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Stoichiometry and the Mole
The Scenario
• You are planning to do a reaction using
hydrochloric acid. Suddenly, you drop the bottle
and spill all 13grams of your very dangerous acid
on the floor!
• You know it can be neutralized (so that it stops
eating through the floor) by sodium hydroxide,
but you need to know how much sodium
hydroxide you need to completely get rid of all of
the acid. How do you do this??????
The acid spill…
• You know that:
– HCl + NaOH ----> NaCl + H2O
– Very dangerous ---> Very harmless
• You also know you have 13 g of HCl
• Wouldn’t it be nice to know how much one
atom of an element weighed!
• The big problem is that elements have
different isotopes.
Isotopes?
• What is an isotope?
• Atoms of the same element with different
numbers of neutrons.
• Ex. Hydrogen-1, Hydrogen-2, Hydrogen-3
• 1H, 2H, 3H
Average Atomic Mass
• The average atomic mass for all elements
are shown on the periodic table.
• The periodic table shows the AVERAGE
weights of all isotopes.
Atomic Mass Units
• All masses for atoms on the periodic table
are in amu’s.
• 1 amu = 1.66 x 10-24g
Formula Mass (Weight)
Molecular Mass (Weight)
• Formula mass (aka molecular weight) is
the mass of one molecule.
• Formula mass =
atomic masses of elements x subscripts
Find the formula mass of water.
•
•
•
•
•
Water is H2O
From periodic table H --> 1.008 amu’s
From periodic table O --> 16.00 amu’s
BUT there are 2 H’s and only 1 O in H2O
Solution:
– H --> 2 x 1.008 = 2.016 amu
+
– O --> 1 x 16.00 = 16.00 amu
18.02 amu
Add these together
Find the formula masses for the
following:
•
•
•
•
•
CO2
K2CrO4
NH4
(NH4)3PO4
H3PO4
Percent Composition
Percent composition is the percent by weight of each element
in a compound.
Step 1: Find the formula mass of the compound..
Step 2:
% Composition = Total mass of element X 100
Formula mass of cmpd
Percent Composition
• What percentage of the total mass is
taken up by one type of atom?
• % comp = [mass of atoms in questions]
[formula mass]
X 100
• What percentage of total mass do EACH of
the types of atoms in water contribute?
– Hint: Water has 2 types of atoms, hydrogens
and oxygens, so you are doing 2 separate
problems here.
Percent Comp. Of Water
• 1. For each element, multiply atomic mass by subscript:
– H2
2 x 1.008 = 2.016
–O
1 x 16.00 = 16.00
• 2. Find formula mass:
– 2.016 + 16.00 = 18.02 amu
• 3. Plug in to percent composition for each
element present.
• H --> [2.016] / [18.02] = 11.19 %
• O --> [16.00 / 18.02] = 88.79 %
• Notice they add up to nearly 100% (only nearly because
of error)
What is percent comp. For
OXYGEN in each of the
following?
•
•
•
•
•
CO2
K2CrO4
NH4
(NH4)3PO4
H3PO4
A Mass Challenge
• 1. How many grams of oxygen are in 237g of
H3PO4?
• 2. How many grams of hydrogen are in 360g of
(NH4)3PO4?
• 3. How many grams of copper are in 296g of
Cu(OH)2?
Introducing…The Mole!!!
• 1 mole = approximately
602,204,531,000,000,000,000,000
• That is…
•
The Mole
• 1 mole = 6.022 x 1023 particles.
• A mole is the “unit if measure.
• 6.02 x 1023 is called Avogadro’s Number.
Avogadro’s Number
No! Not avocado - Avogadro!
Mole Conversions
• How much does one mole of hydrogen
atoms weigh in grams?
• Recall: 1 amu = 1.66 x 10-24g
So, 1 atom of H weighs 1.008 amu
AND 1 mole of H atoms weighs 1.008 g
Conversion options with the mole
1 mole = 6.022 x 1023 particles
1 mole of atoms = mass on per. table in
grams
1 mole = 22.4 L (at STP & for gases only!!)
STP is standard Temperature (273 K) &
Pressure (1 atm)
Molar Mass (aka Gram formula
Mass)
1 mole CO2 = 6.02 x 1023 molecules CO2
1 mole CO2 = 22.4 L CO2 if at STP
How many grams does 1 mole of CO2 weigh?
