1. The Mole
The mole
The number of atoms of any element that is equal to the number of atoms in exactly 12.0 g of carbon-12.
The mole
There are 6.022x1023 particles in 1 mole.
Particles: atoms, molecules, ions, formula units etc.
 The word “Mole” is a homonym because it sounds the same and is spelled the same, but has different meanings. For example: a mole is an animal, a darkened growth on the skin, a spy who has infiltrated and been assimilated into the ranks of an enemy agency and now we will use it as a scientific description of a specific quantity used in stoichiometry.

2. Learning Objectives 9.1 Introduction to Stoichiometry
Define stoichiometry and describe the strategy required to
solve problems based on chemical equations.
9.2 Mole-Mole Calculations
Solve problems where the reactants and products
are both in moles.
Solve problems for “Percent Composition”
Determine Emperical and Molecular Formuls.
9.3 Mole-Mass Calculations
Solve problems where known mass is given and the answer
is to be determined in moles or the moles of known are
given and mass is determined.

3. Learning Objectives 9.4 Mass Mass Calculations (and more)
Solve problems where mass is given and the desired
unit to be determined is mass.
Mass molecules; Mass Volume
9.5 Limiting Reactant and Yield Calculations
Solve problems involving limiting reactants and yield.

4. The balanced equation describes the ratio in which the substances combine. The ratio obtained from the coefficients of a balance equation is called the Mole Ratio.
Stoichiometry:
v Calculations that relate quantities of substances using a
balanced chemical equation.
v “The study of the quantitative, or measurable, relationships
that exist in chemical formulas and chemical reactions.”
v “Stoichion” in Greek means element; “metron” in Greek
means measure.

5. Atomic Mass & Formula Mass
Facts
1.   Atoms of different elements have different atomic masses
2.  amuThe atomic mass unit is used to describe the mass of an atom since an atoms mass is so small.
3.  The scale for expressing atomic masses is based on the mass of carbon-12.
4. In the lab we replace” amu” with grams.
Atomic Mass
The mass of an atom expressed relative to the mass assigned to carbon-12.

6. Formula Mass and/or Molar Mass .
The sum of the atomic masses of all the atoms in a compound. (SI Units = g/mol)
Formula mass describes ionic compounds
Molar mass describes molecules or covalently bonded compounds
The formula mass and/or molar mass is found by referring to the periodic table, finding the atomic masses all the elements in the compound and then adding them up.
What is the atomic mass (in g/mol) for the following elements?
Aluminum Titanium Tungsten Iodine
47.867
g/mol
183.84
g/mol
g/mol
g/mol

7. Finding the formula mass and/or molar mass for the following compounds to hundreths?
H2O = 2H + O
= 2(1.01g/mol) g/mol
= 2.02 g/mol g/mol
= g/mol
CaCl2 =
g/mol
Formula Unit: A single ionic compound formula; such as NaCl, PbSO4 etc.
Molecule: A single covalently bonded compound; such as CO2, P2O5, etc

8. Avogadro's Number Mole Conversions
 Avogadro’s Principle: Amadeo Avogadro determined that equal volumes of gases at the same temperature and pressure contain equal number of particles.
Avogadro’s Number: A value that indicates the number of particles in a mole is 6.022x1023.
       A particle can be an “ion”, “formula unit”, “atom”,
“molecule” “particle”, etc.
Mole Conversions
1 mole = 6.022x1023 particles = formula or molar mass = 22.4L(g)
At STP = 0°C and 1 atm

9. How many moles are in 10.0 grams of NaCl Given mass (g) 1 mole
1 mole of substance is equal to it’s formula or molar mass.
1 mole of NaCl = g
1 mole of O2 =
32.00 g
1 mole of KC2H3O2 =
98.15 g
How many moles are in 10.0 grams of NaCl
  Given mass (g)  
1 mole
= moles of substance
FM or MM (g) of given
10.0 g NaCl
1 mole NaCl
= moles NaCl
58.44 g NaCl
Find the moles in 121 g of NaCl
Ans….2.07 moles NaCl

10. FM or MM (g) of substance
Now reverse the process, if given the number of moles find the mass of substance that is present.
Given moles
FM or MM (g) of substance
1 moles Given
Determine how many grams are in 6.21x10-2 moles of BaCl2
6.21x10-2moles BaCl2
g BaCl2
= 12.9 g BaCl2
1 mole BaCl2
Determine how many grams are in 1.21x10-3 moles of BaCl2
Ans… g BaCl2

