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# Mole

## on Jun 23, 2011

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## MolePresentation Transcript

• The Mole Q: how long would it take to spend a mole of \$1 coins if they were being spent at a rate of 1 billion per second?
• Background: atomic masses
• Look at the “atomic masses” on the periodic table. What do these represent?
• E.g. the atomic mass of C is 12 (atomic # is 6)
• We know there are 6 protons and 6 neutrons
• Protons and neutrons have roughly the same mass. So, C weighs 12 u (atomic mass units).
• What is the actual mass of a C atom?
• Answer: approx. 2 x 10 -23 grams (protons and neutrons each weigh about 1.7 x10 -24 grams)
• Two problems
• Atomic masses do not convert easily to grams
• They can’t be weighed (they are too small)
• The Mole
• With these problems, why use atomic mass at all?
• Masses give information about # of p + , n 0 , e –
• It is useful to know relative mass
• E.g. Q - What ratio is needed to make H 2 O?
• A - 2 : 1 by atoms, but 2 : 16 by mass
• It is useful to associate atomic mass with a mass in grams. It has been found that
• 1 g H, 12 g C, or 23 g Na have 6.02 x 10 23 atoms
• 6.02 x 10 23 is a “mole” or “Avogadro’s number”
• “ mol” is used in equations, “mole” is used in writing; one gram = 1 g, one mole = 1 mol.
• Read 4.3 (167-9). Stop after text beside fig 2.
• Do Q1-6. Challenge: 1 st slide (use reasonable units)
• A mole is a number (like a dozen). Having this number of atoms allows us to easily convert atomic masses to molar masses.
• 6.02 x 10 23
• 602 000 000 000 000 000 000 000
• 3.00 x 6.02 x 10 23 = 18.06 x 10 23 or 1.81 x 10 24
• (note: there are 3 moles of atoms in one mole of CO 2 molecules. In other words, there are 5.42 x 10 24 atoms in 3.00 mol CO 2 )
• 3.01 x 10 23
• a) 1.43 kg  12 = 0.119 kg per orange
• b) 1.01 g  6.02 x 10 23 = 1.68 x 10 – 24 g
• Mollionaire
• Q: how long would it take to spend a mole of \$1 coins if they were being spent at a rate of 1 billion per second?
• A: \$ 6.02 x 10 23 / \$1 000 000 000
• = 6.02 x 10 14 payments = 6.02 x 10 14 seconds
• 6.02 x 10 14 seconds / 60 = 1.003 x 10 13 minutes
• 1.003 x 10 13 minutes / 60 = 1.672 x 10 11 hours
• 1.672 x 10 11 hours / 24 = 6.968 x 10 9 days
• 6.968 x 10 9 days / 365.25 = 1.908 x 10 7 years
• A: It would take 19 million years
• Comparing sugar (C 12 H 22 O 11 ) & H 2 O No, sugar has more (45:3 ratio) Yes (6.02 x 10 23 in each) Yes. No, molecules have dif. masses No, molecules have dif. sizes. 1 mol each Yes, that’s what grams are. mass? No, they have dif. molar masses # of moles? No, they have dif. molar masses # of molecules? No # of atoms? No, they have dif. densities. volume? 1 gram each Same
• Molar mass
• The mass of one mole is called “molar mass”
• E.g. 1 mol Li = 6.94 g Li
• This is expressed as 6.94 g/mol
• What are the following molar masses?
• S SO 2
• Cu 3 (BO 3 ) 2
32.06 g/mol 64.06 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl 2 (NH 4 ) 2 CO 3 O 2 Pb 3 (PO 4 ) 2 C 6 H 12 O 6 Cu x 3 = 63.55 x 3 = 190.65 B x 2 = 10.81 x 2 = 21.62 O x 6 = 16.00 x 6 = 96.00 308.27
• Molar mass
• The mass of one mole is called “molar mass”
• E.g. 1 mol Li = 6.94 g Li
• This is expressed as 6.94 g/mol
• What are the following molar masses?
• S SO 2
• Cu 3 (BO 3 ) 2
32.06 g/mol 64.06 g/mol 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl 2 (NH 4 ) 2 CO 3 O 2 Pb 3 (PO 4 ) 2 C 6 H 12 O 6 110.98 g/mol (Ca x 1, Cl x 2) 96.11 g/mol (N x 2, H x 8, C x 1, O x 3) 32.00 g/mol (O x 2) 811.54 g/mol (Pb x 3, P x 2, O x 8) 180.18 g/mol (C x 6, H x 12, O x 6)
• Converting between grams and moles
• If we are given the # of grams of a compound we can determine the # of moles, & vise-versa
• In order to convert from one to the other you must first calculate molar mass
• g = mol x g/mol
• mol = g  g/mol
• This can be represented in an “ equation triangle ”
9.1 36.46 mol= g  g/mol 0.5419 98.08 g= g/mol x mol 207 58.44 mol= g  g/mol 0.0200 63.55 g mol g/mol g= g/mol x mol 0.25 HCl 53.15 H 2 SO 4 3.55 NaCl 1.27 Cu Equation mol (n) g g/mol Formula
• Simplest and molecular formulae
• Consider NaCl (ionic) vs. H 2 O 2 (covalent)
• Chemical formulas are either “simplest” (a.k.a. “empirical”) or “molecular”. Ionic compounds are always expressed as simplest formulas.
• Covalent compounds can either be molecular formulas (I.e. H 2 O 2 ) or simplest (e.g. HO)
• Q - Write simplest formulas for propene (C 3 H 6 ),
• C 2 H 2 , glucose (C 6 H 12 O 6 ), octane (C 8 H 14 )
• Q - Identify these as simplest formula, molecular
• formula, or both H 2 O, C 4 H 10 , CH, NaCl
Cl Na Na Cl Cl Cl Na Na H O O H H O O H H O O H