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Chapter 3
Stoichiometry
Chapter 3
Chemical Stoichiometry
Stoichiometry – The study
of quantities of materials
consumed and produced
in chemical reactions.
• Since atoms are so
small, we must use the
average mass – we will
count by weighing!
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2
Section 3.1
Counting by Weighing
Ex:A pile of marbles weighs 394.80g. You randomly
count out 10 marbles and determine their weight to
be 37.60 g. A) What is the average mass of 1
marble? B) How many marbles are in the pile?
Avg. Mass of 1 Marble =
37.60 g
= 3.760g / marble
10 marbles
394.80 g = 105.0marbles
3.760g
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Section 3.3
The Mole by Weighing
Counting
•
By observing proportions in which elements
combine, 19th century chemists calculated relative
atomic masses.
•
The modern system is based on 12C
•
12C
•
The most accurate method for comparing masses of
atoms involves the use of the mass spectrometer.
is assigned a mass of exactly 12 atomic mass
units (amu), and the masses of all other atoms are
given relative to this.
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Section 3.2
Atomic Masses
Counting
by Weighing
Schematic Diagram of a Mass Spectrometer
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Section 3.2
Atomic Masses
Counting
by Weighing
Since elements occur in nature
as mixtures of isotopes,
chemists use the average
atomic mass (aka atomic
mass or atomic weight)
Atomic mass can be used to
“weigh out” large numbers
of atoms
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Section 3.4
The
Mole
Molar
Mass
Avogadro’s number
• The number equal to the number of carbon atoms in
exactly 12 grams of pure 12C.
• 1 mole = 6.022 x 1023 units of anything!
• Named in honor of Avogadro (although he did not
discover it!)
Measuring moles
Ex: 1 mole C = 6.022 x 1023 C atoms = 12.01 g C
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Section 3.4
Molar Mass
Molar Mass is the mass in grams of one mole of a
substance:
Ex: Molar Mass of H = 1.008 g/mol
Molar Mass of O = 16.00 g/mol
(Use 4 significant figures when
determining molar mass)
Practice: Determine the molar mass of:
1. H2O
(2 × 1.008 g) + 16.00 g = 18.02 g/mol
2. Ba(NO3)2
137.33 g + (2 × 14.01 g) + (6 × 16.00 g) = 261.35 g/mol
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Section 3.4
Molar Mass
1 mol
= 6.022 x 1023 particles
= molar mass (g)
= 22.42 L of any gas at STP
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Section 3.4
Mole
Map
Molar
Mass
2
2
2
2
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Section 3.3
Concept
Check
The Mole
Counting
by Weighing
1. Calculate the number of iron atoms in a 4.48 mole sample
of iron.
(4.48 mol Fe) × (6.022×1023 Fe atoms / 1 mol Fe) =
2.70×1024 Fe atoms
2. Calculate the number of aluminum atoms in a 10.0 g
sample of aluminum.
(10.0 g Al) × (1 mol Al / 26.98 g Al) x (6.022×1023 Al atoms / 1 mol Al)
2.23×1023 Al atoms
3. Calculate the mass of 5.00 x 1020 atoms of cobalt.
(5.00 ×1020 Co atoms) x (1 mol Co / 6.022 ×1023 atoms Co) x (58.93 g Co / 1 mol Co) =
4.89 x 10-2 g Co
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Section 3.3
Concept Check
The Mole by Weighing
Counting
4. Calculate the number of liters in 6.8g of carbon dioxide at
STP (assuming ideal gas behavior).
(6.8 g CO2) × (1 mol CO2 / 44.01 g CO2 ) x (22.42 L CO2 / 1 mol CO2 ) =
3.5 L CO2
5. Determine the density of carbon dioxide at STP (assuming
ideal gas behavior).
D = m/v *assume 1 mol
= 44.01 / 22.42 1.963 g/mol CO2
6. What is the mass of the carbonate ions present in a
sample of calcium carbonate containing 4.86 mol?
(4.86 mol CaCO3) × (1 mol CO3 / 1 mol CaCO3) x (60.01 g CO3 / 1 mol CO3 ) =
292 g CO32Copyright © Cengage Learning. All rights reserved
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Section 3.3
The Mole by Weighing
Counting
Done for today!