11. Knowing your conversion possibilities, do the following: Find the volume of and mass of 2.33x1021 molecules of hydrogen gas.
2.33x1021 molecules H2
2.02 g H2
= g H2
6.022x1023 molec. H2
2.33x1021 molecules H2
22.4 L H2
= L H2
6.022x1023 molec. H2

12. Moles vs Particles 1 mole or FM or MM (g) FM or MM (g) 1 mole
1 mole = 6.022x1023 particles
soooo……
1 mole = 6.022x1023 particles = formula or molar mass(g) = L(gas)
The above statement can lead to several conversion factors
Goody..Goody..Let’s take a look!
1 mole or FM or MM (g)
FM or MM (g) mole
1 mole or x1023 particles
6.022x1023 particles mole
FM or MM (g) or x1023 particles
6.022x1023 particles FM or MM (g)

13. 1 mole or L gas
22.4 L gas mole
FM or MM (g) or L___
22.4 L FM or MM (g)
22.4 L (gas) or x1023 particles
6.022x1023 particles L gas
Each one of those conversion factors can help solve a problem bases on what you are given and what is wanted as a solution. Get the white boards!
 Given mass, find particles: what conversion factor should be used?
6.022x1023 particles
FM or MM (g)

14. Given moles, find atoms: what conversion factor should be used?
6.022x1023 atoms
1 mole
Given mass, find moles: what conversion factor should be used?
1 mole
FM or MM (g)
Given Liters, find formula units: what conversion factor should be used?
6.022x1023 Formula Units (FU)
22.4 L gas
Given molecules, find moles: what conversion factor should be used?
1 mole
6.022x1023 molecule

15. [ ] [ ] [ ] Percent Composition
The mass of an element in a compound compared to the entire mass of the compound.
Determine the percent composition of each element in the following compound.
NaCl:
% Composition
[ ]
Determine the formula mass
Na = g/mol
Cl = g/mol
= g/mol
Element’s mass
Compounds mass
X 100
[ ]
% Na
22.99 g/mol
x 100
58.44 g/mol
= 39.34%
% Cl
35.45 g/mol
x 100
58.44 g/mol
= 60.66%
[ ]

16. Find the percent of hydrogen and oxygen, to tenths, in ammonium phosphate.
(NH4)3PO4
3N = g/mol
12H = g/mol
P = g/mol
4O = g/mol
g/mol %H
12.01
149.12
= 8.0% %O
64.00
149.12
= 42.9%
[ ]
[ ]
x 100
x 100

17. Empirical Formula vs Molecular Formula
Empirical Formula: A formula where all components are reduced to the lowest whole number ratio.
· Ionic compounds are ALL reduced to the lowest
whole number ratios
· Example Ca2O2  CaO
MolecularFormula: A compound where the formula is written as it exists in nature. Some covalent compounds are empirical formulas but most are molecular formulas which are not reduced to the lowest whole number ratio.
Ethane: C2H6
Empirical formula
CH3
Molecular formula
C2H6

18. Finding the Empirical Formula
1)  If given %, change the % to grams
2)  Determine moles of each substance.
3)  Do a mole to mole ratio between each substance and the
lowest mole value found.
a.   If a ratio turns out to be1.5, 2.5, 3.5 etc, multiply all ratios by 2 to get the correct whole numbers.
b. If a ratio turns out to be 1.25, 2.25, etc, multiply all ratios by 4 to get the correct whole numbers etc.
4) The number found from the mole:mole is the subscript
for each element.

19. Find the (a) percent composition and (b) the empirical formula of a compound that contains 2.30 g of sodium, 1.60 g oxygen, and g of hydrogen in a 4.00 g sample.
2.30 g Na = 57.5%
4.00   g cmd
 1.60 g O = 40.0% O
4.00 g cmd
 0.100 g H = 2.5% H
4.00 g cmd
57.5 g Na 1 mole = mols
22.99g  Na
2.30 g Na 1 mole = mols
22.99g Na
40.0 g O 1 mole = mols
16.0g O O
1.60 g O 1 mole = mols
16.00g O
2.50 g H 1 mole = mols
1.0079g H
0.100 g H 1 mole = mols
1.0079g H
OK…we have a 1:1:1 ratio, thus the empirical formula is NaOH.
Is there another way we can get the same answer?