C = 1 x 12.01 = 12.01g
O= 2 x 16.00 = + 32.00g
44 . 01 g / mole of CO2
Molar Mass Practice
• 1. What is the gram formula mass for
K2CrO4?
• 2. Molar mass for (NH4)3PO4?
149.12 g/ mole (NH4)3PO4
Volume and the Mole
• 1. How many moles are in 38L of carbon
dioxide at STP?
• 2. How many liters are in 6.4moles of
oxygen at STP?
Particles and the Mole
• 1. 1.20 x 1025 atoms of P = How many
moles of P?
• 2. 0.750 moles Zn = How many atoms of
Zn?
• 3. How many moles are in 9.2x 1020
molecules of CO2?
Mole Conversions Practice
• 4. How many grams are in 9.2 x 1020
molecules of CO2?
• 5. How many grams are in 57.2 L of
hydrogen gas at STP?
So what is “MoleAmerican
Airlines” Representing?
• Interpreting Chemical Equations
• What does a chemical equation ACTUALLY tell
you? What does it all mean?
• Ex: N2(g) + 3H2(g) ---> 2NH3(g)
• This says that 1 molecule of N2 reacts with 3
molecules of H2 to give 2 molecules of NH3 gas.
• Let’s say you have 1 mole of nitrogen gas and an
unlimited supply of hydrogen gas
– 1. How much of the hydrogen gas will react with the
nitrogen gas?
3 moles
– 2. How much ammonia gas can you make? 2 moles
I. Intro to Stoichiometry
Stoichiometry is the process of calculating the amount of
substances produced in a chemical reaction.
When you know the amount of one substance, you can
determine the amount of the other reactants or products.
II. Mole Ratios
N2
+ 3H2
2NH3
Coefficient represent the number of moles of that substance.
Mole ratios are written in fraction form.
N2
+ 3H2
2NH3
• Mole ratios:
Left side
Reactants:
1 mol N2
or
1 mol N2
3 mol H2
Right side 1 mol N2
Products:
2 mol NH3
Right side 3 mol H2
Products:
2 mol NH3
3 mol H2
or
2 mol NH3
1 mol N2
or
2 mol NH3
3 mol H2
Mole Road Map
Grams A
Moles A
Use Molar
Mass
Moles B
Use Mole
Ratio
Grams B
Use Molar
Mass
IV. Mole to Mole Stoichiometry
(Moles A to Moles B on roadmap)
Ex. Determine the number of oxygen moles required to burn 1.20 moles
of ethyl alcohol, C2H5OH.
C2H5OH
+
3O2 
2 CO2
+
3H2O
1.2 mol A
?mol B
Step 1: Balance equation. (This is already balanced!)
Step 2: Write the given amount of the chemical – this is your “moles A.”
Step 3: Use the mole ratio from the balanced equation.
1.20 mol C2H5OH
3 mol O2
1 mol C2H5OH = 3.60 mol O2 Answer
Mole Road Map
Grams A
Moles A
Use Molar
Mass
Moles B
Use Mole
Ratio
Grams B
Use Molar
Mass
V. Mole to Mass Stoichiometry
(Moles A to Grams B on roadmap – Just one more step to the
problem we just did!)
Ex. Determine the grams of oxygen required to burn 1.20 mol of
ethyl alcohol, C2H5OH.
Step 1: Write a complete balanced equation.
Step 2: Write the given amount.
Step 3: According to the mole road map, you need to go from:
Moles A  Moles B
 Grams B
1.20 mol C2H5OH
3 mol O2
1 mol C2H5OH
32 g O2
1 mol O2 = 115 g O2 Answer
Mole-Mole Problems
• Balanced chemical equations, through their
coefficients indicate the ratio of moles of reactants
and products.
• 1. Aluminum and oxygen react to form aluminum
oxide. If you have 2.3 moles of aluminum oxide
after the reaction how many moles of aluminum
were used?
• STEP 1: Write the balances equation:
• 4Al + 3O2 ---> 2Al2O3
Mole-Mole Problem cont.
• STEP 2: Do dimensional analysis starting with
what you know to convert moles of aluminum
oxide to moles of aluminum.
– Aluminum and oxygen react to form aluminum oxide.
If you have 2.3 moles of aluminum oxide after the
reaction how many moles of aluminum were used?