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Section 3.6
Percent Composition of Compounds
Calculating Percent Composition
• Calculating any percentage:
"The part, divided by the whole, multiplied by 100"
• Percentage Composition
Calculate the percent of each element in the total
mass of the compound
mass of element in compound
mass % =
× 100%
mass of compound
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Section 3.6
Percent Composition of Compounds
Example: Determine the percent of iron, by mass, in
iron(III) oxide, (Fe2O3):
mass of element in compound
mass % =
× 100%
mass of compound
2( 55.85 g)
111.70 g
mass % Fe =
=
× 100% = 69.94%
2( 55.85 g)+ 3( 16.00 g) 159.70 g
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Ex: Calculate the percent composition by
mass of magnesium carbonate.
Molar mass of MgCO3 (remember, 4 SF!):
24.31 g + 12.01 g + 3(16.00 g) = 84.32 g
100.00  Check!
Section 3.7
Determining the Formula of a Compound
Formulas
Empirical formula
• Formula written as the simplest whole-number
ratio
Molecular formula
 Actual formula of the compound
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• Formulas for ionic compounds are ALWAYS
empirical (lowest whole number ratio).
Ex: NaCl MgCl2 CaO Al2(SO4)3
• Formulas for molecular compounds MIGHT be
empirical (lowest whole number ratio).
Molecular:
H2O
C6H12O6
C12H22O11
Empirical:
H2O
CH2O
C12H22O11
Section 3.7
Determining the Formula of a Compound
The Poem: Percent to mass, Mass to mole
Divide by small, Multiply ‘til whole!
Example:
The composition of adipic acid is 49.3% C,
6.9% H, and 43.8% O (by mass). The molar
mass of the compound is 146 g/mol.
a. What is the empirical formula?
b. What is the molecular formula?
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A) To determine Empirical Formula,
Use the poem!
Percent to mass, Mass to mole:
Divide by small . . .
Carbon:
Hydrogen:
Oxygen:
Multiply ‘til whole . . . (If the answer is already
whole, skip this step!)
Carbon: 1.50
x 2
3
Hydrogen: 2.50
x 2
5
Oxygen: 1.00
x 2
2
Empirical formula: C3H5O2
The composition of adipic acid is 49.3% C, 6.9% H,
and 43.8% O (by mass). The molar mass of the
compound is about 146 g/mol.
b) Determine the Molecular Formula
Use the empirical formula. Divide the molecular mass
by the empirical formula mass. It’s all about M/E
Mass of Molecular – Given (146 g/mol)
Mass of Empirical: C3H5O2
3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g
(C3H5O2) x 2 = C6H10O4
Section 3.8
Chemical Equations
• Done for today . . .
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Warm up: A compound of B and H is 81.10% B.
What is its empirical formula?
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• Percent to mass
– In 100.0 g of the compound there are 81.10 g of B
and 18.90 g of H.
• Mass to mole
1 mol B
81.10 g B •
= 7.502 mol B
10.81 g B
18.90 g H •
1 mol H
= 18.75 mol H
1.008 g H
• Divide by small
7.502 mol B
= 1mol B
7.502 mol
• Multiply til whole
18.72 mol H
= 2.499 mol H = 2.5 mol H
7.502 mol
2.5 mol H  2 = 5 mol H
Answer : B2 H5
1 mol B  2 = 2 mol B
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Section 3.8
Chemical Equations
•
A representation of a balanced chemical
equation:
C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
reactants
•
•
products
Reactants on the left, products on the right.
Since atoms are neither created nor destroyed, all
atoms present in the reactants must be present in
the products, in the same number.
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Section 3.8
Chemical Equations
1C2H5OH(l) + 3O2(g)  2CO2(g) + 3H2O(g)
•
The coefficients (numbers in front) give the
atomic/molecular/mole ratios
Ex: 1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon dioxide
and 3 moles of water.
(s) = solid
(g) = gas
(l) = liquid
(aq) = aqueous: dissolved in water
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Section 3.9
Balancing Chemical Equations
1. Determine what reaction is occurring.
What are the reactants, the products,
and the physical states involved?
