Reaction Stoichiometry
Much like atoms combine in specific ratios to make compounds, elements and
compounds can undergo reactions in specific ratios. These ratios come from a balanced
chemical equation.
Consider the following balanced chemical equation.
2Mg(s) + O2(g)  2MgO(s)
In it, we can see that two atoms of magnesium combine with one molecule of diatomic
oxygen to make two molecules of magnesium oxide. Isn’t that nice. We can clearly see
this on the scale of atoms molecules to see the ratio of 2:1:2. However, what would
happen if we looked at this on a scale of moles? The ratio would still apply. That means
we would have two moles of magnesium atoms combining with one mole of diatomic
oxygen molecules to yield two moles of magnesium oxide. The ratio still applies, and it
comes from the coefficients of the balanced chemical equation.
Since we can easily convert grams of an element or compound into moles, we can use the
same ratios to compare amounts on a gram scale.
Problem: Starting with 4.50g of Mg(s), how many grams of O2(g) would be required
for a complete reaction?
The question is asking, if we have 4.50g of magnesium, how many grams of oxygen
would react with it?
We know that the reaction calls for 2 moles of magnesium reacting with 1 mole of
oxygen to make 2 moles of magnesium oxide. So we start with 4.50g of Mg. How many
moles does that represent?
4.50 g Mg 
1 mole Mg
 0.185 moles Mg
24.30 g
Knowing how many moles this represents, we can then use the ratio to determine how
many moles of oxygen are needed for a complete reaction. For every 2 moles of Mg, 1
mole of O2 is required.
0.185 moles Mg 
1 mole O2
 0.0925 moles O 2
2 mole Mg
OK. So know we know that 0.185 moles (4.50g) of Mg need 0.0925 moles of O2 to react
completely. How many grams does that represent?
0.0925 moles O 2 
32.00 g O 2
= 2.96g O 2
1 mole O 2
That in a nutshell is the process. It is simple
1. Convert the given units to moles
2. Multiply by the ratio from the balanced chemical equation
3. Convert the moles back to grams or the desired unit
With that in mind, let us answer the following question.
1. When sand (SiO2) reacts with elemental carbon at very high temperatures
silicon carbide and poisonous carbon monoxide are given off, as seen in the
unbalanced chemical reaction below. Starting with 100.0 kg of SiO2, what
mass (in grams) of silicon carbide will be made?
SiO2(s) + C(s)  SiC(s) + CO(g)
In this problem we are given 1.00kg of SiO2(s) and we are trying to determine how
much SiC(s) this will make. Not too terrible. Lets first see how much 1.00kg of
SiO2 is in moles.
1.00kg SiO2 
1000 g 1mole SiO2

 16.64moles SiO2
1kg
60.09g
Now our equation is not balanced so the ratio between silicon oxide and silicon
carbide is unclear. The equation must be balanced.
SiO2(s) + 3C(s)  SiC(s) + 2CO(g)
Looks like the ratio between silicon oxide and silicon carbide is 1:1.
16.64moles SiO2 
1mole SiC
 16.64moles SiC
1mole SiO2
Now, we can see how many grams this represents.
16.64moles SiC 
40.09 g
 667 g SiC
1mole SiC
So, if all of the silicon dioxide reacts, we will have a 667g pile of silicon carbide.
Try a question like this on your own. It combines things from chapters 2 and 3.
2. When potassium nitrate is heated it thermally decomposes producing solid
potassium oxide gaseous nitrogen (N2) and gaseous oxygen (O2). In order to
produce 88.6g of gaseous oxygen, how much potassium nitrate must be
heated? How much gaseous nitrogen will also be produced? (Answer 224g
potassium nitrate. 31.0g gaseous nitrogen.)
Percent Yield
The idea of stoichiometry gives us the ability to predict the yields of various reactions.
For instance in the second example, we cold easily predict the mass of silicon carbibe
created from a set mass of silicon dioxide.
What if you were in the lab, and you had to perform this very reaction? What if you did
this reaction, and only got 558g of silicon carbide (amount chosen randomly). Even
though you thought you did everything right, it looks like you didn’t get 100% of what
you expected. The percent you did get, called the percent yield is shown below.
558 g
100%  83.7%
667 g
Lets look at this problem.
When molecular bromine is added to benzene it produces bromobenzene and
hydrobromic acid. What is the theoretical yield of bromobenzene if 5.62g of
benzene completely reacts with bromine? What is the percent yield if you do this
reaction, and only get 9.88g of bromobenzene?
C6H6 + Br2  C6H5Br + HBr
First lets calculate the number of moles of bromobenzene this reaction makes starting
with 5.62g of the starting material.
5.62 g C6 H 6 
1mole C6 H 6 1mole C6 H 5 Br

 0.07205molesC6 H 5 Br
78.00 g
1mole C6 H 6
Converting to grams
156.9 g
 11.3g C6 H 5 Br
1mole C6 H 5 Br
So theoretically, 5.62g of benzene should make 11.3 g of bromobenzene. This is the
theoretical yield.
0.07205molesC6 H 5 Br 
The percent yield, or the percent that we actually got in the lab would be
9.88 g
100%  87.4%
11.3g
Answer this question.
Lithium nitride can be heated to produce solid lithium and nitrogen gas as shown
below.
Li3N(s)  Li(s) + N2(g)
If you heat 15.0g of lithium nitride in the lab, what is the theoretical yield of solid
lithium? What is the percent yield if you only isolate 6.84g of solid Li? (Answer
8.96g Li and 76.3%)