Download Moles

The Mole
Chapter 11
By: Kathryn Johnson 2008
& BuggLady 2006
What is a mole?

Not that one….
A Mole is…..
(p.310)
… an SI unit of measure used for
large numbers of particles
Like the following:
 1 dozen = 12
 1 gross = 144
 1 ream = 500
 1 thousand = 1,000
 1 billion = 1,000,000,000
A Mole (mol) =
= 602,213,670,000,000,000,000,000 particles
= 6.02x1023 particles (scientific notation)
= 6.02E23 particles (calculator notation)
Particles - atoms, molecules, formula units, ions,…
This number is also called ‘Avogadro's number’
Why is the number so large?

Because atoms are so small!
Why do we use the mole?

Because it is easier to talk of 4 moles
than to say we have
2,410,000,000,000,000,000,000,000
or 2.41x1024 molecules of something
We use it the same way we talk of
miles instead of feet when we travel
Remember…
Conversion Factors (Chapter 2)
 They
allow us to decide whether to
divide or multiply by a value
 Also



called:
Dimensional Analysis
Unit Cancellation
Factor Label
How to set one up…


Set up a fraction
Place the units first





the unit you have on the bottom
the other unit from the relationship on top
Scratch off (Cancel) units that are on both the
top and the bottom
Repeat until you only have the units you want
Then place the numbers from the relationships
into the fractions
Last…Do the Math



Multiply all the numbers
on the top together
Divide by all the numbers
on the bottom
Round off to correct
significant figures
USE YOUR
CALCULATOR!
Mole ↔ Molecules

(p.311)
Set up a conversion factor!
Moles x Molecules = Molecules
Moles
Molecules x
Moles = Moles
Molecules
1 mol = 6.02E23 molecules
6.02E23 molecules
1 mol
1 mol
.
6.02E23 molecules
multiplication
division
Be Careful!
This will be easy at first and you
will want to skip steps….
But we will be setting up lots of
these fractions and you have to
keep them straight
Now, lets add
another step!
Molar Mass is
the mass (g)
of one mole (mol)
of a substance
Molar Mass
of Elements

The mass (g) of a mole (mol) of a
particular element is equal to its
Atomic Mass on the Periodic Table


(p.313)
Use two decimal places
Carbon: 1 mol C = 12.01 g C
12.01 g/mol
Molar Mass
of Compounds
(p.322)

The mass (g) of a mole (mol) of a
molecule is found by adding the
atomic masses of all the elements,
multiply by the subscripts

Water… H2O
2(1.01) + 16.00 = 18.02 g/mol
or
1 mol H2O = 18.02 g
Be Careful….
Polyatomic ions with parenthesis
complicate the math!
 Ammonium Phosphate… (NH4)3PO4

= 3(14.01+4(1.01)) + 30.97 + 4(16.00)
= 3(14.01+4.04) + 30.97 + 64.00
= 3(18.05) + 30.97 + 64.00
= 54.15 + 30.97 + 64.00
= 149.12 g/mol
Another way to do it…

Ammonium Phosphate… (NH4)3PO4
N: 3(14.01) = 42.03
H: 4∙3(1.01) = 12.12
P: (30.97) = 30.97
O: 4(16.00) = 64.00
Molar Mass = 149.12 g/mol
Mass ↔ Mole

(p.314)
Now we have a new conversion factor!
Mass x
Mole = Moles
Mass
Moles x
Mass = Mass
Moles
1 mol = (Molar Mass) g
(Molar Mass) g
1 mol
multiplication
1 mol
.
(Molar Mass) g
division
Mass ↔ Molecules
(the 2 step!)
(p.316)
Now we set up two conversion factors!
 You always have to go to Moles first!

Mass x Mole x Molecules = Molecules
Mass
Mole
Molecules x
Mole x
Molecules
Mass = Mass
Mole
Ratio of Moles of an Element
in 1 mole of a Compound (p. 320)

equal to the subscript beside the element

Water… H2O
2 mol H
1 mol H2O
and
1 mol O .
1 mol H2O
It gets complicated…

Watch out for those polyatomic parenthesis!

Ammonium Phosphate… (NH4)3PO4
3 mol N
1 mol (NH4)3PO4
12 mol H .
1 mol (NH4)3PO4
and
1 mol P
1 mol (NH4)3PO4
4 mol O .
1 mol (NH4)3PO4
Mass Compound ↔ Element
(the 3 step!)
Now we set up many conversion factors!
 Always go to moles!

Mass compound x Mole compound x Mole element…
Mass compound Mole compound
… x Mass element = Mass element
Mole element
Yes… it keeps going!


