Transcript Stoichiometry: Calculations with Chemical Formulas and

Chemistry, The Central Science, 11th edition
Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten
Chapter 3
Stoichiometry
Definition: Mathematical calculations for
chemical formulas & equations.
Mrs. Deborah Amuso
Camden High School
Camden, NY 13316
Stoichiometry
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Chemical Equations
Chemical equations are concise
representations of chemical reactions.
(example)
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Stoichiometry
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Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Reactants:
Products:
appear on the
left side of the
equation
appear on the
right side of the
equation
The states are written in parentheses.
Coefficients are inserted to balance the equation.
Stoichiometry
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Subscripts vs Coefficients
• Subscripts tell the number of atoms of each
element in a molecule.
H2O (The 2 means 2 atoms of H)
• Coefficients tell the number of molecules.
3 H2O (The 3 means 3 molecules of water)
Stoichiometry
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Counting Atoms
Total # atoms = Coefficient x Subscript
Practice:
3 H2SO4
6-H 3-S 12-O
or 3 – SO4
2 Ca3(PO4)2
6-Ca 4-P 16-O
or 4 – PO4
Stoichiometry
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Balancing Equations
When balancing equations use
coefficients to get the # of each type of
atom on both sides of the equation the
same.
Consult your teacher for balancing tips.
Stoichiometry
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ReactionTypes
1.
2.
3.
4.
5.
Combination
Decomposition
Combustion
Single Replacement
Double Replacement
Stoichiometry
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Combination Reactions
• In a
combination
reaction two
or more
substances
react to form
one product.
Examples:
2 Mg (s) + O2 (g)  2 MgO (s)
N2 (g) + 3 H2 (g)  2 NH3 (g)
C3H6 (g) + Br2 (l)  C3H6Br2 (l)
Stoichiometry
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Decomposition Reactions
• In a decomposition
reaction one reactant
breaks down into two
or more substances.
Examples:
CaCO3 (s)  CaO (s) + CO2 (g)
2 KClO3 (s)  2 KCl (s) + O2 (g)
2 NaN3 (s)  2 Na (s) + 3 N2 (g)
Stoichiometry
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Combustion Reactions
• In a combustion
reaction a carbon
compound reacts
with oxygen in the
air to produce
carbon dioxide &
water.
Examples:
• These are generally rapid
reactions that produce a
flame.
CH4 (g) + 2 O2 (g)  CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (g)
Stoichiometry
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Single Replacement Reactions
In a single replacement
reaction, an element &
compound react to
form a different
element & compound.
Examples:
Zn (s) + CuSO4 (aq)  Cu(s) + H2SO4 (aq)
2 HCl (aq) + Mg(s)  MgCl2 (aq) + H2(g)
Stoichiometry
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Double Replacement Reactions
In a double replacement
reaction, two compounds
in solution react by
exchanging ions to form
two different compounds.
Examples:
ZnCl2 (aq) + 2 AgNO3 (aq)  2AgCl(s) + Zn(NO3)2(aq)
Ba(NO3)2(aq) + Na2SO4(aq)  BaSO4(s) + 2 NaNO3 (aq)
Stoichiometry
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Formula Mass (FM)
• A formula mass is the sum of the
atomic masses for the atoms in a
chemical formula.
• So, the formula mass of calcium
chloride, CaCl2, would be
Ca: 1(40.1 amu) = 40.1
+ Cl: 2(35.5 amu) = 71.0
111.1 amu
• Formula masses are generally reported
for ionic compounds.
Stoichiometry
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Molecular Mass (MM)
• A molecular mass is the sum of the
atomic masses for the atoms in a
molecule.
• For the molecule ethane, C2H6, the
molecular mass would be
C: 2(12.0 amu) = 24.0
+ H: 6(1.0 amu) = 6.0
30.0 amu
• FM & MM are found the same way!
Stoichiometry
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Mass Percent or Percent Composition
The mass percent of each element in a
compound is found by using this
equation:
% element =
(number of atoms)(atomic mass)
(FM of the compound)
Or simply:
x 100
% = Part/Whole x 100
Stoichiometry
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Percent Composition
So the percentage of carbon in ethane,
C2H6, is…
(2)(12.0 amu)
%C =
(30.0 amu)
24.0 amu
x 100
=
30.0 amu
= 80.0%
Stoichiometry
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Moles & Avogadro’s Number
•
•
•
•
1 mole = 6.02 x 1023
The mass of 1 molecule of water is 18.0 amu.
