Class Notes: Unit 7 – Stoichiometry (Ch. 12)
When you’re cooking, you usually use a recipe. It tell you how much of each ingredient
is necessary and then how many servings it makes. If you need to serve more or less
people, you can change the overall amount of ingredients you use (double, halve, etc)
depending on what you need.
Balanced chemical equations are your recipes in chemistry. They tell you exactly how
much of each ingredient you need and you can predict how much product you’ll have at
the end. This is important because anytime you are doing a reaction in the real world you
are usually doing it to get a specified amount of product(s). Imagine a company just
mixing things together that they know react without knowing exactly how many things
they will be able to have to use/sell.
Stoichiometry is the calculation of amounts of substances involved in chemical reactions.
It comes from the Greek words stoikheioin, meaning element, and metron, meaning to
measure. Stoichiometry allows chemists to track reactants and products using ratios of
moles or particles. It is all based off of balanced equations.
Interpreting Balanced Equations
Coefficients in a balanced chemical equation state the relationships (or ratios) among
substances involved in the reaction. Because it is just a ratio, a balanced chemical
equation can be interpreted in different ways; similar to how we can express chemical
quantities of substances in different ways. Each interpretation says exactly the same thing
just in terms of different chemical quantities.
Balanced chemical equations can be interpreted potentially in 5 different ways:
Number of (1) Atoms
(2) Molecules or Compounds
(3) Moles
(4) Mass
(5) Volume
Ex: N2(g) + 3H2(g) → 2NH3(g)
Atoms:
2 N atoms : 6 H atoms → 2 N atoms : 6 H atoms
(*conserved)
Molecules:
1 nitrogen molecule : 3 hydrogen molecules → 2 ammonia molecules
Moles:
1 mole nitrogen : 3 moles hydrogen → 2 moles ammonia
Mass:
28.02 g N2 : 6.048 g hydrogen → 34.07 g ammonia
(*conserved)
Volume:
22.4 L nitrogen : 67.2 L hydrogen → 44.8 L ammonia
Mole Ratios
The most important and useful interpretation of balances chemical equations is the mole.
Once you know the mole ratio between your substances, you can use that to find other
chemical quantities. (Remember the mole is the intermediate step between your three
chemical quantities – mass, volume, and particles - on the mole road map)
Ex: What are the mole ratios involved in this chemical reaction?
N2(g) + 3H2(g) → 2NH3(g)
1 mole N2 : 3 moles H2 : 2 moles NH3
These molar ratios written this way can also be expressed in terms of fractions:
1molN 2
1molN 2
3molH 2
3molH 2
2molNH 3
2molNH 3
All of these three ratios are also true inverted (flipped).
Mole to Mole Calculations
You can use the molar ratios to go from quantities of one thing to quantities of another.
The thing you are finding can always be either a reactant or a product.
Ex: N2(g) + 3H2(g) → 2NH3(g) How many moles of ammonia are produced when 0.60
moles of nitrogen reacts with hydrogen?
 0.60molN 2  2molNH 3 
  1.2molNH 3


1

 1molN 2 
Ex: N2(g) + 3H2(g) → 2NH3(g) How many moles of hydrogen are needed to react with 0.60
moles of nitrogen to produce ammonia?
 0.60molN 2  3molH 2 
  1.8molH 2


1

 1molN 2 
Ex: 4Al(s) + 3O2(g) → 2Al2O3(s) What are the mole ratios in this equation? How many
moles of aluminum are needed to form 3.7 moles of aluminum oxide?
 3.7molAl2 O3  4molAl 
  7.4molAl


1

 2molAl2 O3 
Ex: 4Al(s) + 3O2(g) → 2Al2O3(s) How many moles of oxygen react completely with 14.8
moles of aluminum?
 14.8molAl  3molO2 
  11.1molO2


1

 4molAl 
Ex: 4Al(s) + 3O2(g) → 2Al2O3(s) How many moles of aluminum oxide are formed with
0.78 moles of oxygen react with aluminum?
 0.78molO2  2molAl2 O3 
  0.52molAl2 O3


1

 3molO2 
Mass to Mass Calculations
A reaction’s chemical equation is set up based on the ratio in which moles of reactants
combine with each other to form moles of products. However, in the lab and industry
reactants are always bought, sold, and used by mass. For this reason, it is useful to know
how to carry out mass to mass stoichiometric calculations.
The general steps for mass to mass calculations are:
Mass A → Moles A → Moles B → Mass B
The conversion from mass to moles or moles to mass is the basic calculation involving
molar mass (from Ch. 10). The conversion from moles A to moles B is performed using
the mole ratio of the two substances.
Because the number of grams in one mole varies depending on the substance (with its
molar mass), the mole ratio must be used to convert from one substance to another when
q mass quantity is involved.
Ex: N2(g) + 3H2(g) → 2NH3(g) How many grams of ammonia are produces when 5.40 g of
hydrogen reacts with nitrogen?
 5.40 gH 2  1molH 2  2molNH 3  17.03gNH 3 
  30.41gNH 3