You can set up these conversion factors in
infinite combinations to transform from one
value to another
The only trick is… to go from one ‘thing’
(compound) to another ‘thing’ (element in
that compound) you have to go through
the mole ratio based on the subscript
... Don’t Skip Steps!
Mass of Element
Atoms of Element
(atomic mass) g = 1 mol
1 mol = (6.02E23) atoms
Moles of Element
(subscript) mol of Element = 1 mol of Compound
Moles of Compound
(molar mass) g = 1 mol
Mass of Compound
1 mol = (6.02E23) molecules
Molecules of Compound
And we still
have it easy…
wait until we
start dealing with
Chemical Reactions!
(Chapter 12)
Percent Composition (p.328)

percent (%) by mass of each element in a
compound


Review: Name → Formula (Chapter 8&9)
Watch out for those polyatomic parenthesis!

% Element = (subscript)x(Atomic Mass)x100
Molar Mass Compound

All the % should add up to 100!
Example: Water…. H2O

% H = 2(1.01) g H x 100 = 11.2%
18.02 g H2O

% O = 16.00 g O x 100 = 88.8%
18.02 g H2O

Check your work:
11.2% + 88.8% = 100%
Another way to do it…

Ammonium Phosphate… (NH4)3PO4
N: 3 (14.01) = 42.03 → 28.2%
H: 12 (1.01) = 12.12 → 8.1%
P: (30.97) = 30.97 → 20.8%
O: 4 (16.00) = 64.00 → 42.9%
Molar Mass = 149.12 g/mol
Practice



Determine the percent by mass of each
element in calcium chloride.
Calculate the percent composition of sodium
sulfate.
Which has the larger percent by mass of
sulfur: H2SO4 or H2S2O8?
Empirical Formula (p.331)


The smallest whole-number mole ratio of the
elements in a compound
Derived from %comp of the elements




Turn Percent Composition (%) into Mass (g)
Convert Mass→Mole with the Molar Mass
Divide by the smallest value to get the whole
number ratios
Like a %comp in reverse
Example: 59.95% O & 40.05% S



0: 59.95 g x 1 mol = 3.747 mol = 3
16.00 g
1.249 mol
S: 40.05 g x 1 mol = 1.249 mol = 1
32.07 g
1.249 mol
Therefore: SO3… Sulfur trioxide!

If the ratios are still decimals, multiply them
by a small factor to make them whole



(EX: 0.5 = ½, multiply by 2)
(EX: 0.33 = 1/3, multiply by 3)
(EX: 0.25 =1/4, multiply by 4)
Example:
48.64% C, 8.16% H, & 43.20% O




0: 43.20 g x 1 mol = 2.700 mol = 1 x 2 = 2
16.00 g
2.700 mol
C: 48.64 g x 1 mol = 4.050 mol = 1.5 x 2 = 3
12.01 g
2.700 mol
H: 8.16 g x 1 mol = 8.10 mol = 3 x 2 = 6
1.01 g
2.700 mol
Therefore: C3H6O2
Two or more substances with
different properties can have
the same % composition and
empirical formula!
The simplest ratio in the
empirical formula does not
indicate the actual number of
elements in the compound
Molecular Formula (p.333)
• The actual numbers of atoms in a compound
• Identifies the compound
• A multiple of the empirical formula
• Molecular Formula = (Empirical Formula) n
n=
Measured molar mass of compound_ _.
Calculated molar mass of empirical formula
Example: 46.68% N & 53.32% O
Measured Molar Mass = 60.01 g/mol


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0: 53.32 g x 1 mol = 3.333 mol = 1
16.00 g
3.332 mol
N: 46.68 g x 1 mol = 3.332 mol = 1
14.01 g
3.332 mol
Empirical Formula: NO … nitrogen monoxide
Molar Mass = 14.01 + 16.00 = 30.01 g/mol
But, 60.01 g/mol = 2 therefore…
30.01 g/mol
Molecular Formula: N2O2 … dinitrogen dioxide!
Review

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
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Mol ↔ Molecules/Atoms (Avogadro's #)
Mol ↔ Mass (Molar Mass)
Mass ↔ Molecules/Atoms (The 2 step)
Element ↔ Compound (Subscript, 3 step)
% Composition
Empirical/Molecular Formula
Practice…
Practice…
Practice…
Work Smart,
Not Hard!
You have to be
fast enough
to finish the test!
Mole Ratios
Coefficients indicate the number of moles
 Mole Ratios – the ratio between the numbers of moles of any
2 substances in a balanced equation

Ex: 2 Al + 3 Br2  2 AlBr3

All possible mole ratios:
2 mole Al : 3 mole Br2
3 mole Br2 : 2 mole Al
2 mole AlBr3 : 2 mole Al



