# Ralph & Bev Shock ### Chapter 8

Stoichiometry

Stoichiometry is the mathematical relationship between the reactants and the products in a chemical reaction. Stoichiometry allows for mass and energy accounting using the molar ratios between any combination of reactant to product, product to reactant, reactant to reactant or product to product.

There are two Laws of nature which you must keep in mind as you balance equations and work with the mathematical relationships:

LAW OF DEFINITE COMPOSITION - Every compound has a definite compositon by weight (mass). This law is valid because: (a) Atoms of an element always have the same average mass, and (b) elements combine in definite proportions by number of atoms to form compounds, because of definite ability to gain, lose, or share electrons.

LAW OF CONSERVATION OF MATTER - Matter is not created or destroyed in a chemical reaction:

(a) Atoms are conserved - the number of atoms of each element in the reactants will be the same as the number of atoms of each element in the products, and

(b) Mass is conserved - the total mass of all the reactants used will equal the total mass of all the products produced.

TO DO STOICHIOMETRY YOU MUST:

(A) WRITE THE BALANCED CHEMICAL EQUATION FOR THE REACTION, AND

(B) UNDERSTAND THE MOLE CONCEPT.
A MOLE IS
(1) 6.02 x 1023 particles (Avogadro's number)
(2) The gram atomic weight of atoms (Tha atomic weight expressed in grams)
(3) The gram molecular weight of molecular compounds.
(4) The gram formula weight of ionic compounds.

READING BALANCED EQUATIONS: The coefficients in a balanced equation indicate how many moles of each reactant and product would theoretically be involve in the reaction. From the mass of the substance used you calculate the number of moles actually involved in the reaction.

EX. How many moles of water would be produced by reacting 0.5 moles of oxygen with an excess of hydrogen? The balanced equation: 2H2 + O2 ---> 2H2O Reads - 2 moles of hydrogen molecules reacts with one mole of oxygen molecules to yield two moles of water molecules.
Since: 1 mole O2 ---> 2 moles H2O
Then: 0.5 mole O2 ---> 1 mole H2O

What weight of water is this? H2O = 18.0 g/mole x 1 mole = 18.0 g

I. SETTING UP THE PROBLEM: (these step are required in all problems no matter what method of solution is used.)
1. Write the balanced equation for the reaction.(With correct formulas !)

2. Above the appropriate symbols in the equation give the problem specifications:
a. What value(s) is given?
b. What does the problem ask you to solve for?

3. Set up a factor label problem starting with the values and units given in the problem for the substance known.

4. Convert these values to moles by determining the mass of one mole (formula weight in grams) of each substance involved in the problem.

5. Using a the mole-mole ratio between the unknown substance and the known convert the moles of known substance into moles for the substance you are solving for. This ratio is the "bridge" that gets you from what you started with to what you want. The moles of what you want is always written in the numerator and the moles of what you started with is always written in the denominator.

6. Convert the moles of the substance your solving for into the units asked for in the problem.

Example: How many grams of rust (iron(III) oxide) will be produced by the complete oxidation of 15.0 g of iron? The Iron metal is the know substance and you are solving for the mass(grams) of the Iron(III)oxide.

1. Balance the equation:
4Fe + 3O2 ----> 2Fe2O3
2. Show the problem Specifications:
4Fe + 3O2 ----> 2Fe2O3

15.0 g mass Fe2O3?
3 & 4. Set up factor label problem starting with amount and units of known substance and converting to moles by determining the molar mass.
4Fe + 3O2 ----> 2Fe2O3 5. Next using the mole-mole ratio from the balanced equation: 4 moles of Fe to 2 moles of Fe2O3 as your bridge to determine the moles of Fe2O3.
4Fe + 3O2 ---> 2Fe2O3 NOTE! moles of what you want is on the top.

6. Finally convert moles of what you want: (Fe2O3) into the units of asked for in the problem. In this case; grams by using the gram formula mass for Fe2O3. Many stoichiometry problems give you amounts for more that one reactant, if the amounts of the reactants are not in exact stoichiometric ratios then there will be some of one of the reactants left over and the other is controlling how much of each product is made. These are limiting reactant or limiting reagent problems.

