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Conversions Factors: Stoichiometry
Dr. MJ Patterson
To work stoichiometry problems, there are two types of conversion factors that you need more often than any others:
 molar mass (or molecular weight)  to convert between grams and moles of a single chemical
 coefficients from balanced equation  to convert between moles of one chemical and moles of a different chemical
To use conversion factors, remember to set up the factor so that the units in the bottom of the factor cancel with the units you currently have.
Why do we have to keep converting between grams and moles? The answer is extremely practical. When you go into the chemistry lab, you measure out reactants and products in grams with a balance. You do not spend all afternoon or evening counting individual molecules until you have 6.02 x 10^23 of them. Instead, you go to a balance and weigh out the molecular weight''s worth of the compound.
However, when molecules react, they do so on a simple molecular or molar basis, instead of on a gram basis. For instance, in the Haber Process:
3H_{2(}_{g)} + N_{2(g)} => 2NH_{3(g)}
3 moles of hydrogen react with 1 mole of nitrogen to produce 2 moles of ammonia. The balanced equation does not mean that 3 grams of hydrogen react with 1 gram of nitrogen to produce 2 grams of ammonia.
So, we are stuck with converting between grams and moles to match up what we can do experimentally in the lab with what theory tells us is happening on a molecular scale.
Stringing Together Conversion Factors
The multimedia lesson approaches stoichiometry problems in a stepbystep manner. To solve one problem, there are typically 3 calculations: Take grams of a reactant and convert to moles of reactant; take moles of reactant and convert to moles of product; take moles of product and convert to grams.
I prefer an alternate approach where all of the conversion factors are strung together in one calculation. Sometimes you can solve problems this way even if you initially do not see how to proceed. All you have to do is write down conversion factors relevant to the problem until you see a path to go from the information you were given to the quantity you are asked to find. The guiding principle is to get the units in the denominator to cancel.
To illustrate this approach, the following two examples are worked in the multimedia lesson in a stepbystep approach. Here, they are worked as a single calculation. Compare the two approaches. What way are you most comfortable working problems?
Example 1: from Section 6
If 0.209 g of P_{4} reacts, how many moles of P_{4}O_{10} can be formed?
P_{4(s)} + 5O_{2(}_{g)} => P_{4}O_{10(s)}
Solution 1:
In a single calculation, convert grams of P_{4(s)} to moles using the molecular weight, and then convert moles of P_{4(s)} to moles of P_{4}O_{10} using the coefficients from the given balanced equation.
(0.209 g P_{4})  x  (1 mol P_{4})  x  (1 mol P_{4}O_{10})  = 1.69 x 10^{3} mol P_{4}O_{10}  
(1)  (123.88 g P_{4})  (1 mol P_{4}) 


 








Example 2: from Section 7
How many grams of calcium hydroxide are required to react with 1.00 kg of sulfur trioxide?
SO_{3(}_{g)} + Ca(OH)_{2(s)} + H_{2}O_{(l)} => CaSO_{4}.2H_{2}O_{(s)}
Solution 2:
This problem starts with a mass in kg, requiring one extra conversion step from kg to g. It also requires that you convert the final answer into grams, where Example 1 stopped at moles.
(1.00 kg SO_{3})  x  (1000 g SO_{3})  x  (1 mol SO_{3})  x  (1 mol Ca(OH)_{2})  x  (74.09 g Ca(OH)_{2})  = 925 g Ca(OH)_{2} 
(1)  (1 kg SO_{3})  (80.06 g SO_{3})  (1 mol SO_{3})  (1 mol Ca(OH)_{2}) 