Chapter 3
Stoichiometry
Atomic Mass
• Carbon-12 is assigned a mass of exactly
12.00 atomic mass units (amu)
• Masses of other elements are compared
to Carbon-12
• Comparisons are made using a
mass spectrometer
– Compares ratios of isotopes present
Mass Spectrometer
Average Atomic Mass
• An average of the atomic masses of all
naturally occurring isotopes of an element
• Average atomic mass is the mass value on
the periodic table
• No atom has the exact mass listed on the
table
• Carbon = 12.01 amu
– This means the majority of C is Carbon-12
• Used to weigh large numbers of atoms
Example
• Lead occurs in 4 main isotopes 1.40% has
a mass of 203.973amu. 24.10% has a
mass of 205.975amu. 22.10% has a mass
of 206.976, and 52.40% has a mass of
207.977amu. What is average atomic
mass of Lead?
The Mole
• 1 mole = 6.022x1023 particles
• Avogadro’s Number
– He did not determine the number!
• The average atomic mass of an element
expressed in grams contains 1 mole of
atoms of that element.
• 22.99 g of Sodium is one mole of sodium
• 40.08 g of Calcium is one mole of calcium
Example
• How many moles of sodium atoms are in
2.83 grams of sodium?
Example
• How many atoms of sodium are there in
2.83 grams?
Molar Mass
• The mass in grams of one mole of an
element or compound
– Units of grams/mole (g/mol)
• Molar mass of sodium = 22.99 g/mol
• Molar mass of compound is the sum of all
atoms present
• Molar Mass of Water (H2O)
• 2(1.01) + 16.00 = 18.02 g/mol
Example
• Determine the molar mass of
• Mg3N2
• C2H5O2N
• Ca(NO3)2
Example
• 18.1g Calcium Nitrate = ? moles
Example
• How many nitrate ions are present in
18.1 grams of Calcium Nitrate?
Percent Composition
• The percent by mass each element
contributes to the compound
(# of atoms of element)(atomic mass) X 100
(molar mass of compound)
Example
• What is the percent composition of
Calcium Nitrate?
Homework
• P. 123 #’s 21,34,36,40,44,49,52,53
Objectives
• Compare Empirical and Molecular
Formulas
• Calculate Empirical and Molecular
Formulas
• Balance and Interpret chemical equations
• Perform stoichiometry calculations
Empirical and Molecular Formulas
• Empirical Formula
– A chemical formula with the simplest whole
number ratio
• CH2O (30.03 g/mol)
• Molecular Formula
– The exact chemical formula for a
compound
• C3H6O3 (90.09 g/mol)
Calculating Empirical Formulas
• Assume 100 g of compound
– (Percents become grams)
•
•
•
•
Or use given masses
Covert grams to moles
Divide each by the smallest # of moles
If you don’t have whole #’s multiply by an
integer to get whole numbers
– 3.33 * 3
4.5 * 2
8.25 *4
Example
A compound contains 49.31%C, 43.79%O
and the rest H. The molar mass is
146 g/mol. What is the empirical and
molecular formula?
Combustion Analysis
• Compounds containing C,H, and O are
often burned to determine the chemical
makeup.
– The products of combustion are CO2 and
H 2O
• These are collected and massed
– Carbon and Hydrogen come from
compound
– Oxygen comes from compound and air
Combustion Analysis
Example
The combustion of 17.27 grams of a
compound which contains only C,H and O
yields 44.06 grams of CO2 and 11.27
grams of H2O. What is the empirical
formula of the compound?
Balanced Chemical Equations
•
•
•
•
•
Gives the recipe for a reaction
Reactants on left / Products on the right
Coefficients tell us the atomic/mole ratios
State symbols tell what state of matter
(s) solid, (l) liquid, (g) gas, (aq) solution
2H2(g) + O2(g)  2H2O(l)
Means – 2 mol of gaseous Hydrogen reacts with 1
mol of gaseous Oxygen to form 2 mol of liquid
Water
Balancing Equations
• Use coefficients to get the same number of
atoms on both sides of the equation
– Due to Law of Conservation of Mass
– Coefficient – number in front of a formula
• Balance the equation
Fe2S3 + HCl

FeCl3 +
H 2S
Cont.
• Combustion Equations
– When there are an even number of Carbons
in the reactant you may have to double the
reaction.
C2H6
+
O2  CO2
+ H2O
Stoichiometry
• A process for comparing amounts in
reactions
• Compares reactant to reactant or reactant
to product
• Start with a balanced equation
• Process
Grams A  Moles A  Moles B  Grams B
Molar Mass A
Molar Ratio
Molar Mass B
Cont.
• Could Include
– Density
– Percent
• Could start or end with moles
• Read the problem and determine the best
way to proceed
Example
6.00 grams of hydrochloric acid is reacted
with barium hydroxide to form barium
chloride and water. How many grams of
barium chloride are formed?
Example
4.82 grams of propane (C3H8) is burned with
excess oxygen to form carbon dioxide and
water. What volume of water vapor is
formed if the density is .8044 g/L?
Homework
• P. 125 #’s 66,74,79,86
Objectives
•
•
•
•
Describe limiting and excess reactants
Be able to calculate the limiting reactant
Compare theoretical and actual yield
Describe and calculate percent yield
Limiting Reactant
• The reactant that runs out in a chemical
reaction.
• When one reactant is used up the reaction
stops
• Based on mole ratio NOT on mass
Example
What is the limiting reactant when 16.00g of
Hydrogen gas reacts with 64.00g of
oxygen gas to produce water
Example
What mass of carbon dioxide gas is
produced when 4.00g of methane (CH4)
reacts with 32.00g of oxygen gas
Theoretical and Actual Yield
• Theoretical yield is the amount of product
you should get based on calculations
• Actual yield is the amount of product you
get when the reaction is run
– Actual yield cannot be more than the
theoretical yield. (Conservation of mass)
– Actual yield is usually less than theoretical b/c
of product loss
Percent Yield
• Measures how much product a reaction
produces
• High percent yields are good
Percent Yield = Actual X 100
Theoretical
Example
1000. grams of calcium phosphate reacts
with 1000. grams of 98% sulfuric acid to
produce calcium sulfate and phosphoric
acid. When the reaction is run 1100. g of
calcium sulfate was produced. What is the
percent yield of the reaction?
Homework
• Stoichiometry Worksheet