2. Write the unbalanced equation.
3. Balance the equation systematically,
starting with the most complicated
molecule(s).
4. Check - The same number of each type
of atom needs to appear on both
reactant and product sides.
Balancing equations takes practice!
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Section 3.9
Balancing Chemical Equations
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Section 3.9
Balancing Chemical Equations
Important
•
•
•
•
The number of atoms of each type of
element must be the same on both sides of
a balanced equation.
Subscripts can NOT be changed to balance
an equation.
A balanced equation tells us the ratio of the
number of molecules/formula units which
react and are produced in a chemical
reaction.
Coefficients can be fractions, although they
are usually given as lowest integer
multiples.
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31
Section 3.9
Balancing Chemical Equations
Exercise
Which of the following correctly balances the
chemical equation given below? There may be more
than one correct balanced equation. If a balanced
equation is incorrect, explain what is incorrect about
it.
CaO + C  CaC2 + CO2
I.
II.
III.
IV.
CaO2 + 3C  CaC2 + CO2
2CaO + 5C  2CaC2 + CO2
CaO + (2.5)C  CaC2 + (0.5)CO2
4CaO + 10C  4CaC2 + 2CO2
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Section 3.9
Balancing Chemical Equations
1. NH3
+
O2
 NO
+
H2O
2. SnO2
+
H2
 Sn
+
H2O
3. Fe
+
H2S04
 Fe2(SO4)3
+
H2
4. KOH
+
H3PO4

+
H2O
5. C2H6
+
O2
 H2O
+
CO2
6. KNO3
+
H2CO3
 K2CO3
+
HNO3
7. B2Br6
+
HNO3

+
HBr
K3PO4
B(NO3)3
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Section 3.9
Balancing Chemical Equations
Concept Check
Which of the following are true concerning balanced
chemical equations? There may be more than one true
statement.
I. The number of molecules is conserved.
II. The coefficients tell you how much of each
substance you have.
III. Atoms are neither created nor destroyed.
IV. The coefficients indicate the mass ratios of the
substances used.
V. The sum of the coefficients on the reactant side
equals the sum of the coefficients on the product
side.
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34
Section 3.10
Stoichiometric Calculations: Amounts of Reactants and Products
Stoichiometry is the calculation of
chemical quantities from balanced
equations.
The four quantities involved are:
• particles (atoms, ions, unit formulas or molecules)
• moles
• mass
• volume - of gaseous reactants or products
Remember: Atoms and mass are always conserved in
chemical reactions.
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Section 3.10
Stoichiometric Calculations: Amounts of Reactants and Products
1. Begin by balancing the equation!
2. Convert the unit given to moles of that substance.
3. Use the balanced equation to set up a mole ratios.
Every stoich problem has a mole-mole ratio!
4. Convert moles of given to moles of the wanted. If
needed, convert moles to unit wanted.
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Volume
1 mol A = 22.42 L A
Mass
1 mol A =
(molar mass) g A
Mol A
(given)
Molecules
1 mol A = 6.022 x 1023 particles A
Get x & y from the coefficients
in the balanced equation
1 mol A
x
x mol B = __ mol B
y mol A
Volume
1 mol B = 22.42 L B
Mol B
(wanted)
Mass
1 mol B =
(molar mass) g B
Molecules
1 mol B = 6.022 x 1023 particles B
Stoichiometry
Calculations Practice
1. How many moles of ammonia are
produced from 0.35 moles of nitrogen?
3H2  2NH3
N2 +
0.35 mol N2
X
2
mol NH3
1
mol N2
= 7.0 x 10-1 mol NH3
2. How many moles of aluminum are needed to
react completely with 49.0 moles of chlorine?
2Al + 3Cl2  2AlCl3
49.0 mol Cl2 x
2
mol Al
3
mol Cl2
= 3.27 x 101 mol Al
3. How many moles of aluminum are needed to
react completely with 49.0 g of chlorine?
2Al + 3Cl2  2AlCl3
49.0 g Cl2
X
1 mol Cl2
70.90 g Cl2
X
2 mol Al
3 mol Cl2
= 4.61 x 10-1 mol Al
4. How many grams of chlorine are needed to
react completely with Al to produce 8.1 mol
of aluminum chloride?
