Stoichiometry

Stoichmetry is a valued part of chemistry that allows you to estimate the amount of product formed in a reaction based on a number of factors. Stoichmetry is perform using either dimension analysis or a stoichiometry table.

Mole Ratio: Solving any reaction-stoichiometry problem requires the use of a mole ratio to convert moles or grams of one substance in a reaction to moles or grams of another substance. A mole ratio is a conversion factor that relates the amounts in moles of any two substances involved in a chemical reaction.

Let’s find the mole ratios of the synthesis reaction of antimony and chlorine to produce antimony chloride.

2Sb + 3Cl = SbCl3

The coefficients in a chemical equation must agree with the law of conservation of matter and represent the relative amounts in moles of reactants and products. Thus, 2 moles of antimony and 3 moles of chlorine produce 1 mole of antimony chloride. The relationships between these substances are expressed as the mole ratios of:

Sb:Cl – __2 moles__ 3 moles

Sb:SbCl3 – __2 moles__ 1 mole

Cl:SbCl3 – __3 moles__ 1 mole

Mole-to-Mole conversions: We can combine the idea of converting from grams to moles with mole ratios to figure out mole to mole conversions. Let’s begin with the simple combustion of H2.

2 H2 + O2 → 2 H2O

For this reaction, the mole ratios show that there are 2 moles of H2 and 2 moles of H2O for every 1 mole of O2 used. As a result, if we can find the number of moles for any one of these we can determine how many moles of the others are present. For example-

If we know that there are .45 moles of H2 present, there are .45 moles of H2O and .45/2 or .225 moles of O2 present.

If we are given the grams of a substance present, we can then use the molar mass to convert grams to moles. Using the same example, if there were 10 grams of hydrogen to start, then there were 5 moles present, because the molar mass of H2 is 2 g/mol. We could then use mole ratios to determine how many moles of the other compounds were present. In table form, this information is shown as follows:


 * || H2 || O2 || H20 ||
 * Mass || 10 ||  ||   ||
 * Molar Mass || 2 ||  ||   ||
 * Moles || 5 || 2.5 || 5 ||

As a result, by using the mole ratios there are 2.5 moles of O2 for 5 moles of H2. This is one example of a mole-to-mole calculation.

One type of conversion is gram to gram. Gram to gram conversions require the skills from mole to mole conversions because in order to do this conversion you must convert the grams to moles and then back to grams at the end. The train of thought for gram to gram conversions is gram->mole->gram. Ex. CH4 + 2O2 → CO2 + 2H2O Given this balanced equation, if we had 48 grams of O2, how many grams of CO2 would be produced? - First the grams of O2 have to be converted to moles so that the mole ratio can be used. CH4 + O2 → CO2 + H2O


 * || CH4 || O2 || CO2 || H2O ||
 * Mass ||  || 48 ||   ||   ||
 * Molar Mass || 16 || 32 || 44 || 18 ||
 * Moles ||  ||   ||   ||   ||

- I filled in the given information and the molar masses into the table. Now by using the table I will divide as I move down to get the moles of O2.


 * || CH4 || O2 || CO2 || H2O ||
 * Mass ||  || 48 ||   ||   ||
 * Molar Mass || 16 || 32 || 44 || 18 ||
 * Moles ||  || 1.5 ||   ||   ||

- Now that the moles of O2 is figured out, by using the 2:1 mole ratio of O2 to CO2 the moles of CO2 can be inserted in the table. - By multiplying up the table the grams of CO2 can be figured out to be 33 grams.
 * || CH4 || O2 || CO2 || H2O ||
 * Mass ||  || 48 ||   ||   ||
 * Molar Mass || 16 || 32 || 44 || 18 ||
 * Moles ||  || 1.5 || .75 ||   ||
 * || CH4 || O2 || CO2 || H2O ||
 * Mass ||  || 48 || 33 ||   ||
 * Molar Mass || 16 || 32 || 44 || 18 ||
 * Moles ||  || 1.5 || .75 ||   ||

Percent Yield

Percent yield is calculated by first dividing the actual mass recorded by the predicted mass of the product. That decimal is then multiplied by one hundred and the percent yield is found.

The predicted yield is determined by the masses used in a reaction and the mole ratios in the balanced equation. This predicted yield is the "ideal". It is not always possible to get this amount of product. Reactions are not always simple. There often are competing reactions. For example, if you burn carbon in air you can get carbon dioxide and carbon monoxide formed.

Example: After performing a reaction the predicted (ideal) mass of the product is 36 grams. However, the lab group only recorded 24 grams of actual product. What is the percent yield of the reaction?

Actual divided by predicted times one hundred equals percent yield.

24 divided by 36 equals .6667 .6667 times one hundred equals 66.67 percent Percent yield equals **66.67%**

Percent Error

Percent error is 100 minus the percent yield

Example: After performing a reaction the predicted (ideal) mass of the product is 48 grams. However, the lab group only recorded 36 grams of actual product. What is the percent error of the reaction?

36 divided by 48 equals .75 .75 times 100 equals 75 percent Percent Yield equals 75%

Percent error equals 100 minus percent yield

100 – 75 equals 25 percent Percent error equals **25%**

Limiting Reagent**
 * Stochiometry Section E.

Limiting Reagent- Opposite of excess reactant; the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed.



There are 8 car bodies here, each one of them using 4 tires, and there are 48 tires. After 4 tires go on each car, the excess tires are of no use. The 8 car bodies in total use 32 tires, leaving 16 tires not in use. No matter how many tires there are, no more than 32 tires can be used on the 8 car bodies. In this example the car bodies are the //limiting reagent// due to the fact that the car bodies limit the amount of cars that can be made when there are over 32 tires. However, in another example, the car tires can be the limiting reagent if there are 8 car bodies and less than 32 tires. This is because less than 32 tires would mean that 8 cars could not be made, thus the tires would be //limiting// the creation of cars.

Limiting Reagent Calculations- A 25.0 g sample of Zinc is mixed with 30.0 g of sulfur. Find the limiting reactant and how much excess reactant remains after the reaction has stopped (Find all molar masses on table of elements)

1.) First create a balanced equation 1Zn(s) + 1S(s) > 1ZnS

2.) Next we use Stochiometry to calculate the amount of moles used for all reactants


 * || Zn || S ||>  || ZnS ||
 * Grams || 25.0 || 30.0 ||  ||   ||
 * Molar Mass || 65.4 || 32.1 ||  ||
 * Moles || **.38** || **.94** ||  ||

With this you can see which reactant is the limiting reagent because it is the lesser amount, in this case the limiting reagent is Zinc Zn S > ZnS


 * || Zn || S ||>  || ZnS ||
 * Grams || 25.0 || 30.0 ||  ||   ||
 * Molar Mass || 65.4 || 32.1 ||  ||
 * Moles || **.38** || .94 ||  ||

3.) Now use .38 moles as the limiting reagent to find the amount of ZnS created Zn S > ZnS


 * || Zn || S ||>  || ZnS ||
 * Grams || 25.0 || 30.0 ||  || **36.7** ||
 * Molar Mass || 65.4 || 32.1 ||  || **96.5** ||
 * Moles || .38 || .94 ||  || **.38** ||

37.6 grams of Zns is created