Example : When hydrazine N2H4 (a type of rocket fuel) reacts with oxygen O2 to from nitrogen and water according to the balanced equation: N2H4 + O2 --> 2H2O + N2

One mole of hydrazine reacts with one mole of oxygen; therefore their mole ratio is 1 to 1.

If we react 3.0 moles of hydrazine with 2.5 moles of oxygen is should be obvious that 0.5 moles of hydrazine will be unreacted and 2.5 moles of the hydrazine will react with 2.5 moles of oxygen.

This means the oxygen was the limiting reactant (reagent) because it was "limiting" how much product is made.

The steps for solving limiting reactant problem are as follows if you cannot determine who the limiting reactant is from inspection:

1. Write down a balanced equation:
2. Determine the limiting reagent(reactant) if there is one
a. pick one reactant and solve for the other using stoiciometry:
b. write down what amount of a substance you know
c. convert moles
d. use the mole-mole ratio from the coefiecients from the balanced equation at the center of your problem.
e. convert to the units to for the other reactant as given in the problem. This lets you do a direct comparison based on what is stated in the problem.
f. if the amount you solved for is greater that the amount stated it the problem then you don't have enough of that reactant and therefore is limiting, if the amount is less than stated in the problem you have enough and the other reactant is limiting.
3. Once the limiting reactant is determined start with this substance and solve for the unknow:
a. write down what amount of a substance you know
b. convert moles
c. use the mole-mole ratio from the coefiecients from the balanced equation at the center of your problem.
d. convert to the units to the unit you want.

Example: How many liters of nitrogen gas at STP is produced by the combusiton of 10.0 grams of hydrazine with 8.00 grams of oxygen.

If you can't just look at the data and determine which reactant is limiting by inspection then.

1. pick one reactant at random and solve for the other.

10.0g 8.00g

N2H4 + O2 --> 2H2O + N2

Lets pick the 8.00 g O2 and solve for the hydrazine. The question we are really asking ourselves is how much hydrazine is needed to completely react with 8.00g of O2. This shows that 8.00g hydrazine are needed to react with 8.00 grams oxygen therefore, the oxygen is limiting and we have 2.00 grams of hydrazine in excess or unreacted. This shows that 10.00g of oygen is needed to react with 10.0 g of hydrazine, however the problem states we only have 8.00 g of oxygen therefore oxygen is still determined to be the limiting reagent.

So now that you have determined the limiting reagent, solve for what the problem is really asking for; the liters of N2 formed.

Start with the amount of limiting reagent (O2) and solve for the N2. We now have determined the amount of nitrogen produced based on the limiting reagent O2.

Remember! First determine limiting reactant, then solve for what the problem is asking for.

Another Example:

Given 20.0 grams of Fe reacting with 16.0 grams of sulfur to form Iron(III)sulfide, how many grams of Iron(III)sulfide are formed.
2Fe + 3S ------------->Fe2S3

It's a limiting reactant problem! (because there are amount for more than one reactant)

pick one: 20.0g Fe x mole(Fe)/55.8g(Fe) x (3moles S/ 2moles Fe) x 32.0 g(S)/mole(S) = 17.2 grams of sulfur needed.

Since the problem states that we only have 16.0 grams S, Sulfur is limiting and we start our problem with sulfur and solve to Iron(III)sulfide in grams.

16.0 g(S) x mole(S)/32.0g(S) x(1mole Fe2S3/3moles S) x 207.6 g(Fe2S3)/mole(Fe2S3)=34.6 grams of Fe2S3

Percent Yield Problems:

Percent yield is just the amount of product that was collected(measured) experimentally divided by the theoretical amount that should of been produced times 100%. Theoretical amount is calculated from the stoichiometry equation. The experimental amount is given in the problem or determined experimentally in the lab. In other words it the is the fraction of a product you got experimentally or given compared to what you should of produced mathmatically represented as a percent.

Example:

The experimental decompostion of 413.0 grams of Aluminium chlorate produces 198.0 liters of oxygen at STP. What is the percent yield?

1. balanced equation: 2Al(ClO3)3 --> 2AlCl3 +9O2

2. stoichiometry: The theoretical yield is 201.6 Liters and the experimental as given in the problem is 198.0 liters, therefore the percent yield is: This means we produced 98.2% of the theoretical amount.