The mass of 1 mole of water is 18.0 grams.
Stoichiometry
Why? 1 gram = 6.02 x 1023 amu.
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Molar Mass or GFM
• Molar mass is the mass of 1 mole of a
substance in g/mol.
– The molar mass of an element is the mass
number that we find on the periodic table.
example: Cu = 63.5
O2 = 2(16.0) = 32
– The molar mass of a compound is the same
number as its formula mass or molecular
mass.
Stoichiometry
example: C2H6 = 30.0
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Using Moles
Moles provide a bridge from the molecular scale to
the real-world scale.
# moles = given mass
GFM
GFM is the same as Molar Mass
(It’s the mass in grams of 1 mole of the substance)Stoichiometry
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Three Helpful Equations
Use these 3 equations to convert between masses,
moles, molecules (or units) , & atoms (or ions):
#moles = mass/GFM
#molecules = (#moles) (6.02 x 1023)
(or # units)
#atoms = (#molecules) (#atoms/molecule)
(or # ions)
Stoichiometry
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Mole Relationships:
Check Your Understanding
How many atoms of oxygen are in 10.0 grams of CH3COOH?
Think: Mass  Moles  Molecules  Atoms
#moles = mass/GFM =
10.0g / 60.0 g/mol = 0.167 moles
#molecules = (#moles)((6.02 x 1023) =
(0.167 moles) (6.02 x 1023) =
1.00 x 1023 molecules
#atoms = (#molecules) (#atoms/molecule) =
(1.00 x 1023 molecules)(2 oxygen atoms/molecule) =
2.00 x 1023 oxygen atoms
Stoichiometry
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Mole Relationships:
Check Your Understanding
How many hydroxide ions are in 10.0 grams of Ca(OH)2
Think: Mass  Moles  Units  Ions
#moles = mass/GFM =
10.0g / 74.1g/mol = 0.135 moles Ca(OH)2
#units = (#moles) (6.02 x 1023) =
(0.135 mol) (6.02 x 1023) =
8.12 x 1022
#ions = (#units) (#ions/unit) =
(8.12 x 1022) (2 OH-/unit) =
1.62 x 1023 OH- ions
Stoichiometry
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Calculating Empirical Formulas
One can calculate the empirical formula from the
percent composition of each element in the formula.
Steps:
1) Assume percent of each element = its mass
2) Divide mass of each element by that element’s
atomic mass to get # moles
3) Calculate the mole ratio by dividing by the smallest
# of moles
4) These are the #’s in the empirical formula
Stoichiometry
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Calculating Empirical Formulas
The compound para-aminobenzoic acid (you may have seen it
listed as PABA on your bottle of sunscreen) is composed of carbon
(61.31%), hydrogen (5.14%), nitrogen (10.21%), and oxygen
(23.33%). Find the empirical formula of PABA.
C:
61.31 g / 12.00 g/mol
= 5.105 mol C
H:
5.14 g / 1.01 g/mol
= 5.09 mol H
N:
10.21 g / 14.01 g/mol
= 0.7288 mol N
O:
23.33 g / 16.00 g/mol
= 1.456 mol O
Stoichiometry
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Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number
of moles:
C:
5.105 mol
0.7288 mol
= 7.005  7
H:
5.09 mol
0.7288 mol
= 6.984  7
N:
0.7288 mol
0.7288 mol
= 1.000
O:
1.458 mol
0.7288 mol
= 2.001  2
Stoichiometry
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Calculating Empirical Formulas
These are the subscripts for the empirical formula:
C7H7NO2
Stoichiometry
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Calculating Empirical Formulas
Find the empirical formula of a compound whose mass
percentage is 50.0% sulfur and 50.0% oxygen.