2 mole Al : 2 mole AlBr3
3 mole Br2 : 2 mole AlBr3
2 mole AlBr3 : 3 mole Br2
Interpreting Mole Conversions
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
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
4 Fe + 3 O2  2 Fe2O3
You can interpret any balanced equation in
terms of moles, mass, and molecules
Molecules: 4 atoms of iron react with 6
atoms of oxygen to produce 2 molecules of
iron (III) oxide
Moles: 4 moles of iron react with 3 moles of
oxygen to produce 2 moles of iron (III) oxide
Mass: convert to grams first
Practice Mole Ratios
For the following equations:
Balance the equations
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Interpret each in terms of moles, molecules, and
mass
Determine all possible mole ratios
1. CaC2 + N2  CaNCN + C
2. NH3 + CuO  Cu + H2O + N2
3. MnO2 + HCl  MnCl2 + Cl2 + H2O
4. P4O10 + C  P4 + CO
5. ZrI4  Zr + I2
6. PbS + O2  PbO + SO2
Mole to Mole Conversions


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

Ex: Determine the number of moles of H2 produced when 0.040 moles
of K is used.
Steps for solving:
Step 1. Write the balanced equation:
2 K + 2 H2O  2 KOH + H2
Step 2. Identify the known (given) substances and the unknown
substances:
Known: .040 moles K
Unknown: moles of H2
Step 3. Determine the mole ratio of unknown substance to known
substance:
1 mole H2
(use coefficients from balanced equation)
2 mole K
Step 4. Convert using the known substance and the mole ratio:
0.040 moles K x 1 mole H2 = 0.02 mole H2 produced
2 mole K
Practice

Handout: Mole to Mole Stoichiometry
Problems
Mole to Mass Conversions

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Ex: Determine the mass of NaCl produced when 1.25 moles of Cl2 reacts with
sodium
Steps:
1.
Write the balanced equation:
2 Na + Cl2  2 NaCl
2.
Identify the known (given) substances and the unknown substances:
Known: 1.25 moles Cl2
Unknown: mass of NaCl
3.
Determine the mole ratio of unknown substance to known substance:
2 moles NaCl
(use coefficients from balanced equation)
1 mole Cl2
4.
Convert using the known substance and the mole ratio:
1.25 moles Cl2 x 2 moles NaCl = 2.5 moles NaCl
1 mole Cl2
5.
Convert moles of unknown to grams of unknown: (use molar mass)
2.5 mole NaCl x 58.44 g = 146 grams NaCl
1 mole
Example: Mole to Mass



Reactants:
4 mole Fe x 55.85 g Fe = 223.4 g Fe
1 mole Fe



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

3 mole O2 x 32.00 g O2 = 96.0 g O2
1 mole O2
Product:
2 mole Fe2O3 x 158.7 g Fe2O3 = 319.4 gFe2O3
1 mole Fe2O3
Example, cont.

** Law of Conservation of Mass – the sum of
the reactants should ALWAYS equal the sum
of the product – check your work!!!

223.4 grams of Fe react with 96.0 grams of
O2 to produce 319.4 grams of Iron (III) oxide
Mass to Mass Conversions

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





Ex: Determine the mass of H2O produced from the decomposition of 25.0
grams of NH4NO3
Steps:
1.
Write the balanced equation:
NH4NO3  N2O + 2H2O
2.
Identify the known (given) substances and the unknown substances:
Known: 25.0 grams of NH4NO3
Unknown: mass of H2O
3. Convert grams of known to moles of known:
25.0 g NH4NO3 x 1 mole NH4NO3 = 0.312 mole NH4NO3
80.04 g
4.
Determine the mole ratio of unknown substance to known substance:
2 moles H2O
(use coefficients from balanced equation)
1 mole NH4NO3
5.
Convert using the known substance and the mole ratio:
0.312 mole NH4NO3 x 2 moles H2O = 0.624 mole H2O
1 mole NH4NO3
6.
Convert moles of unknown to grams of unknown: (use molar mass)
0.624 mole H2O x 18.0 g = 11.2 grams H2O
1 mole
Hydrates (p.338)



Compound that has a specific number of
water molecules (H2O) bound to its atoms
Solids in which water molecules are trapped
Like a sponge, different amounts of water can
be trapped
Formulas and Names

1 formula unit of compound
with a dot and then the
attached water molecules
Ex: Na2CO3∙10H2O


Name the molecule
Use the prefixes in front of the
word ‘hydrate’
Ex: sodium carbonate decahydrate
1
2
3
4
5
6
7
8
9
10
MonoDiTriTetraPentaHexaHeptaOctaNonaDeca-
Anhydrous



“without water”
Water is removed from the hydrate by
heating
Desiccants are used to absorb excess
moisture in packages


Calcium chloride
Calcium sulfate
How much water attached?


How many moles of water attached to one
mole of the molecule (BaCl2∙xH2O)
compare the mass before and after heating



Mass hydrate – Mass anhydrous = Mass water
Convert the Mass water and Mass anhydrous
to moles using the molar mass of each
Calculate the ratio x = mol H2O / mol BaCl2