Q: How Long Would It Take To Spend A Mole

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 1. Atomic masses do not convert easily to grams 2. They can’t be weighed (they are too small)

The Mole

With these problems, why use atomic mass at all? 1. Masses give information about # of p+, n0, e– 2. 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)

The Mole: Answers 1. A mole is a number (like a dozen). Having this number of atoms allows us to easily convert atomic masses to molar masses. 2. 6.02 x 1023 3. 602 000 000 000 000 000 000 000 4. 3.00 x 6.02 x 1023 = 18.06 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) 5. 3.01 x 1023 6. a) 1.43 kg ÷ 12 = 0.119 kg per orange b) 1.01 g ÷ 6.02 x 1023 = 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 1023 / $1 000 000 000 = 6.02 x 1014 payments = 6.02 x 1014 seconds 6.02 x 1014 seconds / 60 = 1.003 x 1013 minutes 1.003 x 1013 minutes / 60 = 1.672 x 1011 hours 1.672 x 1011 hours / 24 = 6.968 x 109 days 6.968 x 109 days / 365.25 = 1.908 x 107 years

A: It would take 19 million years

Comparing sugar (C12 H22 O11 ) & H2O Same

1 gram each 1 mol each No, they have dif. No, molecules volume? densities. have dif. sizes. Yes, that’s what No, molecules mass? grams are. have dif. masses No, they have dif. Yes. # of moles? molar masses

No, they have dif. Yes (6.02 x 1023 in # of each) molecules? molar masses No, sugar has # of atoms? No more (45:3 ratio)

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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 32.06 g/mol 2 64.06 g/mol Cu3(BO3)2 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl2 Cu x 3 = 63.55 x 3 = 190.65 (NH4)2CO3 B x 2 = 10.81 x 2 = 21.62 O x 6 = 16.00 x 6 = 96.00 O2 308.27 Pb3(PO4)2 C6H12 O6

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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 32.06 g/mol 64.06 g/mol 2 Cu3(BO3)2 308.27 g/mol Calculate molar masses (to 2 decimal places) CaCl2 110.98 g/mol (Ca x 1, Cl x 2) (NH4)2CO3 96.11 g/mol (N x 2, H x 8, C x 1, O x 3) O2 32.00 g/mol (O x 2) Pb3(PO4)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) C6H12 O6

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 g mol = g ÷ g/mol mo g/mo • This can be represented in an “equation triangle”

Formula HCl H2SO4 NaCl Cu

g/mol 36.46 98.08 58.44 63.55

l l g mol (n) Equation 9.1 0.25 g= g/mol x mol 53.15 0.5419 mol= g ÷ g/mol 207 3.55 g= g/mol x mol 1.27 0.0200 mol= g ÷ g/mol

Simplest and molecular formulae

Consider NaCl (ionic) vs. H2O2 (covalent)

Na

Cl

Cl

Na

Na

H

O O O H H O H

H O

Na

Cl

O H

Cl

• 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 (C6H12 O6), octane (C8H14 ) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10 , CH, NaCl

Answers

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Q - Write simplest formulas for propene (C3H6), C2H2, glucose (C6H12 O6), octane (C8H14 ) Q - Identify these as simplest formula, molecular formula, or both H2O, C4H10 , CH, NaCl

A - CH2

CH

CH2O

C4H7

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