20. A compound is made up of 80. 0% carbon and 20. 0% Hydrogen
A compound is made up of 80.0% carbon and 20.0% Hydrogen. What is the empirical formula? 1
80.0% C = 80.0 g C
20.0% H = 20.0 g H 2
Lowest mole value
  80.0 g C 1 mole = g 
6.66 moles C
20.0 g H 1 mole = g 
19.8 moles H 3
6.66 moles C
6.66 moles
19.8 moles H =
6.66 moles
= 1C
2.97 = 3 H 4
1C + 3H ===CH3

21. Determine Molecular Formula
1. Determine empirical formula
2. Find the formula or molar mass of the empirical formula generated.
3. Create a ratio between the molar mass given and the mass found in step two.
4. If the ratio generates a value greater than 1, multiply all subscripts by the ratio number. 

22. Example: Earlier, we found an empirical formula to be CH3
Example: Earlier, we found an empirical formula to be CH3. If we were given that the molar mass of a substance with this empirical formula to be 90.5 g/mol, what is the molecular formula.
Step B#1 Completed earlier.
Step #2: CH3 mass = g/mol
Step #3: g/mol = = 6
15.03 g/mol
Step #4: 6(CH3) = C6H18
-carotene, a compound found in carrots, can be broken down to form vitamin A. The empirical formula for -carotene is C5H7. The molar mass of -carotene is 536 g/mol. What is the molecular formula for -carotene?
 Ans…C40H56

23. Find the molecular formula for a compound that contains 4
Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol
4.90 g N 1 mole = moles N
14.01g 
= 1
0.350 moles
11.2 g O 1 mole = moles O g 
= 2
0.350 moles
Empirical Formula = NO2
NO2 molar mass = 46.0 g/mol
Molecular mass
Molar mass
= 92.0 g/mol
46.0 g/mol
= 2, thus 2(NO2) =
N2O4

24. General Set-ups
Stoichiometry simply uses the same method of solving as conversions do.
1. Write and balance (if not given) the equation that
describes the chemical reaction.
o You need the equation balanced because you will retrieve your mole:mole ratio from the coefficients of the substances in the balanced equation.
 2. Identify the given and what you are asked to solve for.
 3. Set-up your T-diagram and fill in the values you need
to solve the problem.
o Be sure to align units and formulas so the ones you begin with cancel and the one(s) you are solving for remain.

25.  Mole-to-Mole
Moles of Given Unk. Moles from Equation Given Moles from Equation
Magnesium burns in oxygen to produce magnesium oxide. How many moles of oxygen are needed to burn 0.53 moles of Mg?
Write Equation:
Mg + O2  MgO
Balance Equation:
2Mg O  2MgO
Identify givens
and unknowns
0.53 mols ?? mols
Set-up “T” and solve…
  0.53 mols Mg mols O2 = mols O2  mols Mg

26. Follow the same general procedure for all stoichiometry problems.
The coefficient of any substance in an equation is equal to the moles of that substance in the chemical reaction.
Mole-to-Mass
problem
Mole:mole ratio
From equation
1 mol = FM (g)
conversion ratio
Coefficient of Unknown
(moles) from equation
Formula Mass (g)
of unknown
Moles of Given
1 mole of
unknown
Coefficient of Given
(moles) from equation
Notice, the answer will be grams of unknown you wanted to find.
All units and compound labels must be in these equations since
we are now using more than one substance.

27. Mass-to-Mass Unk (moles) from eq. Mass of Given 1 mole Given
Mole:mole ratio
from equation
1 mol = FM (g)
of given
1 mol = FM (g)
of unknown
Unk (moles)
from eq.
Mass of
Given
1 mole
Given
Formula Mass (g)
of unknown
FM or MM
(g) of given
Given (moles)
from eq.
1 mole of
unknown
Notice, the answer will be grams of unknown you wanted to find.
What would I change the last conversion factor to if I wanted to
Solve for molecules or volume instead of grams?

28. Mass-to-Molecules Mass of Given Mole:mole ratio From equation
Mass of
Given
Mole:mole ratio
From equation
1 mol = FM (g)
given
1 mol = 6.022x1023
molecules
Unk (moles)
from eq.
1 mole
Given
6.022x1023 molecules
unknown
FM or MM
(g) of given
Given (moles)
from eq.
1 mole of
unknown
Notice, the answer will be molecules of unknown you needed.
How would I change the last conversion factor if I wanted to
solve for liters gas instead of grams or molecules?