2Al
3Cl2  2AlCl3
+
8.1 mol AlCl3
X
3
mol Cl2
2 mol AlCl3
= 8.6 x 102 g Cl2
X
70.90 g Cl2
1 mol Cl2
5. How many grams of aluminum chloride are
produced by the reaction of 5.4 g of Al?
2Al
5.4 g Al
+
3Cl2  2AlCl3
1 mol Al
X
26.98 g Al
X
2 mol AlCl3 X 133.3 g AlCl3
1 mol AlCl3
2 mol Al
= 2.7 x 101 g AlCl3
6. How many liters of carbon monoxide are
needed to react with 4.8 g of oxygen?
2CO + O2  2CO2
4.8 g O2
1 mol O2
32.00 g O2
2 mol CO
1 mol O2
= 6.7 L CO
22.42 L CO
1 mol CO
7. What mass of ammonium is necessary to react
with 2.1 x 1024 molecules of oxygen?
NH4+ + 2O2  2H2O + NO2
2.1 x 1024
molec O2
1 mol O2
1 mol NH4 18.04g NH4
6.022 x 1023
2 mol O2
molec O2
= 3.1 x 101 g NH4
1 mol NH4
8. How many liters of ammonium are necessary
to produce 3.0 mol of water?
NH4+ +
2 O 2  2 H2 O +
3.0 mol H2O
NO2
1 mol NH4 22.42 L NH4
2 mol H2O 1 mol NH4
= 3.4 x 101 L of NH4
9. How many molecules of water are produced
from 222 moles of oxygen?
NH4+ + 2 O2  2 H2O + NO2
222 mol O2 2 mol H2O
2 mol O2
6.022 x 1023
molec H2O
1
mol H2O
= 1.34 x 1026 molec H2O
10. How many moles of nitrogen dioxide are
produced from 46 moles of oxygen?
NH4 + +
2 O 2  2 H2 O +
46 mol O2
NO2
1 mol NO2
2
mol O2
= 2.3 x 101 mol NO2
Section 3.11
The Concept of Limiting Reagent
Done for today
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Limiting and Excess reagents
The limiting reagent is consumed first
and therefore “limits” the amount of
product that can be formed.
The excess reagent is the reactant
that is left over or remaining
The limiting and excess reagents are
always reactants (located on the left
side of the equation!)
Section 3.11
The Concept of Limiting Reagent
Ex: Mixture of CH4 and H2O Molecules Reacting
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Section 3.11
The Concept of Limiting Reagent
CH4 and H2O form H2 and CO – notice that the methane
is limiting, the water was in excess.
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Section 3.11
The Concept of Limiting Reagent
Limiting Reactants
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How to identify a limiting problem
Two amounts of reactants are given – You
need to solve for the amount of product.
Find which reactant you will run out of
first (the limiting reagent) and use this
reactant to solve the problem.
Sample Problem 1
If 10.0 g of aluminum is combined with 10.0g
of oxygen gas, how many grams of aluminum
oxide can be made?
Step One: Predict & balance the equation.
4Al+ 3O2  2Al2O3
+

Step 2: Use mol/coefficient ratio
• Convert reactants to moles.
Aluminum : 10.0g x 1 mol = 0.371 mol Al
26.98 g
Oxygen:
10.0g x 1 mol
= 0.313 mol O2
32.00 g
• Divide each moles by its coefficient.
Aluminum:
0.371 / 4 = 0.0928
Oxygen:
0.313 / 3 = 0.104
Step 3: Identify the Limiting Reagent.
The limiting reagent is the smallest number.
Aluminum is the limiting reagent!
Step 4: Use the limiting reagent to solve the
problem.
10.0gAl x 1 mol Al x 2 mol Al203 x 101.98 g Al203
1 mol Al203
26.98 g Al 4 mol Al
Answer: 1.89 x 101 g Aluminum Oxide
If the problem asks how much excess
reagent is left:
• Use the limiting to see how much of the
excess you need (do a stoich problem
from limiting reagent to excess reagent)
• Subtract the amount you need from the
original amount given in the problem.
