Stoichiometry

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Stoichiometry Ch. 3

• AP/IB Chemistry
• Chandler High School
• Mr. Root

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Stoichiometry

• A. Stoichiometry - the study of quantities of
  substances consumed and produced in chemical
  reactions.
• 1. Atomic mass - the average mass of an atom of
  an element in atomic mass units (amu)
• a. Atomic masses are based on 12C ("carbon
  twelve"), which is assigned a value of exactly
  12 atomic mass units.
• b. The atomic masses of other elements are
  determined by comparison to 12C.

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1. Atomic Mass

• c. Mass spectrometer - currently best method for
  determination of atomic masses of atoms
• - Procedure
• - atoms or molecules are placed in a beam of
  high-speed electrons (knocks electrons off the
  atoms or molecules giving them a positive charge
  - cations)
• - an electric field is then applied accelerating
  these cations which causes each cation to create
  a magnetic field
• - cations pass through an applied magnetic field
• - cations with the least mass are deflected more
  than heavier cations and masses are determined by
  comparing amounts of deflection
• e.g. If , in a mass spectrometer, 13C is found to
  have a mass 1.0836 times that of 12C, then the
  atomic mass of 13C 1.0836 (12 amu) 13.003 amu

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1. Atomic Mass

• The mass spectrometer can also be used to
  determine the isotopic composition of an element.
  When a sample of an element is placed in a mass
  spectrometer, a mass spectrum can be obtained
  which indicates the relative amounts of the
  various isotopes present in the sample.

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1. Atomic Mass

• The average mass of an atom (or mole) of an
  element is the sum of the fractions of each
  isotope times their mass.
• e.g. Naturally occurring carbon is composed of
  two isotopes, 12C (98.89) and 13C (1.11). The
  average atomic mass for an atom of carbon (
  .9889 x 12 amu) ( .0111 x 13.003) 12.01 amu.

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2. The Mole

• 2. The Mole (abbrev. mol) - SI unit for the
  amount of a substance and is equal to the number
  of atoms in exactly 12 grams of pure 12C.
• - Mass spectrometry has determined this number
  to be 6.02214 x 1023 (Avogadro's number)
• (We will use 4 sig.fig 6.022 x 1023).

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2. The Mole

• a. Calculations involving the mole
• - calculate moles from mass and vice versa
• - calculate mass from atoms and vice versa

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3. Molar Mass

• 3. Molar mass - the mass in grams of one mole of
  a substance( equals the atomic mass in grams)
• - be able to calculate the molar mass of atoms,
  ions or molecules or numbers of atoms, ions or
  molecules from moles or vice versa

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4. Mass Percent Composition

• 4. Mass percent composition - the percent
  composition of a substance by mass
• Sample Problem What is the percent composition
  of glucose (C6H12O6)

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5. Determining the Formula of a Compound

• Empirical formula - lowest whole number ratio of
  the atoms in a compound
• - Determination from percent composition data
• Step 1 Determine mass composition (if not
  given) from percent composition (assume 100 g
  and then percents convert to grams)
• Step 2 Determine moles of each element in the
  compound.
• Step 3 Determine the lowest whole number
  ratio of each element in the compound
  (empirical formula)

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Empirical Formula
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Empirical Formula

• - Determination of empirical formula from
  combustion data
• - Use stoichiometry to determine the grams of
  C and H from the balanced chemical equation (if
  the combusted substance contains oxygen,
  determine the mass of oxygen from the mass of
  the original compound)

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Empirical Formula

• Sample problem A substance contains only
  carbon, hydrogen and oxygen. When 0.510 g of this
  substance is burned, 1.122 g of CO2 and 0.612 g
  of H2O are produced. What is the empirical
  formula of this compound?
• Step 1 Determine moles of C and H
• mol C 1.122 g CO2 (1 mol CO2/44.01g)
  0.02549 mol CO2 (1mol C/1 mol CO2) .02549 mol C
• mol H 0.612 g H2O (1 mol H2O/18.02g) .0354
  mol H2O(2 mol H/1mol H2O) .0708 mol H
• Step 2 Determine mass of C, H and O
• Mass C .02549 mol C (12.01 g/mol C) 0.3061
  g C
• Mass H .0708 mol H(1.01g/mol H) .0715 g H
• Mass O total mass of cpd - (mass C H)
  0.510 - (0.3061 .0715) .0784 g O
• Step 3 - Determine empirical formula as in above
  problem

