STOICHIOMETRY
BASICS
Unit 12
Chemistry
Langley
**Corresponds to Chapter 12 in the Prentice Hall Chemistry Textbook
INTRODUCTION
 Stoichiometry
 Dervived from two Greek words meaning measured
and element
 4 types




Mole-Mole
Mass-Mole
Mass-Mass
Volume-Volume
 Mathematically based
 Balanced equations very important; first place to
start
Interpreting Chemical
Reaction
 From a balanced chemical equation, you
can describe a chemical reaction in terms
of quantities of products and reactants.
These quantities include the number of
molecules, atoms, moles, mass, and
volume of the compounds used as well
as each element used to make the
respective compounds
Interpreting a Chemical
Reaction
 N2 + 3H2  2NH3 (Figure 12.3, pg. 357)
 Number of atoms: 2 atoms of nitrogen, 6 atoms of
hydrogen, 8 atoms of ammonia
 Number of molecules: 1 molecule of nitrogen, 3
molecules, 2 molecules of ammonia (1:3:2 ratio)
 Number of relative moles: 1 mole of nitrogen, 3
relative moles of hydrogen, 2 relative moles of
ammonia
 Molecules and moles are not the same thing
 Mass: 28.0 g of nitrogen, 6.0 g of hydrogen, 34 g of
ammonium
 Volume: 22.4L of nitrogen, 67.2L of hydrogen,
44.8L of ammonia
 Because 1 mol of any gas at STP occupies a volume of
22.4L
Molar Mass of Compounds
 The molar mass (MM) of a compound is
determined the same way, except now you add
up all the atomic masses for the molecule (or
compound)




Ex. Molar mass of CaCl2
Avg. Atomic mass of Calcium = 40.08g
Avg. Atomic mass of Chlorine = 35.45g
Molar Mass of calcium chloride =
40.08 g/mol Ca + (2 X 35.45) g/mol Cl
 110.98 g/mol CaCl2
20
Ca
40.08
17
Cl
35.45
Molar Mass Calculation
 Calculate the Molar Mass of calcium
phosphate
 Formula = (NH4)3(PO4)
 Masses elements:
 Molar Mass =
Atoms or
Molecules
Divide by 6.02 X 1023
Multiply by 6.02 X 1023
Moles
Multiply by
atomic/molar mass
from periodic table
Divide by
atomic/molar mass
from periodic table
Mass
(grams)
Calculations
molar mass
Grams
Avogadro’s number
Moles
particles
Everything must go
through Moles!!!
Chocolate Chip Cookies!!
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
How many eggs are needed to make 3 dozen cookies?
How much butter is needed for the amount of chocolate chips used?
How many eggs would we need to make 9 dozen cookies?
How much brown sugar would I need if I had 1 ½ cups white sugar?
Recipes and Equations
 Just like chocolate chip cookies have recipes,
chemists have recipes as well
 Instead of calling them recipes, we call them
reaction equations
 Furthermore, instead of using cups and
teaspoons, we use moles
 Lastly, instead of eggs, butter, sugar, etc. we
use chemical compounds as ingredients
Chemistry Recipes
 Looking at a reaction tells us how much of
something you need to react with
something else to get a product (like the
cookie recipe)
 Be sure you have a balanced reaction
before you start!
 Example: 2 Na + Cl2  2 NaCl
 This reaction tells us that by mixing 2 moles of
sodium with 1 mole of chlorine we will get 2 moles
of sodium chloride
 What if we wanted 4 moles of NaCl? 10 moles?
50 moles?
Starting Point-Balanced
Equation
 Write the balanced reaction for hydrogen gas
reacting with oxygen gas.




2 H2 + O2  2 H2O
How many moles of reactants are needed?
What if we wanted 4 moles of water?
What if we had 3 moles of oxygen, how much
hydrogen would we need to react and how much
water would we get?
What if we had 50 moles of hydrogen, how much
oxygen would we need and how much water
produced?