– 4Al + 3O2 ---> 2Al2O3
Now you try a mole-mole
problem
• 2. Hydrogen gas and oxygen gas react to
make water. If you have 3.24 moles of
hydrogen gas and an unlimited supply of
oxygen gas, how many moles of water can
you make?
• H2 + O2 ---> 2H2O
Mole-Mass Problems
• You have 2.4 moles of Al reacting with oxygen.
How many grams of Al2O3 are produced?
• STEP 1: Write your balanced equation
• 4Al + 3O2 ---> 2Al2O3
• STEP 2: Do dimensional analysis starting with
what you know to convert moles of aluminum
oxide to moles of aluminum.
Mole-Mass Problems- you try:
• You have 6.7 moles of oxygen gas reacting
with unlimited supply of hydrogen. How
many grams of water are produced?
• H2 + O2 ---> 2H2O
Mole Road Map
Grams A
Moles A
Use Molar
Mass
Moles B
Use Mole
Ratio
Grams B
Use Molar
Mass
Mass-Mass Problems
• Dicarbon Dihydride reacts with oxygen gas
to produce carbon dioxide and water. How
many grams of oxygen gas will react
completely with 13.0g of C2H2?
• STEP 1: Write the balanced equation.
• 2C2H2 + 5O2 ------> 4CO2 + 2H2O
• STEP 2: Do dimensional analysis.
You try: Mass-Mass Problem
• CaC2 reacts with water to produce C2H2
and calcium hydroxide. If 51.6g of CaC2
are reacted, how many grams of calcium
hydroxide are produced?
• CaC2 + 2H2O ---> C2H2 + Ca(OH)2
Theoretical, Experimental, &
Percent Yields
Theoretical yield is the amount of product that you mathematically
determine (using stoich) from a chemical reaction.
Experimental yield is the amount of product that you actually get
when you perform the reaction in lab.
Percent yield =
Experimental Yield x 100
Theoretical Yield
Calculating Chemical Formulas
• Empirical Formulas: Simplest whole
number ratio of atoms in a compound
• Ex: C6H12O6 is a molecular formula. The
empirical formula is CH2O.
Determining Empirical Formulas
• Step One:Convert grams to moles.
• Step Two: Divide each by the smallest number of moles
to get subscripts
• Step Three: If you do not have a whole number,or one
that is reasonably close, multiply by an integer to
obtain a whole number.
Empirical Formula
• Hg--> 67.6g x 1mole
= .337 moles Hg / . 337 = 1
200.59g
• S --> 10.8g x 1mole
= .337 moles S / . 337 = 1
32.06g
• O --> 21.6g x 1mole
= 1.35 moles O / . 337 = 4
16.00g
Now divide each by the smallest number of
moles
These are your subscripts for your formula! HgSO4
Empirical Formula from %
cont…
• Problem 2: After doing several experiments
on an unknown substance it was found that
the percent compositions for the compound
were: 3.09% H, 31.6% P, and 65.3 % O.
What is the empirical formula for this
substance?
H3PO4
Empirical Formulas if given
mass.
• STEP 1: Turn mass into moles
• STEP 2: Divide by lowest number of moles to get
subscripts.
• Problem 1: A 19.12 gram sample of an unknown
chemical was discovered. After much
experimentation it was determined that the
unknown was made of 3.03g of hydrogen and
16.09g of oxygen. What is the empirical formula?
H3O
Molecular Formulas
• This is the exact formula of the molecule
giving types of atoms and numbers of
each type. This formula represents how
the compound actually appears in nature.
• You will need to be able to determine the
molecular formula using data.
Determining Molecular Formula
• Example problem: A compound has a molar mass
(gfm) of 120.00 g/mole with an empirical formula
of CH2O. What is the molecular formula?
• STEP ONE: Find the molar mass (gfm) of the
empirical formula
• STEP TWO: Divide true molar mass (gfm) by gfm
of the empirical formula. 120.00 / 30.03 = 3.996 ~ 4
• STEP THREE: Write the molecular formula by
multiplying subscripts by answer to step 2.
C[1x4]H[2x4]O[1x4]
C4H8O4
Molecular Formula Problem
• A compound with the empirical formula
C2H5 had a molar mass of 87.21g/mole.
What is the molecular formula?
C6H15
Hydrated Compounds
• A Hydrated Compound is a compound
that crystallized from a water solution with
water molecules adhering to the particles of
the crystal.
