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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?

Agenda

• Day 49 - The Mole -Pea Story

• Lesson: PPT- The Mole & Molar Mass

• Handouts: 1. The MoleHandout; 2. Avogadro’s Number, Moles and Molar Mass

• Text: 1. P. 266-274- Avogadro’s Number and Molar Mass

• HW: 1. Finish all the worksheets 2. P.256 # 1 – 15; P. 275 # 1

Agenda

• Day 50 - Mass to Moles

• Lesson: PPT- The Mole & Molar Mass

• Handouts: 1. The MoleHandout; 2. Mole Day; 3. Fun with Moles

• Text: 1. P. 275 - 282- Avogadro’s Number and Molar Mass

• HW: 1. Finish all the worksheets 2. P.277 # 1- 2 pract. # 4-12; P. 280 # 1 – 4; P.283 # 4-6

ODE TO A MOLE I find that my heart beat goes out of control

Just thinking how useful to man is the mole! So perfectly compact. What could be neater? Only occupying twenty-two and four-tenths of a litre. What kind of equations could one hope to equate Without calculating a formula weight? And could a solution be kept for posterity Without ever knowing its molarity? Though all of these findings may set you to slumber I myself am aroused by Avogadro's number. Life without the mole? Don't be absurd! Count your blessings up to six point zero two times ten to the twenty-third.

So just how much is a mole? Let’s look at

peas! The Mole: The Green Pea Analogy

102

106

109

1012

100 average-sized peas occupy about 25 mL.

A million peas fill an ordinary household refrigerator.

A billion peas fill a 3 bedroom house, from basement to attic.

A trillion peas will fill a thousand houses, the number you would find in a medium-sized town.

1015

1018

A quadrillion peas will fill all the buildings in a city such as Hamilton.

A quintillion peas – suppose these is a blizzard over Alberta but – instead of snow – it snows peas. Alberta is covered with a blanket of peas about one metre deep all the way from Saskatchewan to B.C. and all the way from the U.S. border to the N.W. Territories. The blanket of peas drifts over the roads and banks up against the sides of the houses and covers all the fields and forests. Think of flying across the province with a blanket of peas extending as far as you can see.

1021

1023

A sextillion – Imagine the blizzard of peas falls over the entire land of the globe – North America, South America, Europe, Asia, Africa, Australia, and Antarctica. All are covered with peas one metre deep.

Imagine that the oceans are frozen over and the blanket of peas covers the Earth’s entire land and ocean surfaces one metre deep. Now go out among the neighbouring stars and collect 250 planets the same size of Earth and cover each of these with a blanket of peas one metre deep.

Now you have a mole of peas:

6.02 x 1023

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

Counting to 1 Mole Is that right? A computer counting 10 million atoms every second would need to count for 2 billion years to count just a single mole.

Lets look at the mathematics.

Therefore 1 year has 31,536,000 seconds or 3.1536 x 107 sec. A computer counting 10,000,000 atoms every second could count 3.153 x 1014 atoms every year.

Finally, 6.02 x 1023 atoms divided by 3.1536 x 1014 atoms every year equals 1,908,929,477 years or approximately 2 billion years!

x sec = 1 year

365 days

1 year 1 day

24 hours 60 min

1 hour

60 sec

1 min

= 31,536,000 sec

The Mole

• A counting unit • Similar to a dozen, gross (144), pair (2), ream (500),

except instead of 12,144,5 or 500, it’s 602 billion trillion 602,000,000,000,000,000,000,000

• 6.02 X 1023 (in scientific notation) • This number is named in honor of Amedeo _________

(1776 – 1856), who studied quantities of gases and discovered that no matter what the gas was, there were the same number of molecules present.

• Other Collection Terms 1 trio = 3 singers 1 six-pack Cola = 6 cans Cola drink

Avogadro

= 6.02 x 1023 C atoms

= 6.02 x 1023 H2O molecules

= 6.02 x 1023 NaCl “molecules” (technically, ionics are compounds not molecules so

they are called formula units)

6.02 x 1023 Na+ ions and

6.02 x 1023 Cl– ions

A Mole of Particles Contains 6.02 x 1023 particles

1 mole C

1 mole H2O

1 mole NaCl

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 cannot 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-1, 12 g C-12, or 23 g Na-23 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.