Mole-Mole Calculations
 These mole ratios can be used to
calculate the moles of one chemical from
the given amount of a different chemical
 Example: How many moles of chlorine is
needed to react with 5 moles of sodium
(without any sodium left over)?
2 Na + Cl2  2 NaCl
5 moles Na 1 mol Cl2
2 mol Na
= 2.5 moles Cl2
Mole-Mole Calculations
 How many moles of ammonia are
produced when 0.60 moles of nitrogen
reacts with hydrogen?
 N2(g) + 3H2 (g)  2NH3(g)
0.6 mol N2
2.0 mol NH3
1.0 mol N2
= 1.2 mol NH3
Mass-Mole Calculations
 Sometimes you are going to start with mass
and will have to convert to moles of product or
another reactant
 We use molar mass and the mole ratio to get
to moles of the compound of interest
 Calculate the number of moles of ethane (C2H6)
needed to produce 10.0 g of water
 2 C2H6 + 7 O2  4 CO2 + 6 H20
10.0 g H2O 1 mol H2O
2 mol C2H6
18.0 g H2O 6 mol H20
= 0.185
mol C2H6
Mass-Mole Calculations
 Calculate how many moles of oxygen are
required to make 10.0 g of aluminum
oxide
Mole-Mass Calculations
 Most of the time in chemistry, the amounts are
given in grams instead of moles
 We still go through moles and use the mole ratio,
but now we also use molar mass to get to grams
 Example: How many grams of chlorine are required
to react completely with 5.00 moles of sodium to
produce sodium chloride?
2 Na + Cl2  2 NaCl
5.00 moles Na 1 mol Cl2
2 mol Na
70.90g Cl2
1 mol Cl2
= 177g Cl2
Mole-Mass Calculations
 Calculate the mass in grams of Iodine
required to react completely with 0.50
moles of aluminum.
Mass-Mass Calculations
 Most often we are given a starting mass
and want to find out the mass of a product
we will get (called theoretical yield) or how
much of another reactant we need to
completely react with it (no leftover
ingredients!)
 Now we must go from grams to moles,
mole ratio, and back to grams of
compound we are interested in
Mass-Mass Calculations
 Ex. Calculate how many grams of
ammonia are produced when you react
2.00g of nitrogen with excess hydrogen.
 N2 + 3 H2  2 NH3
2.00g N2 1 mol N2 2 mol NH3 17.06g NH3
28.02g N2 1 mol N2
= 2.4 g NH3
1 mol NH3
Mass-Mass Calculations
 How many grams of calcium nitride are
produced when 2.00 g of calcium reacts
with an excess of nitrogen?
Volume-Volume Calculations
 1 mole of any substance = 22.4L
 Nitrogen monoxide and oxygen gas
combine to form the brown gas nitrogen
dioxide, which contributes to
photochemical smog. How many liters of
nitrogen dioxide are produced when 34 L
of oxygen reacts with an excess of
nitrogen monoxide? Assume conditions
of STP.
Volume-Volume Calculations
 2NO + O2(g)  2NO2(g)
 Know: volume of oxygen = 34L O2
2 mol NO2/1 mol O2
1 mol O2 = 22.4 L O2 (at STP)
1 mole NO2 = 22.4 L NO (at STP)
34 L O2
1 mol O2
2 mol NO2
22.4 L O2
1 mol O2
22.4 L NO2
1 mol NO2
= 64 L O2
Limiting Reactants
1 cup butter
1/2 cup white sugar
1 cup packed brown sugar
1 teaspoon vanilla extract
2 eggs
2 1/2 cups all-purpose flour
1 teaspoon baking soda
1 teaspoon salt
2 cups semisweet chocolate chips
Makes 3 dozen
If we had the specified amount of all ingredients listed, could we make 4
dozen cookies?
What if we had 6 eggs and twice as much of everything else, could we make 9
dozen cookies?
What if we only had one egg, could we make 3 dozen cookies?
Limiting Reactant
 Most of the time in chemistry we have more of
one reactant than we need to completely use
up other reactant.
 That reactant is said to be in excess (there is
too much).
 The other reactant limits how much product
we get. Once it runs out, the reaction.
