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The Mole Concept

The Mole Concept. . . . Quantities in Chemistry. Real Life Quantities. How many shoes do you have? “I have 26 kg of shoes.” What’s wrong with this answer? Mass was given when quantity was asked for; different shoes have different mass.

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The Mole Concept

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  1. The Mole Concept . . . Quantities in Chemistry

  2. Real Life Quantities How many shoes do you have? “I have 26 kg of shoes.” What’s wrong with this answer? Mass was given when quantity was asked for; different shoes have different mass.

  3. How would 26 kg of work boots differ from 26 kg of sandals? Work boots are more massive—fewer boots than sandals in 26 kg What is the fundamental quantity unit associated with shoes? A pair—2.

  4. Why is the mass of a dozen pencils not the same as the mass of a dozen buses? One bus is more massive than one pencil.

  5. What does this have to do with chemistry? • What is the mass of a C atom, in amu? 12 amu • What is the mass of a U atom, in amu? 238 amu • Explain why it is NOT possible to determine the mass, in grams, of an individual atom by simply weighing that atom on a balance.

  6. In chemistry, the MOLE (mol) is the fundamental unit of QUANTITY.

  7. the details . . . • 1.00 mol contains 6.02 x 1023 particles (atoms, molecules, ions . . . it depends what we’re talking about). This value is known as the Avogadro constant (number). Think of this as the “chemist’s dozen”. We use 6.02 x 1023 particles per mole instead of 12 particles per dozen. Avogadro constant = 6.02E23 particles·mol-1

  8. If you want to get technical . . . The mole is the SI base unit for the amount of substance (symbol: mol). The mole is the amount of substance of a system which contains as many elementary entities as there are atoms in 0.012 kilogram of carbon-12. e When thmole is used, the elementary entities must be specified and may be atoms, molecules, ions ...

  9. How Big is the Avogadro Constant? Imagine that you have $ 6.02E23. The catch: You must spend all the $$ at a rate of 1 billion dollars per second. (ie. $1 x 109/second) How long will it take you to spend the $$ ? Take a guess. Answer in units of years. Get busy.

  10. Answer to the question Calculate number of seconds in a year: (60 s/min)*(60 min/hr)*(24 hr/day)*(365 day/yr) = 31 536 000 s/yr (6.02E23 dollars)/(1E9 dollars/s) = 6.02E14 s 6.02E14 s/(31 536 000 s/yr) > 19 million years!!! Start spending!!!

  11. To a chemist, 1 mol of a substance refers to 6.02E23 units (atoms, molecules, formula units) of that substance. It doesn’t matter if that substance is H2O or Fe or KCl. cf. 1 dozen refers to a quantity of 12, whether you’re referring to flower pots or bull elephants. A dozen always means 12. • It gets easier . . .

  12. 2. The mass of 1 mol of C atoms = 12 g (1 mol of C atoms = 6.02 x 1023 atoms) The mass of 1 C atom = 12 amu. The mass of 1 molecule of H20 = 18 amu. The mass of 1 mol of H2O molecules = 18 g. Simply add the masses of the constituent atoms, in grams, to get the molar mass of a compound.

  13. What is the mass, in amu, of 1 CO2 molecule? 44 amu What is the mass of 1 mol of CO2? 44 g How many formula units of NaCl are contained in 1 mol of NaCl? 6.02E23 fu/mol NaCl What is molar mass of NaCl? 58.5 g

  14. What is the mass, in g, of 1 water molecule? 18 g = 6.02E23 molecules ? g = 1 molecule cross multiplying gives ? = (18 g)/6.02E23 molecules ? = 3.0E-23 g/molecule 1 H20 molecule has mass of 3.0E-23 g or 1 H20 molecule has mass of 18 amu.

  15. 3. The molar volume of any gas at 25oC, 100 kPa (Standard Atmospheric Temp and Pressure, SATP) is 24.8 L. This is true for ALL gaseous compounds as well as atomic gases (eg. He), gaseous molecules (eg. CH4) and mixtures (eg. air). This does not apply to liquids, solids or solutions.

  16. For example: • 1.00 mol CO2(g) occupies a volume of 24.8 L under conditions of SATP. (or 22.4 L at STP—0oC, 101.3 kPa) • 1.00 mol of Xe(g) occupies a volume of 24.8 L under conditions of SATP.

  17. Let’s use methane, CH4(g) to put it all together . . . 1.00 mol methane, CH4: • contains 6.02E23 ___________ of CH4 molecules • contains _______ individual atoms 5 atoms/molecule*( 6.02E23) atoms = 30.1E23 atoms because 1 molecule of CH4 is made up of 5 atoms (1C atom + 4H atoms = 5 atoms) • has a mass of _______ 16 g (use periodic table.)

  18. 1.00 mol CH4(g): • occupies a volume of ______ at SATP 24.8L (SATP = 25oC, 100 kPa), [or 22.4 L at STP(0oC, 101.3 kPa)], since it is a gas under these conditions.

  19. Try this: You have 79.3 g of NH3(g). Express this in a) moles b) number of molecules c) number of atoms d) volume at SATP Answers: 4.66 mol; 2.80E24 molecules; 1.12E25 atoms; 115 L (@SATP)

  20. and this . . . You have 8.37E25 molecules of SO2(g). Convert this to a) moles b) mass c) volume @ SATP d) total # atoms Answers: 139 mol; 8.91E3 g; 3.45E3 L (SATP); 2.51E26 atoms total

  21. Homework • Mole practice problem handout (posted) OR • p 174 PP 10, 11 • p 177 PP 13 – 18 (try a few) • p 178 PP 19 – 22 (try a few) • p 179 SR 1 – 8, 10

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