• Since the mole is so large, we use it to count very tiny things – like atoms. Because the mole is so large, (and we now know that we cannot count out a mole of anything), how do we know when we have a mole of anything?

• We determine the mass and relate that to the number of atoms present.

Comparing sugar (C12H22O11) & H2O

No, sugar has more (45:3 ratio)

Yes (6.02 x 1023 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 molar mass is the mass of 1 mole of a pure substance. The pure substance can be an element or a compound.

• The atomic mass is the mass of 1 atom of that element measured in amu’s.

• The molar mass is also equal to 1 mole of atoms measured in grams.

What a coincidence!!!!

Atomic Mass Vs. Molar Mass

• Mass of 1 atom of Pb = 207.2 amu

• Mass of 1 mole of Pb atoms = 207.2 g

• Mass of 1 atom of N = 14.01 amu

• Mass of 1 mole of N atoms = 14.01 g

• Mass of 1 atom of Ba = 137.33 amu

• Mass of 1 mole of Ba atoms = 137.33 g

• Mass of 1 atom of Al = 26.98 amu

• Mass of 1 mole of Al atoms = 26.98 g

Molar Mass for Compounds How to find the molar mass:

• Write a CORRECT formula for the compound • Look up the atomic mass of each element in the

compound • Multiply the atomic mass by the subscripts, if any.

Add all masses of elements together and use the unit, g/mol

Cu3(BO3)2

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 g/mol

Find the molar mass of copper II bromate.

Molar mass

Calculate molar masses (to 2 decimal places)

CaCl2

(NH4)2CO3

O2

Pb3(PO4)2

C6H12O6

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”

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

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

Other Names Related to Molar Mass

• Molecular Mass/Molecular Weight ( Molecular Compounds):

If you have a single molecule, mass is measured in amu’s

instead of grams. But, the molecular mass/weight is the same

numerical value as 1 mole of molecules. Only the units are

different. (This is the beauty of Avogadro’s Number!)

• Formula Mass/Formula Weight ( Ionic Compounds): Same

goes for compounds. But again, the numerical value is the

same. Only the units are different.

• THE POINT: You may hear all of these terms which mean the

SAME NUMBER… just different units

Atoms/Molecules and Grams

• Since 6.02 X 1023 particles = 1 mole AND 1 mole = molar mass (grams)

• You can convert atoms/molecules to moles and then moles to grams! (Two step process)

• You cannot go directly from atoms to grams!!!! You MUST go thru MOLES.

molar mass Avogadro’s number Grams Moles Particles

Everything must go through Moles!!!

number ofrepresent at ive

part icles

mass

1 mol22.4L

22.4L1 mol

6.02 x 10 part icles1 mol

23

6.02 x 10 part icles23

1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol


part icles

mass

1 mol22.4L

22.4L1 mol


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

35 g Al

35 g Al


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol

? moles Al

35 g Al x 1 mol 27 g


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol


35 g Al x 1 mol 27g

=


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol


35 mol Al 27

=

= 1.3 mols Al


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol

0.8 moles SiO2

? mass SiO2


part icles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol

0.8 moles SiO2 x 60.1 g SiO2

1 mole SiO2

= 0.8 x 60.1 g SiO2

= 48.1 g SiO2


part icles

mass

1 mol22.4L

22.4L1 mol


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

number ofrepresentative

particles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol

200 g N2O

? molecules N2O

number ofrepresentative

particles


23


1 mol

1 molmolar mass

molar mass1 mol

moles

volumeat STP

mass

1 mol22.4L

22.4L1 mol

200 g N2O x x

1 mole N2O 6.02 x 1023 molecules N2O 44.0 g N2O 1 mole N2O

200 x 6.02 x 10 23 molecules N2O 44.0

=

= 2.74 x 1024 molecules N2O

this is so

cool!!

