Welcome to the MOLE

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Welcome to the MOLE
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What is a mole?

• This is not such a bad
• mole, but not what we
• need to discuss. . .

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This little stinker is just plain mean and ugly.
. .
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• and the Mole People are in the dark and clueless.
  . . . .

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We are discussing the Mole used in Chemistry

• Avogadros Number (NA)
• or 6.02214179 x 10 23
• shortened to
• 6.022 x 1023
• This amount 1 mole

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The mole is a way to describe the number of
something without writing a huge number.

• It is similar to common terms like dozen, gross
  or even p in geometry
• 1 mole of anything atoms, molecules,
  cockroaches or even galaxies will number 6.022 x
  1023
• We need a number like this since atoms and
  molecules are extremely small and so many take up
  such a small space

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Key Equations KNOW THESE!

• Divide by Molar Mass x by NA
• Grams Number
• (mass) of Moles of of
• Substance Substance Particles
• x by Molar Mass Divide by NA
• Molar Mass Atomic Weight in Grams per Mole
• (g/mol)

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• Molar Mass is the mass (g) of 1 mole of an
  element
• For example Na is 22.989 amu which is 22.989 g,
  and 1 mole of Na 22.989 g
• CO2 is 1 Carbon at 12.011 g and 2 Oxygens at
  15.999 each therefore the molar mass of CO2
  S of 1 C 2 O 44 g
• Mass (m) mols x g
• mols m / g

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• Mols m / Molar Mass
• Composition by Mass m element x 100
• 
  m compound
• Mols Concentration (Molar) x Volume (L)
• Volume mols / Concentration

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Heres a trick to find the needed equation,
just cover up the wanted result and what is left
is the equation!

• Mass
  Mols
• Atomic X Mols Concentration X
  Volume
• Mass

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The Mole Concept

• Atomic Mass (or atomic weight) is the mass of the
  element in amu (µ)
• It is the number under the elemental symbol
• Simply make this number into grams (g)
• This represents the mass (m) of 1 mole of that
  element and/or the m of 6.022 x 1023 atoms of
  that substance!
• 1 mole of any gas 22.4 L

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• Mole (mol) is the of atoms, ions, molecules
  that is equal to NA
• Molar Mass (M) this is the m in g of 1 mol of a
  substance (g/mol)
• Example Manganese 54.94 µ, thus its
• M 54.94 g/mol
• and 54.94 g of Mn will contain 6.022 x 1023 atoms
• and this is equal to 1 mol of Mn

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• Mass to Mole Calculations
• Remember each
• element has a different
• amu and thus, 1 mol of
• each will differ in mass
• The Mass of a Mole
• Uses the C 12 isotope as its standard
• H µ of 1 or 1/12 of 1 atom of C 12
• He has µ of 4 or 4/12 (1/3) of a C 12 atom
• Remember atomic masses use isotopes and their
  abundance in nature to calculate and the closer
  to a whole number the fewer the isotopes

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• I. Molar Mass of Substance Grams Substance
• 
  1 mol Substance
• Therefore
• 1. Mols of A grams of A given x 1 mol A
• 
  gram A
• 2. Mass of B Mols of B given x gram B
• 
  1 mol B

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Examples

• 3 mol Mn ? Grams
• 3 mol Mn X 54.9 g Mn 165 g Mn
• 1 mol Mn
• 25 g Au ? Mols
• 25 g Au x 1 mol Au 25
  0.127 mol Au
• 196.97 g Au 196.97
• 0.127 mols Au ? Atoms
• 0.127 mol Au x 6.022 x 1023 7.65 x
  1022 atoms Au
• 1 mol Au

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II. Moles to Mass

• Mols (given) x grams mass
• 1 mol
• Example
• 2.5 mol of (C3H5)2S has what mass?
• M 1 mol S 32.07 g
• 6 mol C 6 x 12.01 72
  g
• 10 mol H 10 x 1 10 g
• S
  114.07 g/mol
• 2.5 mol x 114.07 g 286 g
• 1 mol

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We just used the bottom left of the diagram!

• Divide by Molar Mass x by NA
• Grams Number
• (mass) of Moles of of
• Substance Substance Particles
• x by Molar Mass Divide by NA
• Molar Mass Atomic Weight in Grams per
  Mole
• (g/mol)

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III. Mass to Moles with Compounds

• Example
• m of Ca(OH)2 325 g (rounded off)
• M ?
• mols ?
• M 1 mol Ca 40.08 g
• 2 mol O 2 x 16 32 g
• 2 mol H 2 x 1 2 g
• S
  74.096 g/mol
• Given m of 325 g Ca(OH)2 x 1 mol Ca(OH)2
  4.3 mol
• 
  M of 74.096 g

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IV. Mass (g) to Particles

• Mols x NA Particles
• Example
• m 35.6 g of AlCl3
• What is the number of Al3 and Cl- ions?
• M Al 26.981 g/mol
• Cl 35.452 g/mol x 3 106.356
• S
  133.337
• Mols Al m given _____ mols x
  NA _____ Al ions
• 26.981 g/mol
• Mols Cl m given _____ mols
  x NA _____ Cl ions
• 106.356 g/mol
• 
  Continued

