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
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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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Practice

• Find the empirical formula of a compound found to
  contain 26.56 potassium. 35.41 chromium, and
  the remainder oxygen.

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Practice

• A compound is analyzed and found to contain
  36.70 potassium, 33.27 chlorine, and 30.03
  oxygen. What is the empirical formula of the
  compound?

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Practice a classic AP question

• Analysis of 20.0g of a compound containing only
  calcium and bromine indicates the 4.00g of
  calcium are present. What is the empirical
  formula of the compound formed?

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Molecular Formulas

• By now you should know that a molecular formula
  is a group of elemental symbols, and possibly
  subscript numbers, which represent the
  composition of a molecule.  The molecular formula
  shows you how many of each atom can be found in a
  certain molecule. 

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Molecular Formula

• - represent the actual numbers of atoms of the
  different elements in one molecule of a compound.
  An example is Ca(NO3)2

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Molecular Formula

• represent the actual numbers of atoms of the
  different elements in one molecule of a compound.
  An example is Ca(NO3)2

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Determine Molecular Formula

• The molecular formula of a compound is always an
  integer multiple (e.g., 1, 2, 3,...) of the
  empirical formula.
• If the empirical formula of a compound is known,
  the molecular formula can be determined by the
  experimental determination of the molecular
  weight of the compound.

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Determine Molecular Formula

• There are two steps to determining the molecular
  formula once the molecular weight of a compound
  has been determined experimentally.

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Determine Molecular Formula

• The first step is to divide the experimentally
  determined molecular weight of the compound by
  the molecular weight of the empirical formula in
  order to determine the integer multiple that
  represents the number of empirical formula units
  in the molecular formula.
• In the second step, the molecular formula is
  obtained by multiplying the subscripts of the
  empirical formula by the integral multiple of
  empirical formula units.

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• A sample of a compound with a formula mass of
  34.00 amu is found to consist of 0.44g H and
  6.92g O. Find its molecular formula.

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• Whats the empirical formula of a molecule
  containing 18.7 lithium, 16.3 carbon, and 65.0
  oxygen? If the molar mass of the compound in is
  73.8 grams/mole, whats the molecular formula?

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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

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Elemental Analyses

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

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Stoichiometric Calculations

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

9/29
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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
Stopped 9/23
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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• If 3.30 moles of Fe2O3 completely reacts
  according to the following single displacement
  reaction, determine the number of moles of
  aluminum needed, the moles of iron formed, the
  moles of aluminum oxide formed.
• Fe2O3 Al ? Fe Al2O3

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• Fe2O3 Al ? Fe Al2O3
• How many grams of Al are needed to completely
  react with 135g of Fe2O3 ?

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• When copper metal is added to silver nitrate in
  solution, silver metal and copper(II) nitrate are
  produced. What mass of silver is produced from
  100 g of Cu?

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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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Limiting Reagent

• A 2.00 g sample of ammonia is mixed with 4.00 g
  of oxygen.  Which is the limiting reactant and
  how much excess reactant remains after the
  reaction has stopped?

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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