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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?

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Presentation on theme: "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?"— Presentation transcript:

1 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?

2 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 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)

3 The Mole With these problems, why use atomic mass at all?
Masses give information about # of p+, n0, e– It is useful to know relative mass E.g. Q - What ratio is needed to make H2O? 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 1023 atoms 6.02 x 1023 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: 1st slide (use reasonable units)

4 The Mole: Answers 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 1023 3.00 x 6.02 x 1023 = x 1023 or 1.81 x 1024 (note: there are 3 moles of atoms in one mole of CO2 molecules. In other words, there are 5.42 x 1024 atoms in 3.00 mol CO2) 3.01 x 1023 a) 1.43 kg  12 = kg per orange b) 1.01 g  6.02 x = 1.68 x 10 –24 g

5 A: It would take 19 million years
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 1023 / $ = 6.02 x 1014 payments = 6.02 x 1014 seconds 6.02 x 1014 seconds / 60 = x 1013 minutes 1.003 x 1013 minutes / 60 = x 1011 hours 1.672 x 1011 hours / 24 = x 109 days 6.968 x 109 days / = x 107 years A: It would take 19 million years

6 Comparing sugar (C12H22O11) & H2O
Same 1 gram each 1 mol each volume? No, they have dif. densities. No, molecules have dif. sizes. mass? Yes, that’s what grams are. No, molecules have dif. masses # of moles? No, they have dif. molar masses Yes. # of molecules? No, they have dif. molar masses Yes (6.02 x 1023 in each) # of atoms? No, sugar has more (45:3 ratio) No

7 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 SO2 Cu3(BO3)2 32.06 g/mol 64.06 g/mol g/mol Calculate molar masses (to 2 decimal places) CaCl2 (NH4)2CO3 O2 Pb3(PO4)2 C6H12O6 Cu x 3 = x 3 = B x 2 = x 2 = O x 6 = x 6 = 308.27

8 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 SO2 Cu3(BO3)2 32.06 g/mol 64.06 g/mol g/mol Calculate molar masses (to 2 decimal places) CaCl2 (NH4)2CO3 O2 Pb3(PO4)2 C6H12O6 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) g/mol (Pb x 3, P x 2, O x 8) g/mol (C x 6, H x 12, O x 6)

9 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” g mol g/mol g= g/mol x mol 0.25 HCl 53.15 H2SO4 3.55 NaCl 1.27 Cu Equation mol (n) g g/mol Formula 36.46 9.1 98.08 0.5419 mol= g  g/mol 58.44 207 g= g/mol x mol 63.55 0.0200 mol= g  g/mol

10 Simplest and molecular formulae
Consider NaCl (ionic) vs. H2O2 (covalent) Cl Na Cl H O Na H O H O Cl Cl Na Na 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. H2O2) or simplest (e.g. HO) Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12O6), octane (C8H14) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10, CH, NaCl

11 For more lessons, visit www.chalkbored.com
Answers For more lessons, visit Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12O6), octane (C8H14) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10, CH, NaCl A - CH2 A - H2O is both simplest and molecular C4H10 is molecular (C2H5 would be simplest) CH is simplest (not molecular since CH can’t form a molecule - recall Lewis diagrams) NaCl is simplest (it’s ionic, thus it doesn’t form molecules; it has no molecular formula) CH CH2O C4H7


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