Honors Chemistry

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Honors Chemistry

• Chapter 3 Mass relationships in Chemical
  Reactions

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3.1 Atomic Mass

• Too tiny to mass individual atoms
• Unit needed to compare masses of atoms
• Atomic Mass Unit
• Early amu had several standards
• Now defined as 1/12 the mass of carbon-12
• Labeled u to signify unified amu
• Average Atomic Mass
• Weighted average of all isotopes of element
• Used for most calculations

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3.2 Avogadros Number

• Amedeo Avogadro
• Looking for a convenient relationship between
  grams and number of particles
• Varies by atomic mass
• 1 atomic mass in grams 6.022 x 1023 atoms
• Avogadros number NA 6.022 x 1023
• The Mole
• 1 mol NA particles
• Unit of counting particles, similar to the unit
  dozen

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3.3 Molecular Mass

• Sum of all atoms in a molecule
• E.g., H2O 2 (1.008) 1 (16.00) 18.02 u
• Try this Find M of NaHCO3
• Mole Conversions
• 1 mol NA particles
• 1 mol (molecular mass) g
• mass ?? moles ?? particles M
  NA
• Solve using dimensional analysis

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3.3 Molecular Mass

• How many grams are in 0.75 mol NH3?
• M 1 (14.01) 3 (1.008) 17.03 g/mol
• Therefore, 1 mol 17.03 g
• 0.75 mol 17.03 g ------------ x
  ----------- 13 g 1 1 mol
• Try this
• How many moles are in 15.0 g N2O?

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3.3 Molecular Mass

• How many molecules are in 125 g CH3COOH?
• M 2 (12.01) 4 (1.008) 2(16.00)
• M 60.05 u
• 125 g 1 mol 6.022 x 1023
  molecule-------- x ---------- x
  ------------------------- 1 60.05 g
  1 mol
• 1.25 x 1024 molecules

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3.4 Mass Spectrometer

• Device that separates particles by mass
• helped verify the existence of isotopes
• Used to identify unknown substances
• Not terribly important to us!

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3.5 Percent Composition

• Percent by mass of each element in the compound
• Find composition of CaCl2
• 1 Ca 40.082 Cl 70.90
• 110.98 u
• Ca 40.08 / 110.98 36.11
• Cl 70.90 / 110.98 63.89

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3.6 Empirical Formula

• Based on laboratory data
• Simplest whole number ratio of elements
• E.g., glucose is C6H12O6
• Empirical formula is CH2O
• Subscripts represent ratio by mole, not mass

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3.6 Empirical Formula

• A compound contains 0.900 g Ca and 1.60 g Cl.
  Find the empirical formula.
• Ca 0.900 g 1 mol ---------- x
  ---------- 0.0225 mol 1
  40.08 g
• Cl 1.60 g 1 mol -------- x
  ---------- 0.0451 mol 1
  35.45 g
• Divide by lowest value to get whole numbers
• 0.0451 / 0.0225 2, so formula is CaCl2

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3.6 Empirical Formula

• Try this
• Find the empirical formula for a compound that is
  composed of 53.73 Fe and 46.27 S.
• Remember that it is percent by mass, so those
  percents can be treated as grams!
• There is a trick at the end of this one.

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3.6 Molecular Formula

• Need to know the molecular mass
• Divide molecular mass by empirical mass to find
  how much bigger the molecular formula is
• Multiply empirical formula by that number
• Try this
• A compound is 40.0 C, 6.67 H, 53.3 O.
• Molecular mass 180.1 u.
• Find molecular formula.

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3.7 Chemical Equations

• Chemical reaction
• Process in which a substance is changed into one
  or more new substances
• Chemical equation
• Arrangement of chemical symbols to represent what
  happens during a chemical reaction
• Example reaction
• Carbon reacts with oxygen to yield carbon dioxide
• C O2 ? CO2

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3.7 Chemical Equations

• States of substances
• (s) solid
• (l) liquid
• (g) gas
• (aq) aqueous solution
• HgO (s) ? Hg (l) O2 (g)
• Note that the number of oxygens is not conserved.
  We have to fix that!

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3.7 Balancing Equations

• Consider H2 O2 ? H2O
• Weve lost an oxygen!
• Can we change it to H2 O2 ? H2O2 ?
• Can we add an O H2 O2 ? H2O O ?
• A better solution H2 O2 ? 2 H2O
• Now fix the hydrogens 2 H2 O2 ? 2 H2O
• Two H2 molecules react with one O2 molecule to
  form two water molecules

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3.7 Balancing Equations

• Try these
• NaClO3 ? NaCl O2
• 2 NaClO3 ? 2 NaCl 3 O2
• CO2 H2O ? C6H12O6 O2
• 6 CO2 6 H2O ? C6H12O6 6 O2
• CaCl2 Na3PO4 ? Ca3(PO4)2 NaCl
• 3 CaCl2 2 Na3PO4 ?Ca3(PO4)2 6 NaCl

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3.8 Stoichiometry

• Quantitative study of chemical reactions
• Realize that reaction coefficients are really
  moles!
• 15.0 g oxygen react with excess hydrogen. How
  much water is produced?
• 2 H2 O2 ? 2 H2O
• 15.0g O2 1 mol O2 2 mol H2O 18.0 g
  H2O------------ x ------------- x --------------
  x ------------- 1 32.0 g O2 1
  mol O2 1 mol H2O
• 16.9 g H2O produced

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3.9 Limiting Reagents

• When reactants are not present in the exact
  stoichiometric ratio, one reactant will be used
  up before the others
• Limiting reagent reactant used up first
• Excess reagent reactant that is present in
  larger quantity than needed
• Convert both to moles to find out which is the
  limiting reagent

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3.9 Limiting Reagents

• Try this
• 5.0 g hydrogen react with 20.0 g nitrogen. How
  much ammonia is produced?
• N2 3 H2 ? 2 NH3
• Find how many mol NH3 can be made from the H2
• Find how many mol NH3 can be made from the N2
• Use the lesser amount (limiting reagent)
• The other substance is present in excess

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3.10 Reaction Yield

• Theoretical yield
• Amount you should have gotten from reaction
• Result of mol calculation
• Actual Yield
• What you actually got from the reaction
• Percent Yield
• Actual / theoretical x 100