Step 1
S:
50.0 g / 32.1 g/mol
O:
50.0 g / 16.0 g/mol
= 1.56 mol S
= 3.13 mol O
Step 2
Step 3
S:
1.56 / 1.56 = 1
O:
3.13 / 1.56 = 2
Empirical Formula is SO2
Stoichiometry
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Calculating Molecular Formulas
From Empirical Formulas & Molar Mass
What is the MF if the EF is C4H5N2O and the MM is 194 g/mol?
Step 1: Determine molar mass of empirical formula
C4H5N2O = 97 g/mol
Step 2: Divide:
Molar Mass given/Molar Mass of empirical formula
194 / 97 = 2
Step 3: Multiply:
Answer from Step 2 * Empirical Formula
(2)(C4H5N2O) = C8H10N4O2
Stoichiometry
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Calculating Molecular Formulas
From Empirical Formulas & Molar Mass
What is the MF if the EF is CH2 and the molecular mass is
84.16 amu?
Step 1: Determine molecular mass of empirical formula
CH2 = 14.03 amu
Step 2: Divide:
Molecular mass given/molecular mass of emp formula
84.16 amu /14.03 amu = 6
Step 3: Multiply:
Answer from Step 2 * Empirical Formula
(6)(CH2) = C6H12
Stoichiometry
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Combustion Analysis
• Compounds containing C, H and O are routinely
analyzed through combustion in a chamber like this.
– C is determined from the mass of CO2 produced.
– H is determined from the mass of H2O produced.
– O is determined by difference after the C and H have been
determined.
– Then you can determine the Empirical Formula of Stoichiometry
the compound
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Combustion Analysis
When 0.255 g of caproic acid is burned, you get 0.512 g CO2
and 0.209 g H2O. What is the emp formula of the acid?
#mole CO2 = mass/GFM = 0.512g / 44.0g/mol = 0.0116 mol
0.0116 mol CO2 * (1 mole C/1 mol CO2) = 0.0116 mol C
0.0116 mol C * (12.0 g/mol) = 0.140 g C
#mole H2O = mass/GFM = 0.209g / 18.0g/mol = 0.0116 mol
0.0116 mol H2O * (2 mole H/1 mol H2O)= 0.0232 mol H
0.0232 mol H * (1.01 g/mol) = 0.0235 g H
Mass O = Mass caproic acid – (mass C + mass H)
Mass O = 0.225 g – (0.140 g + 0.0235 g) = .0615 g
Stoichiometry
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2nd Part of Problem
For carbon: 0.140 g C / 12.0 g/mol = 0.0117 mol
For hydrogen: 0.0235 g H / 1.01 g/mol = 0.0233 mol
For oxygen: 0.0615 g O / 16.0 g/mol = 0.00384 mol
For carbon: 0.0117 / 0.00384 = 3.05
For hydrogen: 0.0233 / 0.00384 = 6.07
For oxygen: 0 0.00384 / 0.00384 = 1
Empirical Formula = C3H6O
Stoichiometry
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Stoichiometric Calculations
The coefficients in the balanced equation
represent the # of molecules or # of moles of
reactants and products.
The coefficients tell the ‘recipe’ for the rxn.
Stoichiometry
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Stoichiometric Calculations
Stoichiometry
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Stoichiometric Calculations
C6H12O6 + 6 O2  6 CO2 + 6 H2O
Starting with 1.00 g of C6H12O6…
we calculate the moles of C6H12O6…
use the coefficients to find the moles of H2O…
and then turn the moles of water to grams.
Stoichiometry
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Stoichiometric Calculation Steps
1) Complete & balance equation using coefficients.
2) Identify Given & Unknown.
3) Convert Given mass into moles:
#molesgiven = massgiven/GFMgiven
4) Use mole ratio to find moles of Unknown.
#molesgiven = #molesunk
Coeffgiven
Coeffunk
5) Convert moles of ‘unknown’ to answer.
massunknown = (#molesunk) (GFMunk)
Stoichiometry
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CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
How many grams of water are produced when 32.0 grams of methane
completely react?