• A way to solve math problems in chemistry • Used to convert

km to miles, m to km, mol to g, g to mol, etc. • To use this we need: 1) desired quantity, 2) given quantity, 3) conversion factors • Conversion factors are valid relationships or equities

expressed as a fraction E.g. for 1 mol= 6.02 x10 23 particles the conversion factor is

The factor label method

1 mol

6.02x1023 entities or

6.02x1023 entities

1 mol

Conversion factors • We have looked at conversion factors that are always

true. There are conversion factors that are only true for specific questions

• E.g. A recipe calls for 2 eggs, 1 cup of flour and 0.5 cups of sugar

• We can use these conversion factors

sugar cups 0.5

eggs 2 ,

flour cup 1

sugar cups 0.5 ,

flour cup 1

eggs 2

• Q - the chemical equation between H2 and O2 involves 2 H2 molecules combining with 1 O2 molecule to make 2 H2O molecules. Write all possible conversion factors

2 molecules H2

1 molecule O2

1 molecule O2

2 molecules H2

2 molecules H2

2 molecules H2O

2 molecules H2O

2 molecules H2

1 molecule O2

2 molecules H2O

2 molecules H2O

1 molecule O2

2 mol H2

1 mol O2

1 mol O2

2 mol H2

2 mol H2

2 mol H2O

2 mol H2O

2 mol H2

1 mol O2

2 mol H2O

2 mol H2O

1 mol O2

2H2 + O2 2H2O

The steps to follow Now we are ready to solve problems using the factor label

method. The steps involved are:

1. Write down the desired quantity/units

2. Equate the desired quantity to given quantity

3. Determine what conversion factors you can use (both universal and question specific)

4. Multiply given quantity by the appropriate conversion factors to eliminate units you don’t want and leave units you do want

5. Complete the math

Factor label example

Q - How many moles are present in 35.4 grams of Cu?

First write down the

desired quantity

# moles


Next, equate desired

quantity to the given

quantity

# moles = 35.4 g



Now we have to

choose a conversion

factor

# atoms = 35.4 g


Q - How many atoms are present in 35.4 grams of Cu?

What conversion

factors are possible?

# moles = 35.4 g 1 mol

63.5 g

63.5 g

1 mol



Pick the one that will

allow you to cancel

out grams

# mol = 35.4g


1 mol

63.5 g

63.5 g

1 mol

Pick the one that will

allow you to cancel

out grams



# mol = 35.4g 1 mol

63.5 g


Multiply given

quantity by chosen

conversion factor


# mol = 35.4g 1 mol

63.5 g


Multiply given

quantity by chosen

conversion factor


# mol = 35.4g x 1 mol

63.5 g


Cross out common

factors


# mol = 35.4g x 1 mol

63.5 g


Cross out common

factors


# mol = 35.4 x 1 mol

63.5


Are the units now

correct?


# mol = 35.4 x 1 mol

63.5


Yes. Both sides have

moles as units.


# mol = 35.4 x 1 mol

63.5


Yes. Both sides have

moles as units.

#

k

m


# mol = 35.4 x 1 mol

63.5


Now finish the math.

= 0.558 mol


# mol = 35.4 x 1 mol

63.5


The final answer is

0.558 moles


= 0.558 mol # mol

Summary

The previous problem was not that hard. In other

words, you probably could have done it faster

using a different method. However, for harder

problems the factor label method is easiest.

Example.


# atoms =

35.4 g Cu 1 mol Cu 6.02 X 1023 atoms Cu

63.5 g Cu 1 mol Cu

= 3.4 X 1023 atoms Cu

Calculating the Number of Molecules in a Mole

• How many molecules of bromine are present in 0.045 mole of bromine gas?

2

23

2

2

2

23

10022.6

10022.6

Brmolecules

Brmoland

Brmol

Brmolecules

2

23

2 Br molecules 10022.6Br mol 1

2

22 Br molecules 107.2

Given: 0.045 mol Br2 Need: molecules of Br2 Avogadro’s number

2

2

23

2 10022.6045.0

Brmol

BrmoleculesBrmol

Conversion factors:

Equality:

Set Up Problem:

Calculating the Moles of an Element in a Compound

• How many moles of carbon atoms are present in 1.85 moles of glucose?

Plan: moles of glucose moles of C atoms subscript

(One) mol C6H12O6 = 6 mols C atoms Equality:

Conversion Factors:

Set Up Problem:

atomsCmoles6

OHCmol

OHCmol

atomsCmoles6 6126

6126

and

6126

6126

OHCmol

atomsCmols6OHCmol1.8511.1 mol C atoms

Converting Grams to Particles

• Ethylene glycol (antifreeze) has the formula C2H6O2. How many molecules are present in a 3.86 × 10-20 g sample?

