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Chapter 3 Stoichiometry- Calculations with Chemical Formulas and Equations



Chemistry, The Central Science, 11th edition Theodore L. Brown, H. Eugene LeMay, Jr., and Bruce E. Bursten

Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equation

s
Stoichiometry

Chemical Equations
Chemical equations are concise representations of chemical reactions.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Stoichiometry
? 2009, Prentice-Hall, Inc.

Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Reactants appear on the left side of the equation.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Products appear on the right side of the equation.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

The states of the reactants and products are written in parentheses to the right of each compound.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Anatomy of a Chemical Equation
CH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O (g)

Coefficients are inserted to balance the equation.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Subscripts and Coefficients Give Different Information

? Subscripts tell the number of atoms of each element in a molecule.
Stoichiometry
? 2009, Prentice-Hall, Inc.

Subscripts and Coefficients Give Different Information

? Subscripts tell the number of atoms of each element in a molecule ? Coefficients tell the number of molecules.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Reaction Types
Stoichiometry
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Combination Reactions
? In this type of reaction two or more substances react to form one product.

? Examples:
– 2 Mg (s) + O2 (g) – N2 (g) + 3 H2 (g) – C3H6 (g) + Br2 (l) 2 MgO (s) 2 NH3 (g) C3H6Br2 (l)

Stoichiometry
? 2009, Prentice-Hall, Inc.

Decomposition Reactions
? In a decomposition one substance breaks down into two or more substances.

? Examples:
– CaCO3 (s) – 2 KClO3 (s) – 2 NaN3 (s) CaO (s) + CO2 (g) 2 KCl (s) + O2 (g) 2 Na (s) + 3 N2 (g)

Stoichiometry
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Combustion Reactions
? These are generally rapid reactions that produce a flame. ? Most often involve hydrocarbons reacting with oxygen in the air.

? Examples:
– CH4 (g) + 2 O2 (g) – C3H8 (g) + 5 O2 (g) CO2 (g) + 2 H2O (g) 3 CO2 (g) + 4 H2O (g)
Stoichiometry
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Formula Weights
Stoichiometry
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Formula Weight (FW)
? A formula weight is the sum of the atomic weights for the atoms in a chemical formula. ? 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. Stoichiometry
? 2009, Prentice-Hall, Inc.

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
C: 2(12.0 amu) + H: 6(1.0 amu) 30.0 amu
Stoichiometry
? 2009, Prentice-Hall, Inc.

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

Stoichiometry
? 2009, Prentice-Hall, Inc.

Percent Composition
So the percentage of carbon in ethane is…
(2)(12.0 amu) (30.0 amu) 24.0 amu 30.0 amu x 100

%C = =

= 80.0%
Stoichiometry
? 2009, Prentice-Hall, Inc.

Moles
Stoichiometry
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Avogadro’s Number

? 6.02 x 1023 ? 1 mole of 12C has a mass of 12 g.
Stoichiometry
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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).

Stoichiometry
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Using Moles

Moles provide a bridge from the molecular scale to the real-world scale.

Stoichiometry
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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. Stoichiometry
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Finding Empirical Formulas
Stoichiometry
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Calculating Empirical Formulas

One can calculate the empirical formula from the percent composition.

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.

Stoichiometry
? 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas
Assuming 100.00 g of para-aminobenzoic acid, C:
H: N: O:

1 mol 12.01 g 1 mol 5.14 g x 1.01 g 1 mol 10.21 g x 14.01 g 1 mol 23.33 g x 16.00 g 61.31 g x

= 5.105 mol C
= 5.09 mol H = 0.7288 mol N = 1.456 mol O

Stoichiometry
? 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas
Calculate the mole ratio by dividing by the smallest number of moles:
C: 5.105 mol 0.7288 mol 5.09 mol 0.7288 mol 0.7288 mol 0.7288 mol 1.458 mol 0.7288 mol = 7.005 7

H:

= 6.984

7

N:

= 1.000

O:

= 2.001

2

Stoichiometry
? 2009, Prentice-Hall, Inc.

Calculating Empirical Formulas
These are the subscripts for the empirical formula: C7H7NO2

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 CO 2 produced. – H is determined from the mass of H 2O produced. – O is determined by difference after the C and H have been determined.
Stoichiometry
? 2009, Prentice-Hall, Inc.

Elemental Analyses
Compounds containing other elements are analyzed using methods analogous to those used for C, H and O.

Stoichiometry
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Stoichiometric Calculations

The coefficients in the balanced equation give the ratio of moles of reactants and products.

Stoichiometry
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Stoichiometric Calculations
Starting with the mass of Substance A you can use the ratio of the coefficients of A and B to calculate the mass of Substance B formed (if it’s a product) or used (if it’s a reactant).
Stoichiometry
? 2009, Prentice-Hall, Inc.

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

? 2009, Prentice-Hall, Inc.

Limiting Reactants
Stoichiometry
? 2009, Prentice-Hall, Inc.

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).
Stoichiometry
? 2009, Prentice-Hall, Inc.

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.

Stoichiometry
? 2009, Prentice-Hall, Inc.

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).

Stoichiometry
? 2009, Prentice-Hall, Inc.

Limiting Reactants
In the example below, the O2 would be the excess reagent.

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
? 2009, Prentice-Hall, Inc.

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
? 2009, Prentice-Hall, Inc.



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