1) Given = 32.0 g CH4
Unk = ? g H2O
2) Find # moles CH4
#moles CH4 = mass/GFM = 32.0 g / 16.0 g/mol = 2.00 moles CH4
3) Find # moles H2O
2.00 moles CH4
1 mole
= ? moles H2O
2 mole
Ans. = 4.00 moles H2O
4) Find mass of H2O
mass H2O = (#moles)(GFM) = (4.00 mol)(18.0 g/mol) =
Stoichiometry
Ans. = 72.0 g H2O
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2 H2 (g) + O2 (g)
2 H2O (g)
How many grams of water are produced when 6.0 grams of oxygen gas
completely react?
1) Given = 6.0 g O2
Unk = ? g H2O
2) Find # moles O2
#moles O2 = mass/GFM = 6.0 g / 32.0 g/mol = 0.19 moles O2
3) Find # moles H2O
0.19 moles O2
1 mole
= ? moles H2O
2 mole
Ans. = 0.38 moles H2O
4) Find mass of H2O
mass H2O = (#moles)(GFM) = (0.38 mol)(18.0 g/mol) =
Stoichiometry
Ans. = 6.8 g H2O
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Limiting Reactants:
How Many Cookies Can I Make?
• You can make cookies until
you run out of one of the
ingredients.
• Once you run out of sugar,
you can’t make any more
cookies.
• In this example the sugar
would be the limiting
reactant, because it will
limit the amount of cookies
you can make.
Stoichiometry
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Limiting Reactants
• The limiting reactant (or limiting reagent) is
the reactant you’ll run out of first
• What is the LR in this reaction?
Stoichiometry
H2
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How to Determine Which Reactant is
the Limiting Reactant
Step 1: Convert the masses given to moles.
Step 2: Use these equations:
# moles reactant A / coefficient A = x
# moles reactant B / coefficent B = y
Step 3: The smaller # (x or y) identifies the
Limiting Reagent.
Step 4: Use the # of moles of the Limiting
Reagent to solve the rest of the
problem.
Stoichiometry
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CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Given 50.0 g of CH4 and 60.0 g of O2
Which is the Limiting Reagent?
Step 1:
#moles CH4 = mass/GFM = 50.0 g/16.0 g/mol = 3.13 moles
#moles O2 = mass/GFM = 60.0 g/32.0 g/mol = 1.88 moles
Step 2: Use these equations:
# moles reactant CH4 / coefficient CH4= 3.13 / 1 = 3.12
# moles reactant O2 / coefficent O2 = 1.88 / 2 = 0.94
Step 3: The smaller # (x or y) identifies the Limiting Reagent
0.94 is smaller, so O2 is the L.R.
Stoichiometry
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2 CHCl3 (g) + 2 Cl2 (g)
2 CClO4 (g) + 2 HCl (g)
Given 15.9 g of CHCl3 and 12.6 g of Cl2
Which is the Limiting Reagent?
Step 1:
#moles CHCl3 = mass/GFM = 15.9 g/119.5 g/mol = 0.133 mol
#moles Cl2 = mass/GFM = 12.6 g/71.0 g/mol = 0.177 mol
Step 2: Use these equations:
# mol reactant CHCl3 / coefficient CHCl3= 0.133/ 2 = 0.0665
# mol reactant Cl2 / coefficent Cl2 = 0.177/ 2 = 0.0885
Step 3: The smaller # (x or y) identifies the Limiting Reagent
0.0665 is smaller, so CHCl3 is the L.R.
Stoichiometry
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Theoretical Yield
• The theoretical yield is the maximum
amount of product that can be made.
– In other words it’s the amount of product
possible as calculated through the
stoichiometry problem.
• This is different from the actual yield,
which is the amount one actually
produces and measures.
Stoichiometry
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Percent Yield
One finds the percent yield by
comparing the amount actually obtained
(actual yield) to the amount it was
possible to make (theoretical yield).
Actual Yield
Percent Yield =
Theoretical Yield
x 100
Stoichiometry
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Percent Yield Problem
CH4 (g) + 2 O2 (g)
CO2 (g) + 2 H2O (g)
Previously, you determined 72.0 g of water should be
produced from 32.0 g of methane. Assume only 68.0 g
of water are produced when this experiment is actually
conducted. What is the percent yield?
% yield = [actual yield/theoretical yield] x 100
% yield = [68.0g / 72.0 g] x 100
% yield = 94.4%
Stoichiometry
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