chapter 3 stoichiometry: calculations with chemical formulas and equations
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Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations. Law of Conservation of Mass. - PowerPoint PPT PresentationTRANSCRIPT
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Stoichiometry
Chapter 3 Stoichiometry:Calculations with Chemical Formulas
and Equations
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Stoichiometry
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Stoichiometry
Law of Conservation of Mass“We may lay it down as an
incontestable axiom that, in all the operations of art and nature,
nothing is created; an equal amount of matter exists both
before and after the experiment. Upon this principle, the whole art
of performing chemical experiments depends.”
--Antoine Lavoisier, 1789
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Stoichiometry
Poor Antoine
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Stoichiometry
Dalton’s Postulate #4
• Atoms of elements are neither created nor destroyed, simply rearranged during a chemical reaction.
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Stoichiometry
Chemical EquationsChemical equations are concise representations of chemical reactions.
Write a balanced equation for this reaction.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
Anatomy of a Chemical Equation
Reactants appear on the left side of the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
Anatomy of a Chemical Equation
Products appear on the right side of the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
Anatomy of a Chemical Equation
The states of the reactants and products are written in parentheses to the right of each compound.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
Anatomy of a Chemical Equation
Coefficients are inserted to balance the equation.
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
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Stoichiometry
Subscripts and Coefficients Give Different Information
• Subscripts tell the number of atoms of each element in a molecule.
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Stoichiometry
Subscripts and Coefficients Give Different Information
• Subscripts tell the number of atoms of each element in a molecule
• Coefficients tell the number of molecules.
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Stoichiometry
Reaction Types
SynthesisDecomposition
Combustion DoubleReplacement
Single Replacement
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Stoichiometry
Combination or Synthesis Reactions
Examples:2 Mg (s) + O2 (g) 2 MgO (s)
N2 (g) + 3 H2 (g) 2 NH3 (g)
C3H6 (g) + Br2 (l) C3H6Br2 (l)
• In this type of reaction two or more substances react to form one product.
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Stoichiometry
• In a decomposition one substance breaks down into two or more substances.
Decomposition Reactions
Examples:CaCO3 (s) CaO (s) + CO2 (g)
2 KClO3 (s) 2 KCl (s) + O2 (g)
2 NaN3 (s) 2 Na (s) + 3 N2 (g)
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Stoichiometry
NaN3 = Sodium Azide• - colorless salt is the gas-forming
component in many car airbag systems.
• - an ionic substance, is highly soluble in water, and is very acutely toxic.
Azide is the anion with the formula N-3.
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Stoichiometry
Combustion Reactions
• Examples:CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)
C3H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (g)
• These are generally rapid reactions that produce a flame.
• Most often involve hydrocarbons reacting with oxygen in the air.
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Stoichiometry
Formula WeightsMolecular Weights
Percent Composition
Molar Mass
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Stoichiometry
Formula Weight (FW)
• A formula weight is the sum of the atomic weights for the atoms in a formula unit.
• So, the formula weight of calcium chloride, CaCl2, would be
Ca: 1(40.1 amu)
+ Cl: 2(35.5 amu)
111.1 amu
• Formula weights are generally reported for ionic compounds.
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Stoichiometry
Molecular Weight (MW)• A molecular weight is the sum of the atomic weights of the atoms in
a molecule.
• For the molecule ethane, C2H6, the molecular weight would be
• Molecular weights are generally reported for covalent compounds.C: 2(12.0 amu)
30.0 amu+ H: 6(1.0 amu)
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Stoichiometry
Percent Composition
One can find the percentage of the mass of a compound that comes from each of the elements in the compound by using this equation:
% element =(number of atoms)(atomic weight)
(FW of the compound)x 100
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Stoichiometry
Percent CompositionSo … the percentage of carbon in ethane (C2H6) is…
%C =(2)(12.0 amu)
(30.0 amu)
24.0 amu
30.0 amu= x 100
= 80.0%
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Stoichiometry
Moles
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Stoichiometry
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Stoichiometry
Avogadro’s Number
• 6.02 x 1023
• 1 mole of 12C has a mass of 12 g.
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Stoichiometry
Molar Mass
• By definition, a molar mass is the mass of 1 mol of a substance (i.e., g/mol).– The molar mass of an element is the mass
number for the element that we find on the periodic table.
– The formula weight (in amu’s) will be the same number as the molar mass (in g/mol).
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Stoichiometry
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Stoichiometry
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Stoichiometry
Using Moles
Moles provide a bridge from the molecular scale to the real-world scale.
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Stoichiometry
Mole Relationships
• One mole of atoms, ions, or molecules contains Avogadro’s number of those particles.
• One mole of molecules or formula units contains Avogadro’s number times the number of atoms or ions of each element in the compound.
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Stoichiometry
Calculating Empirical + MolecularFormulas
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Stoichiometry
Empirical vs. Molecular
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Stoichiometry
Calculating Empirical Formulas
One can calculate the empirical formula from the percent composition.
% to mass mass to mole
÷ by small
× ‘til whole
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Stoichiometry
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Stoichiometry
Calculating Empirical FormulasThe 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.
