Stoichiometry

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Stoichiometry

• AP Chemistry
• Mr. Martin

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Topics

• Law of Conservation of Matter
• Balancing Chem Eq
• Mass Relationships in rxns
• Limiting Reagents
• Theoretical, Act and yld
• Empirical and Molecular formulas

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A. Balancing Equations

• Law of Conservation of Mass matter is neither
  created or destroyed during chem rxns some
  matter is converted to energy why we must bal
  equations
• Stoichiometry the relationship among the
  substances in chem rxns.

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Chemical Equations

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Balancing Equations

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Steps to Balance

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Practice Problems

• Cu O2 ----gt Cu2O
• CaCl2 AgNO3 ----gt AgCl Ca(NO3)2
• H2 O2 ----gt H2O
• Mg P4 ---gt Mg3P2
• H2SO4 NaOH ----gt H2O Na2SO4

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Solutions

• 4Cu O2 ----gt 2Cu2O
• CaCl2 2AgNO3 ----gt 2AgCl Ca(NO3)2
• 2H2 O2 ----gt 2H2O
• 6Mg P4 ---gt 2Mg3P2
• H2SO4 2NaOH ----gt 2H2O Na2SO4

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Additional Practice

• Fe2O3 CO -gt Fe3O4 CO2
• Fe3O4 CO -gt FeO CO2
• C12H22O11(s) O2(g) -gt CO2(g) H2O
• Fe(s) O2(g) -gt Fe2O3
• Ca(s) H2O(l) -gt Ca(OH)2(aq) H2(g)

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B. Atomic and Molecular Weights

• Atomic Mass Scale uses atomic mass units (AMUS)
  relative mass scale originally based on hydrogen
  1 amu
• Now 1 amu 1/12 of the carbon-12 isotope
• Average atomic masses (atomic weight)
• elements are a mixture of isotopes and
  therefore the mass of an element is a weighted
  average of the naturally occurring isotopes.

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Calculation of Atomic Mass

• AM sum (abundance) x (isotope mass)
• of all the naturally occurring
• isotopes
• Ex Ne-20 19.992 amu 90 abund
• Ne-22 21.991 amu 10 abund
• (0.9 x 19.992) (0.1 x 21.991)
• 20.192 amus atomic mass

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Formula Weight

• Sum of the average atomic masses of each atom in
  its chem ormula
• EX C12H22O11
• C (12) 12 144 amus
• H (22) 1 22 amus
• O (11) 16 176 amus
• 342 amus formula weight

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Percent Composition by Weight

• What is the comp of Methane CH4?

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The Mole

• A number 6.02 x 1023 like a dozen 12 , a
  pair 2, or a gross 144
• A mol is the amount of matter that contains as
  many objects (atoms, molecules or formula units)
  as the of atoms in exactly 12g of the carbon-12
  isotope.

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The Mole and Molar Mass

• One mole of Carbon 6.02x1023 atoms of carbon
• One mole of H2O is equal to 6.02 x 1023 H2O
  molecules.
• One mole of HCl is equal to 6.02 x 1023 formula
  units of HCl.

• Molar Mass 12.01 g of Carbon
• Molecular Weight H2O
• H (2) 1 2
• 0 (1) 16 16
• 18 g/mol
• Formula Weight HCl 36.5g/mol

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Conversion Problems

• Gram?? moles ?? atoms, fus,
• molecules
• How many moles are in 200g of sulfuric acid. (
  2.04 mole of H2SO4)

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Conversion Problems

• How many formula units are in 10g of sulfuric
  acid? ( 6.14 x 1022 fus)

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Empirical and Molecular Formulas

• Empirical Formulas gives the lowest whole
  ratio of atoms (moles) in a compound.
• EFs are derived from exp info most notably
  comp.

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Example Problems

• An analysis of sodium dichromate gives the
  following mass percentages, 17.5 Na, 39.7 Cr,
  and 42.8 O. Calc the emp. formula?

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Example Problem Mol. Formula

• Molecular formula is a multiple of the emp.
  formula. n exp mm/ef weight
• The composition of acetaldehyde is 54 C, 9.2
  H, and 36.3 O. Its molar mass is 88 amus. What
  is its molecular formula.

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Combustion Analysis

• A method of analysis where a sample cont. C,H, O
  undergoes complete combustion. The gas stream
  goes through two tubes one absorbs CO2, the other
  H2O.

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Combustion Analysis cont.

• The change of mass in the tubes gives the mass of
  CO2, and H2O.
• A 4.24mg Acedic Acid sample is completely burned
  yielding 6.24mg of CO2 and 2.54mg of H2O. What
  is the mass of each element?

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Quantitive Information from Bal. Eq

• grms ? ? mol ? ? mol ? ? grms
• Ex. A (moles to moles) How many moles of H2O will
  be produced if 17.5 moles of H2 combine with
  excess O2? (17.5)

?
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Example Problems Cont.

• How many grams of KClO3 must decompose to produce
  185.0 moles of O2? (1.51 x 104 g)

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Example Prob Cont.

• How many grams of silver chloride can be produced
  from the reaction of 17.5 g of silver nitrate
  with excess sodium chloride solution? (14.4g)

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Example Problems

• How many mg of aluminum nitrate would be required
  to completely react with 1.75 kg of barium
  hydroxide? (1.45 x 106mg)

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Limiting Reagent Percent Yld

• The reactant that is entirely consumed when a
  reaction goes to completion is the limiting
  reagent.
• The reactant that is not completely consumed is
  said to be in excess.
• The amount of product that is produced is
  determined by the limiting reactant.

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Ex. Limiting Reagent Prob.

• In a lab test reaction 20.0g of CH3CHO and 10.0g
  of O2 were put in a reaction vessel. (a) How many
  grams of acedic acid can be produced (b) How many
  grams of excess reactant remain after the
  reaction is complete.

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Theoretical and Percent Yld.

• The max amount of product that can be obtained is
  the theoretical yield. ( based on stoichiometric
  calc.)
• The actual yield is the mass obtained by the
  actual reaction.
• yield actual yld/theorical yld x 100

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Problem

• The amount of acedic acid produced from the
  previous reaction was 23.8g. What is the percent
  yield of the reaction. (87.2)