DO NOW: (Refresh your memory)

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DO NOW (Refresh your memory)

• Classify the following as an atom, molecule, ion,
  or formula unit
• 1.Fe _________
• 2.F2 _________
• 3.H2O _________
• 4.Na _________
• 5.NaCl _________
• 6.Na1 _________

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DO NOW (Refresh your memory)

• Classify the following as an atom, molecule, ion,
  or formula unit
• 1.Fe ATOM
• 2.F2 MOLECULE
• 3.H2O MOLECULE
• 4.Na ATOM
• 5.NaCl FORMULA UNIT
• 6.Na1 ION

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Chapter 10

• THE MOLE!!!

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What is a mole?
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(No Transcript)
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What is a mole?

• It is a unit of measure.
• 6.02 x 1023
• If we write this out (standard notation),
• we get

• 602,000,000,000,000,000,000,000

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A mole is equal to

• 6.02 x 1023 Representative Particles
• Atoms, molecules, formula units, and ions
• Fe F2 NaCl
  Na1
• Now we have a conversion unit to use
• 1mole
• 6.02 x 1023 r.p. (any of the four choices
  above)

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Lets do some problems

• 2.1 moles of Fe how many r.p. of Fe?
• First what r.p. is Fe?
• (atom)
• 2.1 moles Fe 6.02 x 1023 atoms Fe
• 1mole Fe

• 2.1moles Fe 1.3x 1024 atoms Fe

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• 140 moles of MgCl2 how many r.p. of Cl-?
• First what r.p. is MgCl2 and then Cl-?
• (formula unit, ion)
• 140 moles MgCl2 6.02 x 1023 f.u. MgCl2 2 ions
  Cl-
• 1mole MgCl2 1 f.u. MgCl2
• 140 moles MgCl2 1.7 x 1026 ions Cl-

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Gram Formula Mass Gram Molar Mass (molar mass)

• Use individual atomic masses to determine overall
  mass
• In 1 mole of NaCl, there are 58 g of NaCl
• 23 35 58
• In 1 mole of H2O, there are 18 g of H20
• 2(1) 16 18

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A mole is equal to

• The molar mass of a substance
• Now we have a conversion unit to use
• 1mole
• molar mass of a substance
• EX what is the gfm of C6H12O6
• 180g/mol
• So, in 1 mole of C6H12O6, there are 180 g
• 6(12) 12(1) 6(16) 180

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Lets do some problems

• 11.3 moles of C6H12O6 how many grams?
• 11.3 moles C6H12O6 180 g C6H12O6
• 1mole C6H12O6
• 11.3 moles C6H12O6 2030 g C6H12O6
• 2.03 x 103 g C6H12O6

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A mole is equal to

• The 22.4L of gas at STP
• STP Standard Temperature and Pressure
• 0o C or 273 K and
• (101.3 kPa or 1 atm or 760 mmHg or 760 torr)
• Now we have a conversion unit to use
• 1mole
• 22.4L of a gas _at_ STP

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THE MOLE ISLAND!!!!
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Lets do some problems

• 5.3x1024 molecules of CH4 how many grams CH4?
• !!!!! You must go from the r.p. island then to
  the mole island, and then to the molar mass
  island!!!!!
• 5.3x1024 molecules of CH4 1 mole CH4 16g CH4
• 6.02x1023r.p. CH4 1 mole CH4
• 5.3x1024 molecules of CH4 140 g CH4
• 1.4 x 102 g CH4

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Chapter 10

• Percent Composition

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Calculating Composition

• Chemists use this calculation when new compounds
  are created in the lab and they would have to
  determine the formula of the cmpd.

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Formula

• mass of element grams of element x 100
• grams of cmpd

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Lets do some problems

• An 8.20g piece of magnesium combines completely
  with 5.40g of oxygen to form a compound. What is
  the composition of Magnesium and Oxygen the
  cmpd?

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• Step 1 Add masses to get total
• 8.20g 5.40g 13.60g
• Step 2 Find the of each element.
• Mg (8.20g/13.60g)100 60.3
• O (5.40g/13.60g)100 39.7
• Step 3 Make sure your s add up to 100.

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Chapter 10

• Empirical Formulas

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Calculating Empirical Formulas

• Gives the lowest whole number ratio of the atoms
  of the elements in a cmpd.

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Remember the Poem

• to mass
• Mass to mole
• Divide by the smaller
• And multiply until whole

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Lets do some problems

• What is the empirical formula of the cmpd
  that is 25.9 N and 74.1 O?
• 1. 25.9 N 25.9g N
• 74.1 O 74.1g O
• 2. 25.9g N 1mole N 1.85 mole N
• 14 g N
• 74.1 g O 1 mole O 4.63 mole O
• 16 g O

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• 1.85 1.00 mole for N 4.63 2.50 mole for O
• 1.85 1.85
• Make into whole s by multiplying by 2
• 1 mole N x 2 2 moles N
• 2.5 mole O x 2 5 moles O
• So, the empirical formula is N2O5

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Empirical vs. Molecular Formulas

• (lowest whole ratio) vs (multiple of empirical)

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Calculating Molecular Formulas

• 1. Go through all of the steps of an empirical
• formula problem. (poem) Ex CH3
• 2. Add up the mass of the empirical formula.
• Ex 15g/mol (efm)

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• 3. Divide the mass of the molecular formula
  (gfm), which will be given in the problem
  (30g/mol), by the mass of the empirical formula
  (efm). (gfm/efm)
• Ex 30g/mol 2 15g/mol

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• 4. Multiply all of the subscripts in the
  empirical formula by the 2.
• This will be the new molecular formula.
• CH3 (empirical) x 2C2H6 (molecular formula)
• Try 38 and 39 p. 312.