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• And 35.6 g AlCl3
  0.267 mol
• 133.337 g/mol AlCl3
• 0.267 mol AlCl3 x (6.022 x 1023)
• 1.6 x 10 23
  molecules

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V. Percent Composition of Compounds

• Mass Element (m) x 100 by mass
• Mass Cmpd (M)
• Example
• H2O what percent is H and what percent is O?
• H 2 x 1 (the molar mass of H) 2 x
  100 11.2
• (the molar mass of H2O)
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• Thus, all compounds equal 100, so 100 11.2
  88.8 for O

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• What is the of C and O in CO2?
• g C x 100 12.01 C x 100
  27.29
• total g CO2 44.01 g CO2
• 32 g O x 100 72.71
• 44.01 g CO2

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Example

• H3PO4 (aq) (Phosphoric Acid)
• H 3 g H x 100 3
• M 98 g
• P 31 g P x 100 32
• 98 g
• O 64 g O x 100 65
• 98 g
• m Compound H (3 x 1) P (1 x 31) O (4 x 16)
  98 g

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VI. Mole Ratios

• Given Vitamin C (ascorbic acid) with the
  following percentages, determine formula
• 40.92 C
• 4.58 H
• 54.5 O
• Set up with unknown moles (n)
• nC 40.92 g C / 12.01 g C 3.4 mol C
• nH 4.58 g H / 1.00 g H 4.5 mol H
• nO 54.5 g O / 16 g O 3.4 mol O
• This is the Mole
  Ratio
• 
  

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• Set Mole Ratio values as subscripts
• Divide each by the lowest value
• C 3.4 / 3.4 H 4.5 / 3.4 O 3.4 / 3.4
• C 1 H 1.33 O 1
• The 1.33 on H needs to be ?d into integer
• Do this by multiplying until closest to a whole
  number
• 1.33 x 2 2.66
• 1.33 x 3 3.99 which can be rounded off to 4
• So the magic number is 3 must multiply all
  subscripts by 3
• Result is C1 x 3H1.33 x 3O1 x 3 or C3H4O3!

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What is the molecular formula that has 92.2 C
and 7.8H. The molar mass is 52.1.

• First assume a 100 g sample of the substance
• The elements percentages are assumed to be
  masses (g)
• Determine the moles of elements in compound
• 92.2 g C x 1 mol C 7.68 mol C
• 12.01 g C
• 7.8 g H x 1 mol H 7.72 mol H
• 1.01 g H
• Divide all mols by the lowest value
• 7.68 C 1 mol C 7.72 H 1.01
  mol H
• 7.68 7.68
• Continued

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• The Empirical Formula is C1H1 (this is the basic
  form)
• To get the Molecular Formula
• Molar mass of the Empirical Formula is 12.01
  1.01 13.02 g/mol
• Molar mass of unknown is 52.1 g/mol
• So
• Molar Mass Compound Whole to
  multiply
• Molar Mass Emp. Formula subscripts to get
• 
  molecular formula
• 52.1 g/mol 4
• 13.02 g/mol thus, (CH) x 4 C4H4

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Once again, a molecular formula calculation

• Given 38.7 C, 9.7 H, and 51.6 O with a
  molecular formula mass of 62.0 g. What is the
  true molecular formula?
• First - find the empirical formula (this is
  CH3O)
• Find the formula mass C 1 x 12 12
• H
  3 x 1.01 3.03
• O
  1 x 16 16.0
• 
  S 31.0

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• Divide the molecular mass by the empirical
  formula mass
• 62 g (given) / 31 g (mass of emp. form.) 2
• Multiply each subscript by (n) or 2 in this
  example. . .
• Thus, the molecular formula
• (CH3O)(n) ? (CH3O)(2) ? C2H6O2

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VII. Hydrates

• Hydrates are
• substances that include
• H2O in their
• formulas, but are
• not wet!
• Hydration adding H2O
• Dehydration removing it
• Anhydrous no H2O present

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• Methane Hydrate is found on the oceans floor
• The methane will burn but the water in it keeps
  the skin from burning!

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• The methane molecule (CH4) is in a cage of water
  molecules
• There is 1 mole CH4 per 5.75 mols H2O
• It is found at depths of 300 meters or more
• There is an estimated 1,300 trillion cubic feet
  of methane hydrate in the oceans
• However the problem is that methane is one of
  the major greenhouse gases which contributes to
  global warming so more study on retrieval and
  use is needed

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Example

• Barium Chloride Hydrate
• Mass 5 g. How many H2O per molecule?
• BaCl2____H2O
• The sample is heated and the result is 4.26 g
  anhydrous BaCl2
• The difference between the 5 g hydrate and the
  4.26 anhydrate is .74 g H2O
• So. . . . . . . . .

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A common hydrate is. . .

• MgSo47H2O
• Magnesium Sulfate
• Heptahydrate

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• 4.26 g BaCl2 0.0205 mols BaCl2
• M 208.23 g/mol
• 0.74 g H2O 0.041 mol H2
• 18.02 g/mol
• H2O x mols H2O 0.041 2
• mols cmpd 0.0205
• Thus BaCl22H2O or barium chloride dihydrate

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Summary