Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

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Chapter 3StoichiometryCalculations with
Chemical Formulas and Equations
Chemistry, The Central Science, 10th
edition Theodore L. Brown, H. Eugene LeMay, Jr.,
and Bruce E. Bursten

• John D. Bookstaver
• St. Charles Community College
• St. Peters, MO
• ? 2006, Prentice-Hall

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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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Chemical Equations

• Concise representations of chemical reactions

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Anatomy of a Chemical Equation

• CH4 (g) 2 O2 (g) CO2 (g) 2
  H2O (g)

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Anatomy of a Chemical Equation

• CH4 (g) 2 O2 (g) CO2 (g) 2
  H2O (g)

• Reactants appear on the left side of the equation.

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Anatomy of a Chemical Equation

• CH4 (g) 2 O2 (g) CO2 (g) 2
  H2O (g)

• Products appear on the right side of the equation.

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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.

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Anatomy of a Chemical Equation

• CH4 (g) 2 O2 (g) CO2 (g) 2
  H2O (g)

• Coefficients are inserted to balance the equation.

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Subscripts and Coefficients Give Different
Information

• Subscripts tell the number of atoms of each
  element in a molecule

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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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Reaction Types
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Combination Reactions

• Two or more substances react to form one product

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

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2 Mg (s) O2 (g) ??? 2 MgO (s)
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Decomposition Reactions

• One substance 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)

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Combustion Reactions

• Rapid reactions that produce a flame
• Most often involve hydrocarbons reacting with
  oxygen in the air

• Examples
• CH4 (g) 2 O2 (g) ??? CO2 (g) 2 H2O (g)
• C3H8 (g) 5 O2 (g) ??? 3 CO2 (g) 4 H2O (g)

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Formula Weights
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Formula Weight (FW)

• 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
• These are generally reported for ionic compounds

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Molecular Weight (MW)

• Sum of the atomic weights of the atoms in a
  molecule
• For the molecule ethane, C2H6, the molecular
  weight would be

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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

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Percent Composition

• So the percentage of carbon in ethane is

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Moles
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Avogadros Number

• 6.02 x 1023
• 1 mole of 12C has a mass of 12 g

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Molar Mass

• By definition, these are 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 amus) will be the same
  number as the molar mass (in g/mol)

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Using Moles

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

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Mole Relationships

• One mole of atoms, ions, or molecules contains
  Avogadros number of those particles
• One mole of molecules or formula units contains
  Avogadros number times the number of atoms or
  ions of each element in the compound

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Finding Empirical Formulas
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Calculating Empirical Formulas

• One can calculate the empirical formula from the
  percent composition

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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.
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Calculating Empirical Formulas
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Calculating Empirical Formulas
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Calculating Empirical Formulas
These are the subscripts for the empirical
formula C7H7NO2
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Stoichiometric Calculations

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

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Stoichiometric Calculations

• From 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 its a
  product) or used (if its a reactant)

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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

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Limiting Reactants
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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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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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Limiting Reactants

• The limiting reactant is the reactant present in
  the smallest stoichiometric amount

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Limiting Reactants

• The limiting reactant is the reactant present in
  the smallest stoichiometric amount
• In other words, its the reactant youll run out
  of first (in this case, the H2)