This is called the limiting reactant.
Limiting Reactant
 To find the correct answer, we have to try all of
the reactants. We have to calculate how much
of a product we can get from each of the
reactants to determine which reactant is the
limiting one.
 The lower amount of a product is the correct
answer.
 The reactant that makes the least amount of
product is the limiting reactant. Once you
determine the limiting reactant, you should
ALWAYS start with it!
 Be sure to pick a product! You can’t compare to
see which is greater and which is lower unless
the product is the same!
Limiting
Reactant
Limiting
Reactant:
 10.0g of aluminum reacts with 35.0 grams of
chlorine gas to produce aluminum chloride. Which
reactant is limiting, which is in excess, and how
much product is produced?
2 Al + 3 Cl2  2 AlCl3
 Start with Al:
10.0 g Al
1 mol Al
27.0 g Al
2 mol AlCl3 133.5 g AlCl3
2 mol Al
1 mol AlCl3
= 49.4g AlCl3
 Now Cl2:
35.0g Cl2
1 mol Cl2
71.0 g Cl2
2 mol AlCl3 133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 43.9g AlCl3
Limiting Reactant
 We get 49.4g of aluminum chloride from the given
amount of aluminum, but only 43.9g of aluminum
chloride from the given amount of chlorine.
Therefore, chlorine is the limiting reactant. Once
the 35.0g of chlorine is used up, the reaction
comes to a complete
.
Limiting Reactant
 15.0 g of potassium reacts with 15.0 g of
iodine. Calculate which reactant is
limiting and how much product is made.
Finding the Amount of
Excess
 By calculating the amount of the excess
reactant needed to completely react with
the limiting reactant, we can subtract that
amount from the given amount to find the
amount of excess.
 Can we find the amount of excess
potassium in the previous problem?
Finding Excess
 15.0 g of potassium reacts with 15.0 g of iodine.
2 K + I2  2 KI
 We found that Iodine is the limiting reactant, and
19.6 g of potassium iodide are produced.
15.0 g I2
1 mol I2
2 mol K
39.1 g K
254 g I2
1 mol I2
1 mol K
= 4.62 g K
USED!
15.0 g K – 4.62 g K = 10.38 g K EXCESS
Given amount
of excess
reactant
Amount of
excess
reactant
actually
used
Note that we started with
the limiting reactant! Once
you determine the LR, you
should only start with it!
Limiting Reactant: Recap







Convert ALL of the reactants to the SAME product
(pick any product you choose.)
The lowest answer is the correct answer.
The reactant that gave you the lowest answer is the
LIMITING REACTANT.
The other reactant(s) are in EXCESS.
To find the amount of excess, subtract the amount
used from the given amount.
You can recognize a limiting reactant problem
because there is MORE THAN ONE GIVEN AMOUNT.
If you have to find more than one product, be sure to
start with the limiting reactant. You don’t have to
determine which is the LR over and over again!
Empirical vs. Molecular
Formulas
 Empirical Formula of a compound shows the
smalles whole number ration of the atoms in
the compound
 A compound is analyzed and found to contain
25.9% nitrogen and 74.1% oxygen. What is the
emprical formula of the compound?
 25.9 g N * (1 mol/14.0g) = 1.85 mol N
 74.1 g O * (1 mol O/16.0g) = 4.63 mol O
 Empirical Formula = N2O5
Empirical vs. Molecular
Formulas
 Molecular Formula tells you the actual number
of each kind of atom present in a molecule of
the compound
 Calculate the molecular formula of a compound
whose molar mass is 60.0 g and empirical
formula is CH4N.
 Find empirical formula mass: 12+4+14=30g
 Molar mass/efm = 60/30 = 2
 Molecular Formula = C2H8N2
Percent, Theoretical, and
Actual Yield
 Actual Yield – the amount produced in a
reaction (amount of products)
 Theoretical Yield – the amount that should
have been produced in a reaction
 Maximum amount of product that could be formed
from given amount of reactants
 Percent Yield – (actual/theoretical) X 100%
 Often less than the theoretical