1

 2.016 gH 2  3molH 2  1molNH 3 
Ex: CaC2 + 2H2O → C2H2 + Ca(OH)2 5.00 g of CaC2 reacts with water to produce
what mass of C2H2?
 5.00 gCaC2  1molCaC2  1molC2 H 2  26.04 gC2 H 2 


  2.03gC2 H 2


1

 64.12 gCaC2  1molCaC2  1molC2 H 2 
Ex: CaC2 + 2H2O → C2H2 + Ca(OH)2 How many moles of CaC2 react completely with
49 g of water?
 49 gH 2O  1molH 2O  1molCaC2 

  1.4molCaC2


1

 18.02 gH 2O  2molH 2O 
Stoichiometric Calculations
Using a very similar process, you can convert between any of the three chemical
quantities on the mole road map (mass, volume, or particles) all through the mole ratio
connection.
The basic stoichiometric process is:
1. convert the given quantity to moles
2. use the mole ratio from the balanced chemical equation to convert the moles of
the given substance to the moles of the desired substance
3. convert moles to desired quantity
Ex: How many molecules of oxygen are produced when 29.2 g of water is decomposed?
2H2O → 2H2 + O2
(Mass H2O → Moles H2O → Moles O2 → Molecules O2)
23
 29.2 gH 2O  1molH 2O  1molO2  6.02 x10 moleculesO 2 





  4.88 x1023 moleculesO 2






1
1molO2

 18.02 gH 2O  2molH 2O 

Ex: How many liters of NO2 are produced when 34 L of O2 reacts with NO at STP?
2NO(g) + O2(g) → 2NO2(g)
(Volume O2 → Moles O2 → Moles NO2 → Volume NO2)
 34 LO2  1molO2  2molNO2  22.4 LNO2 


  68LNO2


 1  22.4 LO2  1molO2  1molNO2 
You can notice from this example above that if mass is not involved as either a initial or
final quantity and the beginning and ending quantities are the same then it is not as
necessary to go through the whole process of converting to and from moles for the
different substances; the mole ratio between the two can just be directly applied as either
a volume ratio or a particles ratio:
 34LO2  2LNO2 
  68LNO2


 1  1LO2 
Limiting and Excess Reagents
Whenever a reaction is being carried out in which there is more than one reactant, you
need to consider the amounts of each reactant to produce a specific amount of product.
For example, if you wanted to make 10 ham and cheese sandwiches and you have 10
pieces of cheese but only 5 pieces of ham, you are limited in how many ham and cheese
sandwiches you can make.
You can only make as much product as you have reactants for and whichever reactant
runs out first is your limiting reactant or “limiting reagent”. The reactant(s) that are left
over when your limiting reactant has run out is known as the excess reactant or “excess
reagent”.
Ex:
If we started with 2 molecules of N2 and 3 molecules of H2, how many molecules of NH3
could be produced?
Every new product molecule needs 1 N and 3 H. All of the H2 molecules would get used
up and there would be one N2 molecule left.
Therefore, the H2 is the limiting reagent because it is the lack of more hydrogen that
makes the reaction stop and the N2 is the excess reagent because there are additional
nitrogen molecules left when the reaction has stopped.
It is the limiting reagent that determines the amount of product that can be formed.
Therefore, if quantities are given for each of the reactants in a reaction, the limiting
reagent must be determined first because it is that one that you need to use to determine
the amount of product you can produce.
The limiting reagent can change for any reaction; it all depends on how much of each
reactant you have to begin with.
Ex: If you have 10 water bottles and 5 caps, the caps are your limiting reagent.
If you have 10 water bottles and 10 caps, there is no limiting reagent.
If you have 10 water bottles and 15 caps, the bottles are your limiting reagent.
Method to Determine Limiting and Excess Reagents
You always have to compare moles to determine limiting reagent because masses vary
greatly depending on what your compounds are and just because the mass of one may be
smaller than the other, it may not be the limiting reagent.
Convert the given amount of 1 reactant to moles of the 2nd reactant. This tells you the
amount of the 2nd reactant that is needed to completely react with all of the 1st reactant. If
there is more of the second reactant available than what was calculated to be necessary to
completely react with the 1st reactant, then the 2nd reactant is in excess. If there is less of
the 2nd reactant available than what you calculated to be necessary to completely react
with the 1st reactant, then the 2nd reactant is limiting.
Ex: H3PO4 + 3NaOH → Na3PO4 + 3H2O
If 1.75 mol of H3PO4 reacts with 5 mol of NaOH which is the limiting reagent?
 1.75molH 3 PO4  3molNaOH 
  5.25molNaOH  needed