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Limiting Reactants

• In the example below, the O2 would be the excess
  reagent

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SAMPLE EXERCISE 3.18 Calculating the Amount of
Product Formed from Limiting Reactant
How many moles of NH3 can be formed from 3.0 mol
of N2 and 6.0 mol of H2?
Plan If we assume that one reactant is
completely consumed, we can calculate how much of
the second reactant is needed in the reaction. By
comparing the calculated quantity with the
available amount, we can determine which reactant
is limiting. We then proceed with the
calculation, using the quantity of the limiting
reactant.
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SAMPLE EXERCISE 3.18 continued
Comment The table below summarizes this example
Notice that we can calculate not only the number
of moles of NH3 formed but also the number of
moles of each of the reactants remaining after
the reaction. Notice also that although the
number of moles of H2 present at the beginning of
the reaction is greater than the number of moles
of N2 present, the H2 is nevertheless the
limiting reactant because of its larger
coefficient in the balanced equation.
Check The summarizing table shows that the mole
ratio of reactants used and product formed
conforms to the coefficients in the balanced
equation, 1 3 2. Also, because H2 is the
limiting reactant, it is completely consumed in
the reaction, leaving 0 mol at the end. Because
2.0 mol H2 has two significant figures, our
answer has two significant figures.
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SAMPLE EXERCISE 3.18 continued
PRACTICE EXERCISE
Consider the reaction A mixture of 1.50 mol of
Al and 3.00 mol of Cl2 is allowed to react. (a)
Which is the limiting reactant? (b) How many
moles of AlCl3 are formed? (c) How many moles of
the excess reactant remain at the end of the
reaction?
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SAMPLE EXERCISE 3.19 Calculating the Amount of
Product Formed from a Limiting Reactant
Consider the following reaction
Suppose a solution containing 3.50 g of Na3PO4 is
mixed with a solution containing 6.40 g of
Ba(NO3)2. How many grams of Ba3(PO4)2 can be
formed?
Plan We must first identify the limiting
reagent. To do so, we can calculate the number of
moles of each reactant and compare their ratio
with that required by the balanced equation. We
then use the quantity of the limiting reagent to
calculate the mass of Ba(PO4)2 that forms.
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SAMPLE EXERCISE 3.19 continued
Thus, there are slightly more moles of Ba(NO3)2
than moles of Na3PO4 The coefficients in the
balanced equation indicate, however, that the
reaction requires 3 mol Ba(NO3)2 for each 2 mol
Na3PO4 That is, 1.5 times more moles of Ba(NO3)2
are needed than moles of Na3PO4. Thus, there is
insufficient Ba(NO3)2 to completely consume the
Na3PO4 That means that Ba(NO3)2 is the limiting
reagent. We therefore use the quantity of
Ba(NO3)2 to calculate the quantity of product
formed. We can begin this calculation with the
grams of Ba(NO3)2 but we can save a step by
starting with the moles of Ba(NO3)2 that were
calculated previously in the exercise
Check The magnitude of the answer seems
reasonable Starting with the numbers in the two
conversion factors on the right, we have 600/3
200 200 ? 0.025 5. The units are correct, and
the number of significant figures (three)
corresponds to the number in the quantity of
Ba(NO3)2.
Comment The quantity of the limiting reagent,
Ba(NO3)2 can also be used to determine the
quantity of NaNO3 formed (4.16 g) and the
quantity of Na3PO4 used (2.67 g). The number of
grams of the excess reagent, Na3PO4 remaining at
the end of the reaction equals the starting
amount minus the amount consumed in the reaction,
3.50 g - 2.67 g 0.82 g.
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SAMPLE EXERCISE 3.19 continued
PRACTICE EXERCISE
A strip of zinc metal having a mass of 2.00 g is
placed in an aqueous solution containing 2.50 g
of silver nitrate, causing the following reaction
to occur
(a) Which reactant is limiting? (b) How many
grams of Ag will form? (c) How many grams of
Zn(NO3)2 will form? (d) How many grams of the
excess reactant will be left at the end of the
reaction?
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Theoretical Yield

• The theoretical yield is the amount of product
  that can be made
• In other words its the amount of product
  possible as calculated through the stoichiometry
  problem
• This is different from the actual yield, the
  amount one actually produces and measures

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Percent Yield

• A comparison of the amount actually obtained to
  the amount it was possible to make

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SAMPLE EXERCISE 3.20 Calculating the Theoretical
Yield and Percent Yield for a Reaction
Adipic acid, H2C6H8O4, is used to produce nylon.
The acid is made commercially by a controlled
reaction between cyclohexane (C6H12) and O2
(a) Assume that you carry out this reaction
starting with 25.0 g of cyclohexane and that
cyclohexane is the limiting reactant. What is the
theoretical yield of adipic acid?
(b) If you obtain 33.5 g of adipic acid from
your reaction, what is the percent yield of
adipic acid?
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SAMPLE EXERCISE 3.20 continued
Solve
Check Our answer in (a) has the appropriate
magnitude, units, and significant figures. In (b)
the answer is less than 100 as necessary.
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SAMPLE EXERCISE 3.20 continued
PRACTICE EXERCISE
Imagine that you are working on ways to improve
the process by which iron ore containing Fe2O3 is
converted into iron. In your tests you carry out
the following reaction on a small scale
(a) If you start with 150 g of Fe2O3 as the
limiting reagent, what is the theoretical yield
of Fe? (b) If the actual yield of Fe in your test
was 87.9 g, what was the percent yield?