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Stoichiometry
Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid,
C: 61.31 g x = 5.105 mol C
H: 5.14 g x = 5.09 mol H
N: 10.21 g x = 0.7288 mol N
O: 23.33 g x = 1.456 mol O
1 mol12.01 g
1 mol14.01 g
1 mol1.01 g
1 mol16.00 g
% to mass
mass to mole
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Stoichiometry
Calculating Empirical FormulasCalculate the mole ratio by dividing by the smallest number of moles:
C: = 7.005 7
H: = 6.984 7
N: = 1.000
O: = 2.001 2
5.105 mol0.7288 mol
5.09 mol0.7288 mol
0.7288 mol0.7288 mol
1.458 mol0.7288 mol
÷ by small
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Stoichiometry
Calculating Empirical FormulasThese are the subscripts for the empirical formula:
C7H7NO2 Coefficients become subscripts
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Stoichiometry
Calculating Molecular from Empirical
Mesitylene, a hydrocarbon that occurs in small amounts in crude oil, has an empirical formula of C3H4. The experimentally determined molecular weight of this substance is 121 amu. What is the molecular formula of mesitylene?
1. Calculate the empirical formula weight of C3H4
3 (12 amu) + 4 (1 amu) = 40.0 amu = FW
2. Divide the molecular weight by the formula weight to obtain the factor used to multiply the subscripts in C3H4:
MW = 121 = 3.02 ~ 3FW 40
3. Multiply each subscript in the empirical formula by 3 to give the molecular formula: C9H12
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Stoichiometry
Let’s Try and Example! Practice Exercise 3.14 pg. 94
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Stoichiometry
Combustion Analysis Technique
Lab technique used to determine Empirical Formula contiaining hydrocarbons mainly. Go to p. 95 Figure 3.14!
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Stoichiometry
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.
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Stoichiometry
Elemental AnalysesCompounds containing other elements are analyzed using methods analogous to those used for C, H and O.
Complete Sample Exercise 3.15 pg. 95
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Stoichiometry
Combustion Analysis Problems
1. Use Stoichiometry to find mass of C, H, O
2. Mass (g) to moles of C, H, O
3. Divide by small, multiply ‘til whole
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Stoichiometry
What is the mole ratio of smores?
Where do mole ratios come from?
2 Gc + 1 M + 4 Cp 1 Smore
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Stoichiometry
Stoichiometric Calculations
The coefficients in the balanced equation give the ratio of moles of reactants and products.
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Stoichiometry
Stoichiometric Calculations
Mole Ratio
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Stoichiometry
Ex. How many grams of water are produced in the ocidation of 1.00 g of glucose?
C6H12O6 + 6 O2 6 CO2 + 6 H2O
Circle the known. Box what you are trying to find.
GIVEN
1g C6H12O6
g C6H12O6
mol C6H12O6
C: 6 x 12 = 72H: 12 x 1 = 12O: 6 x 16 = 96 +
180 g/ mol
180
1
mol C6H12O6
mol H2O
1
6
mol H2O
g H2O
H: 2 x 1 = 2O: 1 x 16 = 16 +
18 g/mol
1
18.0= 0.600 g H2O
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Stoichiometry
Limiting Reactants
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Stoichiometry
What’s the max amount of product that I can yield?
2 Gc + 1 M + 4 Cp 1 Smore
7 Gc + 1 M + 8 Cp ? Smore
What is limiting?
What is excess?
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Stoichiometry
Which is limiting the amount of product?
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Stoichiometry
How Many Cookies Can I Make?
• You can make cookies until you run out of one of the ingredients.
• Once this family runs out of sugar, they will stop making cookies (at least any cookies you would want to eat).
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Stoichiometry
How Many Cookies Can I Make?
• In this example the sugar would be the limiting reactant, because it will limit the amount of cookies you can make.
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Stoichiometry
Limiting Reactants• The limiting reactant is the reactant present in the
smallest stoichiometric amount.– In other words, it’s the reactant you’ll run out of first
(in this case, the H2).
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Stoichiometry
Limiting ReactantsIn the example below, the O2 would be the excess reagent.
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Stoichiometry
Limiting Reactants Practice
p. 101 Practice Exercise
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Stoichiometry
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.
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Stoichiometry
Theoretically vs. Actually
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Stoichiometry
Theoretical Yield Example• What is the theoretical yield of CaO if 24.8 g of
CaCO3 is heated?
_____ CaCO3 _____ CaO + ______ CO2
24.8 g CaCO3 x ________ x _________ x __________ = 13.9 g CaO
• Theoretical Yield = 13.9 g CaO
100.1 g CaCO3
1 mol CaCO3
1 mol CaCO3
1 mol CaO
1 mol CaO
56 g CaO
1 1 1
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Stoichiometry
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 YieldTheoretical YieldPercent Yield = x 100
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Stoichiometry
Percent Yield Example
• What is the percent yield if 13.1 g of CaO was actually produced when 24.8 g of CaCO3 was heated?
% Yield = 13.1 g x 100 = 94.2 %
13.9 g