1

 1molH 3 PO4 
You need 5.25 mol of NaOH to react with all 1.75 mol of the H3PO4. You only
have 5 mol of NaOH so therefore, the NaOH is the limiting reagent.
You could have also done this calculation starting with the other reactant:
 5molNaOH  1molH 3 PO4 
  1.67molH 3 PO4  needed


1

 3molNaOH 
You need 1.67 mol of H3PO4 to react with all 5 mol of NaOH. You have 1.75
mol of H3PO4 so therefore, the H3PO4 is the excess reagent.
** The maximum amount of product that can be formed is limited by the number of
moles of the limiting reagent. Therefore, any calculations in which the amount of product
is asked for, you must use the number of moles of your limiting reagent to solve. **
Theoretical, Actual, and Percent Yield
When you are calculating the mass of product you can potentially produce based on a
chemical equation you are predicting the yield of your reaction. Yields are always in
terms of mass. The yield determined based on reacting the entire limiting reagent is the
“theoretical yield.” The theoretical yield is the maximum amount of product that could be
formed from a specified amount of reactants.
In most lab settings however it is very difficult to convert 100% of you reactants to
products. This could be due to several factors, such as: contamination, measurement
errors, some materials being lost during weighing or transfer, etc. The amount of product
that actually forms when the reaction is carried out is the “actual yield.”
The “percent yield” is the ratio between the actual and the theoretical yields and is
expressed as a percent. It is a measure of how efficient a reaction was carried out in the
lab. Because the actual yield is often less than the theoretical yield, the percent yield is
often less than 100% (and should never be more than 100% because that means you
created matter).
ActualYiel d
% yield 
*100%
TheoreticalYield
Ex: CaCO3(s) → CaO(s) + CO2(g)
What is the theoretical yield of CaO is 24.8g of CaCO3 is decomposed?
Since there is only one reactant in this case, you do not need to worry about which is
limiting. However, if there were more than one reactant, you would first need to
determine which is limiting to know which to use to solve for product quantities.
 24.8gCaCO3  1molCaCO3  1molCaO  56.08gCaO 




  13.9 gCaO
1

 100.1gCaCO3  1molCaCO3  1molCaO 
Ex: For the example above, if the actual yield was 13.1 g what is the percent yield?
% yield 
ActualYiel d
13.1g
*100% 
*100%  94.2%
TheoreticalYield
13.9 g
Ex: 3Li + Al(OH)3 → Al + 3LiOH
Find the percent yield of Al when 12.5 g of Li reacts with 8.25 g of Al(OH)3 to
produce 2.50 g of Al.
In order to find the percent yield, you need the actual and theoretical yields. The actual
yield is given in the problem. The theoretical yield must be calculated. In order to
calculate the theoretical yield, you must start with the limiting reactant. Because there is
more than one reactant, you first need to determine which is limiting to know which to
use to solve for product quantities.
So the steps for this problem include:
1. Find the limiting reactant (convert 1 reactant to the other and compare with number
in problem)
2. Use the amount of limiting reactant stated in problem to calculate the theoretical
yield.
3. Plug the actual yield (given in problem) and the theoretical yield (just calculated)
into the percent yield equation.
Find the limiting reactant:
 12.5gLi  1moleLi  1moleAl (OH ) 3  78.00 gAl (OH ) 3 
  46.8gAl (OH ) 3 needed




 1  6.94 gLi  3moleLi  1moleAl (OH ) 3 
46.8 g of Al(OH)3 is needed and there are only 8.25 g of Al(OH)3 available to react,
according in the problem. Therefore, there is not enough Al(OH)3 and it is the limiting
reactant.
Use the limiting reactant to calculate the theoretical yield:
 26.98gAl 
 8.25gAl (OH ) 3  1moleAl (OH ) 3  1moleAl




  2.85gAl
1

 78.00 gAl (OH ) 3  1moleAl (OH ) 3  1moleAl 
Plug the actual and theoretical yield into the percent yield equation:
ActualYiel d
2.50 g
2.50 g
% yield 
* 100% 
* 100% 
 87.7%
TheoreticalYield
TheoreticalYield
2.85g
This reaction had an 87.7% yield.