Chapter 3: Stoichiometry Emily Scheerer

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Chapter 3 Stoichiometry Emily Scheerer

• Section 3.1Chemical Equations
• Section 3.2Patterns of Chemical Reactivity
• Section 3.3Atomic and Molecular Weights
• Section 3.4The Mole
• Section 3.5Empirical Formulas from Analyses
• Section 3.6Quantitative Information from
  Balanced Equations
• Section 3.7 Limiting Reactants

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3.1 Chemical Equations

• Law of conservation of mass total mass of all
  substances after a chemical reaction is the same
  as the total mass before the reaction.

Balancing Chemical Equations (a few tips) ?
above the arrow means the addition of heat Never
change subscripts, only coefficients Use the
symbols (g), (l), (s), (aq)
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3.2 Patterns of Chemical Reactivity

• All alkali metals react with water to form
  hydrogen and hydroxide compounds. (M represents
  an alkali metal)
• 2M(s) 2H2O(l) ? 2MOH(aq) H2(g)
• Combustion reactions are rapid reactions that
  produce flame, generally involving O2.
• Example C3H85O2? 3CO24H2O
• In combination reactions, two or more substances
  form one product.
• Example 2Mg(s) O2(g)?2MgO(s)
• In decomposition reactions one substance produces
  2 or more other substances.
• Example CaCO3(s) ? CaO(s) CO2

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3.3 Atomic and Molecular Weights

• We can measure the masses of individual atoms
  with the atomic mass unit, amu. The amu is
  defined by assigning the mass of exactly 12 amu
  to the 12C isotope of carbon.
• 1 amu 1.66054 x 10-24g and 1g 6.02214 x
  1023amu
• Most elements are a mixture of isotopes. The
  average atomic mass of each element (expressed in
  amus) is known as its atomic weight. The atomic
  weights are in the periodic table.

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3.3 Atomic and Molecular Weights

• The formula weight of a substance is the sum of
  the atomic weights of each atom in its chemical
  formula.
• If the chemical formula is merely the chemical
  symbol of an element, such as Na, then the
  formula weight equals the atomic weight of the
  element. If the chemical formula is that of a
  molecule, then the formula weight is also called
  the molecular weight.
• Percentage Composition from Formulas
• Occasionally we must calculate the percentage
  composition of a compound (that is, the
  percentage by mass contributed by each element in
  the substance). Heres how

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3.4 The Mole

• Mole - The amount of matter that contains as
  many objects as as the number of atoms in 12g of
  12C.
• This number is 6.02214199 x 1023 and is called
  Avogadros number. (use 6. 02 x 1023)
• Molar Mass The mass in grams of 1 mole is
  called the molar mass (in grams) and is
  numerically equal to its formula weight (in amu)

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3.4 The Mole
Example What is the mass in grams of 1.000 mol
of glucose, C6H12O6? Solution Analyze We are
given the chemical formula for glucose and asked
to determine its molar mass. Plan The molar mass
of a substance is found by adding the atomic
weights of its component atoms. Solve
Because glucose has a formula weight of
180.0 amu, 1 mol of this substance has a mass of
180.0 g. In other words, C6H12O6? has a molar
mass of 180.0
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3.4 The Mole

• Calculate the number of H atoms in 0.350 mol of
  C6H12O6
• Solution
• Analyze We are given both the amount of the
  substance (0.350 mol) and its chemical formula
  (C6H12O6). The unknown is the number of H atoms.
• Plan Avogadro's number provides the conversion
  factor between the number of moles of and the
  number of molecules of C6H12O6. Once we know the
  number of molecules of we can use the chemical
  formula, which tells us that each molecule of
  contains 12 H atoms. Thus, we convert moles of
  to molecules of and then determine the number
  of atoms of H from the number of molecules of
  C6H12O6
• Solve

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Helpful Mole Diagram
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3.4 The Mole

• Interconverting Masses, Moles, and Numbers of
  Particles (also called dimensional analysis)
• Calculate the number of moles of glucose(C6H12O6)
  in 5.380 g of C6H12O6.
• Solution

Analyze We are given the number of grams of
C6H12O6 and asked to calculate the number of
moles. Plan The molar mass of a substance
provides the conversion factor for converting
grams to moles. The molar mass of C6H12O6 is
180.0 Solve Using 1 mol C6H12O6 to write the
appropriate conversion factor, we have
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3.5 Empirical Formulas

• The empirical formula tells the relative number
  of atoms of each element it contains. The general
  procedure for finding empirical formula is seen
  below

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• Remember that the formula from percent
  composition is always empirical.
• To transfer to molecular, you need to know the
  molecular weight.
• The subscripts in the molecular formula are
  always whole number multiples of the subscripts
  in the empirical formula.

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• 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?
• Solution
• Analyze We are given the empirical formula and
  molecular weight of mesitylene and asked to
  determine its molecular formula.
• Plan The subscripts in a molecular formula are
  whole-number multiples of the subscripts in its
  empirical formula. To find the appropriate
  multiple, we must compare the molecular weight
  with the formula weight of the empirical formula.
• Solve First we calculate the formula weight of
  the empirical formula, C3H4

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3.5 Empirical Formulas

• Next we divide the molecular weight by the
  empirical formula weight to obtain the factor
  used to multiply the subscripts in C3H4
• Only whole-number ratios make physical sense
  because we must be dealing with whole atoms. We
  therefore multiply each subscript in the
  empirical formula by 3 to give the molecular
  formula C9H12

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3.6 Quantitative information from balanced
equations

• The coefficients in a balanced equation can be
  interpreted as the relative number of molecules
  and as the relative number of moles involved in
  the reaction.
• consider the combustion of butane (C4H10), the
  fuel in disposable cigarette lighters
• the conversion sequence is
• combined in a single sequence of factors

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3.6 Quantitative information from balanced
equations

• Outline of the procedure used to calculate the
  number of grams of a reactant consumed or of a
  product formed in a reaction, starting with the
  number of grams of one of the other reactants or
  products.

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3.7 Limiting Reactants

• The reactant that is completely consumed in in a
  reaction is called the limiting reactant.
• Real World Example
• You have 10 pieces of bread and 7 pieces of
  cheese. You want to make grilled cheese
  sandwiches with two pieces of bread and 1 piece
  of cheese, which will run out first?
• The bread will run out first, because even though
  there is less cheese, you need more bread for
  each sandwich, making bread the limiting reactant.

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3.7 Limiting Reactants

• The quantity of the product that is calculated to
  form when all of the limiting reactant has
  reacted is the theoretical yield. The amount of
  product actually obtained is the actual yield.
  The percent yield relates the actual yield to the
  theoretical.

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Limiting Reactant Example

• Adipic acid, H2C6H8O4 is used to produce nylon.
  It 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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3.7 Limiting Reactants

• Solution
• Plan (a) The theoretical yield is the calculated
  quantity of adipic acid formed in the reaction.
  We carry out the following conversions g C6H12 ?
  mol C6H12 ? mol H2C6H8O4 ? g H2C6H8O4. (b) Once
  we have calculated the theoretical yield, we use
  Equation 3.13 to calculate the percent yield.
• Solve (a)
• (b)

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Key things to remember

• How to balance equations
• common chemical reactions (like decomposition,
  combustion, combination, etc)
• amus, molecular weight, percent composition
• Avogadros number and how to use it
• dimensional analysis
• Empirical formulas how to get them, how to
  get molecular formulas from them, how to get
  theoretical yield from them.
• Limiting reactants and percent yield formula