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 H2   +   O2   ------->  2 H2O which can mean            2 molecules of H2   +  1 molecules of O2   -------->  2 molecules of H2O 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 lab-sized units of each chemical. 2  X (6.02 X 1023 molecules) of  H2  +   1 X (6.02 X 1023 molecules) of O2                               -------->         2 X (6.02 X 1023 molecules) of H2O 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 H2   + 1 moles of O2   --------> 2 moles of H2O 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 H2   +   0.01  moles of O2   -------->   0.02 moles of H2O or         1.36 moles of H2   +   0.68 moles of O2   -------->   1.36 moles of H2O or         88 moles of H2   +   44 moles of O2   -------->   88 moles of H2O In every case, the relative mole quantities of H2 to O2 to H2O are 2:1:2.  We could say that 2 moles of H2, 1 mole of O2, and 2 moles of H2O 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 O2   ------->  2 SO3 How many moles of sulphur react in this way with 9 moles of O2? Solution: From the balanced equation you can see that  2 S react with 3 O2 Set up your ratio like this:    2 S  =   3 O2                                             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 2-3 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 C2H6, combines with oxygen O2 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  C2H6    +  7 O2      -------->  4  CO2   +  6  H2O 2.  Compute the needed formula molecular masses.   We need the molecular mass for water and for      ethane.    H2O = 18.02 g.mol;   C2H6 = 30.08 g/mol 3.   Draw a roadmap outline the math steps that need to be taken:                   moles of water  --------->  moles of ethane                            ^                                        |                            |                                        V                  grams of water                   grams 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                             ^                                        |                              |                                        V                   grams of water                  grams 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 C2H6 is needed to make 6 moles of water.  Setup a ratio: 6 mol H2O = 2 mol C2H6                      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 K2Cr2O7 are needed to make 35.8 grams of I2 according to the following equation. K2Cr2O7  +  6 NaI  +  7 H2SO4   ---> Cr2(SO4)3 + 3 I2 + 7 H2O + 3 Na2SO4 + K2SO4 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, I2, and potassium dichromate, K2Cr2O7.  Calculate the molecular masses for these only.  The other parts of the equation can be forgotten about.                     I2 =  253.82 g/mol      K2Cr2O7 = 294.20 g/mol Draw a roadmap outlining the mathematical steps that need to be taken:                   moles of iodine  --------->  moles of potassium dichromate                            ^                                        |                            |                                        V                  grams of iodine                  grams 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 I2    =   1 K2Cr2O7               0.14 mol          x                              x = 0.05 moles of potassium dichromate 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