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Molecular Formula

• b. Molecular formulas - the actual ratio of the
  atoms of elements in a compound and are whole
  number multiples of their empirical formulas.
• For example if the molecular formula of a
  compound was N2O4, the empirical formula would be
  NO2. In this case the molecular formula is 2
  times the empirical formula ( NO2 x 2 N2O4 ).
  The mass of the molecular formula is necessarily
  also 2 times that of the mass of the empirical
  formula. This relationship can then be used to
  determine the molecular formula of a compound if
  the empirical formula and molecular mass of the
  compound are known. Take a look at a sample
  problem.

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Molecular Formula
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6. Chemical Equations

• 6. Chemical Equations - represent what happens in
  a chemical reaction
• a. reactant ? products (read "Reactants yield
  products")
• b. conservation of atoms (mass) - atoms are
  neither created nor destroyed in chemical
  reactions, they are recombined to form different
  substances
• - mass is neither created nor destroyed chemical
  reactions (as opposed to nuclear reactions)
• - chemical reactions must therefore be balanced
  - have same kinds and numbers of atoms on both
  sides of the yields sign (?)

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6. Chemical Equations

• c. The physical states of the substances involved
  in a chemical reaction are represented by symbols
  
• (aq) - aqueous (dissolved in water)
• (g) - gas
• (l) - liquid
• (s) - solid

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Balancing Chemical Equations

• 7. Balancing chemical equations - usually by
  inspection
• - it helps to do the most complex substance
  first
• - never change the correct formulas of reactants
  or products (change the amounts of the substances
  represented, do not change the substances)
• - balance by placing coefficients in front of
  the reactant or product species

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8. Stoichiometric Calculations

• 8. Stoichiometric Calculations Amounts of
  Reactants and Products
• a. The following diagram represents the basic
  conversions used in stoichiometry

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8. Stoichiometric Calculations

• Calculations between moles and mass and moles
  and numbers of particles have already been
  covered (moles and volume conversion will be
  covered later). The only new thing here is the
  use of the mole ratio in a balanced equation to
  convert moles of one substance (known) to moles
  of another (unknown)

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• Sample From the balanced equation for the
  Haber process (N2 3H2 ? 2 NH3) we see that
  one mole of nitrogen is needed for every three
  moles of hydrogen present. How many moles of
  nitrogen are needed if 1.2 moles of hydrogen are
  present?

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Limiting Reactant

• 9. Calculations Involving a Limiting Reactant.
• a. Stoichiometric quantities - the exact amounts
  of reactants needed so that no amount of any
  reactant will be left unused
• b. Limiting Reactant - a reactant that there is
  proportionally less of so that it is used up
  first and limits the amount of product that can
  be produced
• - you must first determine which reactant is
  the limiting reactant when doing stoichiometry
  problems involving limiting reactants
• - to determine which reactant is limiting
  arbitrarily choose one reactant and use the mole
  ratio of reactants to see if you have an excess
  or not of the other reactant

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Limiting Reactant

• - e.g. Lets say you have 100.0 g of hydrogen and
  200.0 g of nitrogen and you are going to combine
  them to form ammonia according to the Haber
  Process
• N2 3H2 ? 2NH3
• You can start with either the 100.0 g of H2, or
  the 200.0 g of N2 and convert to the other
  reactant. Starting with grams of N2 and
  converting to moles of H2
• Converting grams of H2 to moles H2

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Limiting Reactant

• Since the 200.0 g of N2 equates to a smaller
  quantity of H2 than what is available, N2 is the
  limiting reactant. Or, to put it another way,
  since the 100.0g of H2 present is much more than
  needed for the 200.0 g of N2, H2 is the excess
  reactant.

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Limiting Reactant

• c. Excess reactant -opposite of limiting reactant
  - some will remain at the completion of a
  reaction
• d. Theoretical yield - the amount of product a
  reaction should yield if 100 of the limiting
  reactant is converted into the product
• e. Actual yield - the amount of product that is
  actually produced when the reaction takes place
• f. Percent yield - a measure of the efficiency of
  a chemical reaction or process