• Chemists use heat to dry these compounds
and then calculate the ratio of the compound
to water.
Hydrates Cont…
• Example: NiSO3 6H2O The dot shows
that 6 water molecules adhere to each
formula unit.
• To calculate the molar mass of a hydrate
add the molar mass of the anhydrous
compound to the mass of the water present.
Empirical Formulas of Hydrates
•
Step 1: Find the moles of anhydride (compound
without water)
• Step 2: Find the moles of water.
• Step 3: Divide moles of water by moles of
anhydride. This should turn out to be a whole
number
• Step 4: Place this whole number into the formula
in front of water.
Empirical formula of a hydrate
practice
• What is the formula of a hydrate which consists
of 0.391 g of Li2SiF6 and 0.0903 g of water?
• Your formula will be Li2SiF6 ____H2O
• Step 1: .391 g Li2SiF6 x 1 mole Li2SiF6 = .00250 mole
155.97 g Li2SiF6
• Step 2:
.0903 g H2O x 1 mole H2O = .00500 mole H2O
18.02 g H2O
• Step 3: .005 mole
Li2SiF6
/ .0025 mole ~ 2
2H2O
% water in a Hydrate
• Step 1: Find the molar mass of the hydrated
compound.
• Step 2: Divide the mass of water by the
mass of the hydrate and multiply by 100.
Hydrates: Practice Problem
• What is the formula and % water of a
hydrate which consists of 9.520g of BaI2
and 0.887 g of water?
BaI2
2H2O
8.43%
It’s the Holidays… Lets talk
Cake!
• Let’s say you want to make as many cakes
as you can for the holiday season. You
don’t have time to go to the store, so you
take inventory of what you have in your
pantry.
Cake Inventory
•
•
•
•
•
•
•
You Have:
1 dozen eggs
5 cups oil
12 cups flour
3 cups sugar
4 boxes cake mix
Unlimited supply of
water
•
•
•
•
•
•
•
Each cake needs:
1 egg
1/3 cup oil
1 cup flour
1/2 cup sugar
1 box cake mix
1/2 cup water
How many cakes can we make? 4
What ingredient limits us? Cake mix
A Limiting Reagent
Analogy
So, Cake mix was our limiting reactant (aka limiting reagent)!
This also works with sandwiches!
All other ingredients besides the limiting reactant are said to be
in excess
Limiting Reactants
Limiting reactant is the chemical in a reaction that
restricts how much you can make, it’s the reactant
that runs out first.
STEP 1: Write the balanced equation!
STEP 2: Find desired product in grams. You
Will need to do 2 stoich problems.
STEP 3:The reactant that gives you the
smaller amount of product is the limiting reactant.
Hint: You will know if you are working with a limiting reactant
problem if you are given two quantities of two reactants or if one is
given in excess.
Working Limiting Reactant
Problems
• Ex: How many grams of lithium nitride can
be made by reacting 56.0g of nitrogen with
56.0g of lithium?
• STEP 1: Write the balanced equation!
• N2 + 6Li ----> 2Li3N
• STEP 2: Convert each reactant to what
you are looking for.
Final steps of LR problems
• STEP 3: The smallest amount found is
the maximum amount you can produce
using the materials you have.
• The limiting reactant is the reactant that
gives you the least amount of product
• So, Li is my limiting reactant and I can
make a maximum of 93.7 g Li3N.
LR Sample Problem
• Ethylene, C2H4, burns in air according to
the following unbalanced equation.
• C2H4(g) + O2(g) ----> CO2(g) + H2O(g)
• How many grams of CO2 can form when a
mixture of 2.93g of ethylene and 5.29g of
oxygen are reacted?
•C2H4(g) + 3 O2(g) ---->2 CO2(g) + 2 H2O(g)
•O2--> 4.85g CO2 possible 4.85g CO2 can be
formed based on LR
•C2H4--> 9.19g CO2 possible
which is O2.
LR Sample #2
• If 40.0 g of a Pb(NO3)2 solution are mixed
with 37.0g of a NaI solution, how many
grams of bright yellow PbI2 form?
Pb(NO3)2 + 2NaI ---> PbI2 + 2NaNO3
 Pb(NO3)2 ---> 55.7 g PbI2 possible
NaI ---> 56.9 g PbI2 possible
So, Pb(NO3)2 is our L.R.
We make 55.7 g of PbI2