute the needed formula molecular masses. We need the molecular mass for water and for ethane. H_{2}O = 18.02 g.mol; C_{2}H_{6} = 30.08 g/mol 
3. Draw a roadmap outline the math steps that
need to be taken:
moles of water > moles of ethane 
Therefore this type of question requires a three step
solution. Step #1 Convert 1.61 grams of water to moles of water. Step #2 Using the equation, compare the moles of water made from moles of ethane. Step #3 Convert the moles of ethane back into grams of ethane. 
Draw a roadmap outlining the mathematical steps that need
to be taken:
moles of water > moles of ethane 
Math Step #1 Convert 1.61 grams of water into moles of water 
Moles = g/mm = 1.61 g/ 18.02 g/mol = 0.09 mol of water were used. 
Math Step #2 The chemical equation shows us that 2 moles of C_{2}H_{6} is needed to make 6 moles of water. Setup a ratio: 
6 mol H_{2}O = 2 mol C_{2}H_{6} 0.09 mol x x = 0.03 moles of ethane are needed. 
Math Step #3 Convert the moles of ethane back into grams of ethane. 
g = n * mm = 0.03 mol * 30.08 g/mol = 0.90 grams of ethane 
Finish off the question with a statement. 1.61 grams of water can be made from 0.03 moles of ethane or 0.90 grams of ethane. 
Example Question #2 
Calculate how many grams of K_{2}Cr_{2}O_{7} are needed to make 35.8 grams of I_{2} according to the following equation. 
K_{2}Cr_{2}O_{7} + 6 NaI + 7 H_{2}SO_{4} > Cr_{2}(SO_{4})_{3} + 3 I_{2} + 7 H_{2}O + 3 Na_{2}SO_{4} + K_{2}SO_{4} 
The equation is already balanced but you should check it over just to be sure. 
Next calculate the molecular masses that you need. The
question deals with iodine, I_{2}, and potassium dichromate, K_{2}Cr_{2}O_{7}.
Calculate the molecular masses for these only. The other parts of
the equation can be forgotten about.
I_{2} = 253.82 g/mol K_{2}Cr_{2}O_{7} = 294.20 g/mol 
Draw a roadmap outlining the mathematical steps that need
to be taken:
moles of iodine > moles of potassium dichromate

Math Step #1 Convert grams of iodine into moles of
iodine.
moles = g/mm = 35.8 g / 253.82 g/mol = 0.14 moles of iodine 
Math Step #2 Compare using the equation
coefficients, the number of moles of potassium dichromate is need to
make how many moles of iodine.
3 I_{2} = 1 K_{2}Cr_{2}O_{7}

Math Step #3 Convert the moles of potassium dichromate
back into grams of potassium dichromate.
g = n * mm = 0.05 mol * 294.20 g/mol = 14.71 grams To make 35.8 grams of iodine you must start with 14.71
grams
of potassium dichromate. 
Stoichiometry Gram to Gram Calculations Worksheet 
More Gram to Gram Calculations Worksheet 
Go to the Next Stoichiometry Section on Empirical Formulas 