29. How would I solve for liters gas if I started with Liters of given?
Mass-to-Volume
Mass of
Given
Mole:mole ratio
From equation
1 mol = FM (g)
given
1 mol = 22.4 L gas
Unk (moles)
from eq.
1 mole
Given
22.4 L gas unknown
FM or MM
(g) of given
Given (moles)
from eq.
1 mole of
unknown
Notice, the answer will be liters of unknown you needed.
How would I solve for liters gas if I started with Liters of given?

30. Volume-to-Volume Liters of Given Mole:mole ratio From equation
Liters of
Given
Mole:mole ratio
From equation
Mole:mole ratio
From equation
1 mol = 22.4 L
given
1 mol = 22.4 L gas
Unk (moles)
from eq.
1 mole
Given
Unk. (moles)
from eq.
22.4 L gas unknown
The label for the answer will be
Liters of Unknown.
Given (moles)
from eq.
22.4 L
of given
Given (moles)
from eq.
1 mole of
unknown
Notice, the conversion factor of 1 mol = 22.4 L is seen twice, but as
reciprocals of each other. Therefore, they cancel out.
Would the labels be correct if you didn’t show the 1 mol = 22.4 L
conversions?

31. Limiting Reactant
A limiting reactant is just as it implies, one of the reactants will be used up first thus causing the entire reaction process to come to an end.
 The quantities of products formed in a reaction are always determined by the quantity of the limiting reactant. 
 It will be our job to identify the limiting reactant from data given to us in problems and/or the lab setting.
When two reactants are not present in the exact mole ratio given in the balanced equation, one reactant will be the limiting reactant.
The other reactant is said to be in “excess” Knowing the limiting reactant allows us to determine how much product we could get in a perfect world.

32. Finding the Limiting and Excess reactants
 Write and balance the chemical equation that represents the chemical reaction.
Identify the givens and what you need to find.
Find mols from each given
Do a mol:mol ratio of mols from step 3.
Make a comparison of the mole ratios of the givens to the coefficient ratio of the givens.
Method 2: Do two full stoichiometric calculation and
Pick the smallest answer.
This, however…is way too much work!!

33. Write and balance the equation: Zn + 2HCl --> ZnCl2 + H2
Example: 1.21 grams of zinc and 2.65 grams of hydrochloric acid react to form zinc chloride and hydrogen gas. Determine which reactant is in excess and which is the limiting reactant and determine how much zinc chloride is produced.
Write and balance the equation:
Zn HCl --> ZnCl H2
1.21 g g ? g
Find moles of each given: By this time you should have an understanding that you simply divide the given mass by the formula, atomic or molar mass of the substance to get moles.
Zn HCl --> ZnCl H2
1.21 g g
mols =
= mols
g/mol
36.56 g/mol

34. Do a mol:mol ratio between the mols found.
Zn HCl --> ZnCl H2
1.21 g g
mols =
= mols
mols
g/mol
36/47 g/mol
3.91 1
Smallest
mole value 1 2
Make a comparison of the mol ratio just completed to
the mols found in the equation. 
3.91 HCl vs 2 HCl indicates HCl is in excess because the calculated value is more than is coefficient value.
Thus, Zinc is the limiting reactant.

35. Now, to calculate how much zinc chloride is produced we must use zinc’s data.
 Since we’ve already calculated the moles of zinc, we can simply finish the problem using a mole to mass stoichiometry.
1 mols ZnCl2
136.3 g ZnCl2
moles Zn
= 2.52 g ZnCl2
1 mol Zn
1 mol ZnCl2
Zn HCl --> ZnCl H2
 This calculated value, 2.52 g Zn, is called the theoretical value…which is the amount you can get in a perfect situation.

36.  The amount of a product that should be produced based on stoichiometic calculations is called the “Expected yield” or “theoretical Yield”
Percent
Yield
The amount of a product that is really obtained in the lab from a chemical reaction is called the “actual yield”
 Why should they ever be different?
*As the reactants decrease in quantity, they do not bump into each other often enough, or hard enough, or at the right orientation to proceed with the formation of products.
To calculate percent yield
Actual yield
% yield = x 100
Theoretical yield