** Be careful of units!!
Sample problem 2: If 10.0 g of aluminum is combined
with 10.0g of oxygen gas, how many grams of the excess
reagent remains unreacted?
4 Al + 3 O2  2 Al2O3
(Excess Reagent given) – (What you need) = Excess
USE THE LIMITING REAGENT to figure out how much you
need.
10.0 g Al
1 mol Al
3 mol O2
26.98 g Al
4 mol Al
32.00 g O2
1 mol O2
= 8.90 g O2
10.0g O2
-
8.90 g O2
= 1.1 g O2 remains
unreacted
6.4 mol
2H2 +
3.4 mol
O2 
2 H2O
Problem 3: How many grams of water are
produced given 6.4 mol of hydrogen and 3.4
mol of oxygen?
Step #1: Find the limiting & excess reagent.
6.4 mol H2
/ 2=
3.2
3.4 mol O2
/ 1=
3.4
H2 is the limiting reagent
O2 is the excess reagent
6.4 mol
2H2 +
3.4 mol
O2 
2H2O
Step #2 – Use the limiting reagent to solve the
problem!
6.4 mol H2 X
2 mol H2O X 18.02 g H2O
2 mol H2
1 mol H2O
=1.2 x 102 g H2O
6.4 mol
2H2 +
3.4 mol
O2 
2H2O
How much of the excess reagent remains
unreacted?
(Excess Reagent given) – (What you need) = Excess
USE THE LIMITING REAGENT to figure out how much you
need.
6.4 mol H2
3.4 mol O2
-
x
1 mol O2 = 3.2 mol O2
2 mol H2
3.2 mol O2 = 0.2 mol O2 remains
unreacted
Section 3.11
The Concept of Limiting Reagent
Concept Check
Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2  2H2O
a)
b)
c)
d)
e)
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product.
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Section 3.11
The Concept of Limiting Reagent
Percent Yield
• An important indicator of the efficiency of a particular
laboratory or industrial reaction.
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Section 3.11
The Concept of Limiting Reagent
Calculating Percent Yield
1. Actual yield - what you got by actually performing
the reaction
2. Theoretical yield - what stoichiometric calculation
says the reaction SHOULD have produced
Αctual Yield
Percent Yield
=
Τheoretical Yield
100
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Section 3.2
Atomic Masses
Counting
by Weighing
Calculating % yield video
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Example 1 – Find the percent yield when 68.0 g
silver reacts with oxygen to form 50.0 g silver
oxide.
Step 1: Determine and balance the equation:
68.0 g
4 Ag
50.0 g
+
O2

2 Ag2O
Step 2: Calculate the theoretical yield from
the reactant given.
68.0 g
4 Ag
68.0 g Ag
1
+ O2
50.0 g
2 Ag2O

mol Ag 2 mol Ag2O 231.8 g Ag2O
107.9 g Ag 4 mol Ag
1 mol Ag2O
= 73.0 g Ag2O
Theoretical yield
Step 3: Plug numbers into the equation and... solve!
Actual Yield
(product given)
Theoretical Yield
X 100 = % Yield
(product calculated)
50.0 g
(product given)
73.0 g
X 100 = % Yield
(product calculated)
% Yield = 68.5%
Example 2 – Find the actual product made when
the incomplete reaction between 16.5 g hydrogen
gas and oxygen gas produces a 70.0% yield of
water.
Step 1: Determine and balance the equation:
16.5 g
2 H2
70.0%
+
O2

2 H2O
Step 2: Calculate the theoretical yield from
the reactant given.
16.5 g
2 H2
16.5 g H2
1
+ O2
mol H2
2.016 g H2
2 H2O

2 mol H2O
2 mol H2
18.02 g H2O
1 mol H2O
= 1.47 x 102 g H2O
Theoretical yield
Step 3: Plug numbers into the equation and... solve!
Actual Yield
(product given)
Theoretical Yield
X 100 = % Yield
(product calculated)
Actual Yield
(product given)
1.47 x 102 g
X 100 = 70.0
(product calculated)
Actual Yield = 1.03 x 102 g H2O