Plan: convert g moles molecules of ethylene glycol

Equality 1: Conversion Factor 1

Equality 2:

262

262

262

262

OHCg62.05

OHCmoland

OHCmol

OHCg62.05

molecules106.022OHCmol1 23262

Conversion Factor 2 molecules106.022

OHC1molor

OHC 1mol

molecules106.02223

262

262

23

262

23

262

262262

20

OHCmol

molecules106.022

OHC g62.05

OHCmolOHCg103.86 375 molecules

262262 OHCg62.05OHCmol1

Molar mass

Avog Number

More examples 1. You want to buy 100 U.S. dollars. If the

exchange rate is 1 Can$ = 0.65 US$, how much will it cost?

# Can$ = 100 US$ x 1 Can$

0.65 US$

= 153.85 Can$

2. One mole of a gas has a volume of 22.4 L. How many L will 300 grams of CO2 occupy? (hint: the molar mass of CO2 is ____ g/mol).

# L CO2 =

300 g CO2 x 1 mol CO2

44.01 g CO2

= 152.7 L

CO2

x 22.4 L CO2

1 mol CO2

44.01

More examples 3. There are 12 inches in a foot, 0.394 inches in a

centimeter, and 3 feet in a yard. How many cm are in one yard?

# cm = 1 yd x 3 ft

1 yd = 91.37 cm

x 12 in

1 ft

x 1 cm

0.394 in

4. A chemical reaction requires 3.000 moles of sodium chloride. How many grams is this?

#g NaCl =

3.000 mol NaCl x 58.44 g NaCl

1 mol NaCl = 175.3 g NaCl

Sodium chloride is NaCl (58.44 g/mol)

Assignment Answer questions using the factor label method:

1. How many moles of H2 are in 100 g of H2?

2. 300 g of CuSO4 is needed in an experiment. How many moles does this represent?

3. A chemical reaction requires 23.78 moles of silver chloride. How many grams is this?

4. How many molecules are in 73 grams H2O? (hint: form a conversion factor using Avogadro’s #)

5. 255 g of calcium phosphate are produced in a chemical reaction. How many moles of calcium phosphate does this represent?

1.

2.

# mol H2 = 100 g H2 x 1 mol H2

2.02 g H2

= 49.5 mol H2

# mol CuSO4 =

300 g CuSO4 x 1 mol CuSO4

159.61 g CuSO4

= 1.88 mol CuSO4

# g AgCl =

23.78 mol AgCl x 143.32 g AgCl

1 mol AgCl

= 3408 g AgCl

3.

4. # H2O molecules =

73 g H2O x 1 mol H2O

18.02 g H2O

= 2.44 x 1024 molecules H2O # mol Ca3(PO4)2 =

255 g Ca3(PO4)2 x 1 mol Ca3(PO4)2

310.18 g Ca3(PO4)2

= 0.822 mol

Ca3(PO4)2

5.

x 6.02x1023 molecules

1 mol H2O

Mixed Practice Problems: 1. Prozac, C17H18F3NO, is a widely used antidepressant

that inhibits the uptake of serotonin by the brain. Find its molar mass.

2. How many grams are in 2.50 mols of S? 3. How many atoms are present in 11 grams of Si? 4. The artificial sweetener aspartame (Nutra-Sweet)

formula C14H18N2O5 is used to sweeten diet foods, coffee and soft drinks. How many atoms of hydrogen of aspartame are present in 225 g of aspartame?

5. What is the mass in grams of 1 water molecule.

Mixed Practice Problems: 6. How many moles of atoms are in 100.0 grams of

nitrogen gas, N2? 7. How many ions are there in 28.7 grams of lithium

nitrate, LiNO3? 8. How many moles of cations are there in 1.75 kg of

magnesium chloride, MgCl2? 9. How much would 3.80 x10 25moecules of oxygen

gas weigh? 10. How many electrons are there in 75.0 grams of pure

iron, Fe? 11. What is the formula mass of cupric sulfate

petahydrate, CuSO4 · 5H2O?

the mole - sch3uking - homesch3uking.wikispaces.com/file/view/intro+to+moles.pdfagenda • day 49 -...

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