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Transcript
Chapter 6: Moles, Molar Mass,
Percent Composition and
Formulas
From moles to mass and to the moon!
AMU (Atomic Mass Units)
The mass of Carbon-12 is 12 AMU.
But wait, when I look on the periodic table,
the atomic mass is listed as 12.01078
AMU??? WHY? Why, cruel world?
WwWWwwhhHHHhhyyYYYYyy???
6.1 Atoms and Moles
Amedeo Avogadro
 Avogadro’s Number
6.022 x 1023
Avogadro
discovered that
there are
6.022 x 1023
atoms in 1 gram
of hydrogen.
Count Lorenzo
Romano Amedeo
Carlo Avogadro di
Quaregna e
Cerreto
Be able to explain and use the
concept of the “mole”
 This number is called a “mole.”
 The word “mole”
is just like the
word “dozen”. Dozen
means “12”. You can
have a dozen of
anything. You can
also have a
mole of anything.
Hmmm… I shall call
6.022 x 1023... a “mole”.
Yes…that has a nice ring
to it.
So How Big is a “MOLE”
Ummm… NO!
Don’tout
be
Here it is written
Aaaiiie
cruel now…
e
602,200,000,000,000,000,000,000
That’s 602 billion groups of a trillion!
Let’s just do an example with paper
clips.
If you have a mole of paper clips
and made them into a chain, how
many times could you go to the
moon and back with your chain?
 Assume a paper clip ( still folded) is about 3 cm
long.
 To find the total distance of the paper clips we use
the following equation:
3cm  6.02 x10 clips 
24 cm
  1.806 x 10
x
clip 
1 mole
mole

23
Notice the unit “clips” cancels!!! Isn’t
that Great…
Anyone…
Anyone see the greatness???
Man I love Conversions!
The moon is 382,171 km from Earth, so
to the moon and back would be 764,342
km.
 So we need to convert our cm into km…
… oh how fun… this is a metric conversion
 This of course is a “2-step conversion” because
both units have a prefix


 1m  1 km 
19
1.806 x10 cm 

1
.
806
x
10
km


 100 cm  1000m 
24
I love conversions!
 We’re almost done!!!
1 trip and back
13
1.806 x10 km (
)  2.36 x10 trips
764,342km
19
 That’s 23 trillion trips!! Mole-tastic!
 Marshmallow example: A bed of
marshmallows covering the U.S. would be 776
miles deep
Convert moles to # of atoms
 How many atoms are in 3.2 mol potassium (K)?
 Remember: 1 mol = 6.02 x 1023 atoms
 This can be written as a conversion factor:
 6.02  10 23 atoms 


1 mol


 6.02  10 23 atoms 
  19.264 x 10 23 atoms of K
3.2 mol K 
1 mol


 1.9 x10 24 atoms of K
How do we use the “Mole” in chemistry?
The atomic mass of an element is the
grams of 1 mole of that atom
Why do chemists use moles?
It’s fun.
It’s impossible to count atoms with your hands.
You can easily measure the mass (in grams) of
a chemical.
Atomic mass = grams of
1 mole of this element,
Cobalt
Convert moles of an atom to grams
 I need 2.0 moles of copper (Cu) for an
experiment. How many grams is that?
 Atomic mass of Cu = 63.55 g/mol (round to 2
decimals)
 “mol” is the abbreviation of “Mole”… I know it’s
only one letter different… chemists!!!
 63.55 g 
2.0 mol Cu 
  127.01
 1 mol 
 130g Cu (2 sig figs! )
Converting grams to moles
 I have 302 grams of silver (Ag). How many
moles of silver do I have?
Step 1: Atomic mass of Ag = 107.87 g/mol
Step 2: Calculate
 1 mol 
  2.79966 mol
302g Ag 
 107.87 g 
 2.80 mol (3 sig figs)
6.2 Molar Mass and Percent Composition
 Atomic Mass = mass of one mole of an
atom
 Molar Mass= mass of one mole of a
substance
Calculate Molecular Weights
Example: Calculate the Molecular Weight
(MW) of RbI2
Step 1: Assume you have 1 mole of this
molecule and determine how much each
element weighs from the periodic table.
Step 2: Determine how many of each element
you have
Step 3: Add all the masses together
Step 1: Find how much each element weighs
from the periodic table
 Rb is atomic # 37. How much does each mole
of Rb weigh?
85.47 grams/mol Rb
 I is atomic # 53. How much does it weigh?
126.90 g/mol I
Step 2: Determine how many of
each element you have
Look at the formula:
We have 1 “Rb” atom
and 2 “I” atoms
RbI2
Step 3: Add all the masses
together
You will need to show this work:
(85.47 g/mol Rb)  (1 Rb)  85.47 g/mol
plus
(126.90 g/mol)  (2 I)  253.80 g/mol
Because the units are the same we can add
these two numbers together, so…
253.80 g/mol + 85.47 g/mol = 339.27 g/mol
339.27 g/mol is the “molar mass”
Converting from moles of a compound to
grams
 Example: I need 3.00 mol NaCl for an
experiment. How many grams is that?
 Step 1: Find the molar mass
Molar mass = 22.09g/mol + 35.45g/mol
= 57.54 g/mol
 Step 2: Use the molar mass like a conversion factor.
 57.54 g 
3.00 mol 
  172.62 g
 1 mol 
 173g NaCl
Converting from grams of a compound to
moles
Example: How many moles are in 10.0 g
of Na2SO4?
Step 1: Find the molar mass.
Molar mass = 142.1 g/mol
Step 2: Use the molar mass like a
conversion factor. You need “grams” on the
bottom of the fraction.
 1 mol 
2
moles of Na 2SO4 = 10.0 g  

7.04

10
mol

 142.1 g 
6.3 Formulas of Compounds
Calculate “percent composition”
Just like any other %
 stuff


 x 100  Percent Compositio n
 total stuff 
Stuff = grams of elements
Calculate “percent composition”
Ex: calculate % of Cu and S in Cu2S
 stuff


 x 100  Percent Compositio n
 total stuff 
 Stuff = grams Cu
(63.55 g/mol Cu)(2 mol Cu) = 127.1g Cu
 Total stuff = grams Cu + grams S
= 127.1 g + 32.07 g = 159.17 g = 159.2 g
 127.1 g 

 x 100  79.84%
 159.2 g 
You should be able to…
Identify an “empirical formula” and a
“molecular formula”
Empirical formula – simplest ratio of atoms of
each element in a compound (whole #’s only)
Molecular formula – actual # of atoms of each
element in a compound
Molecular
H2O2
C3H6
N2O3
C2H6
C3H9
Empirical
HO
CH2
N2O3
?
?
Using % composition to determine a
formula
Law of Definite Composition – Any amount of a
pure compound will always have the same ratio
of masses for the elements that make up that
compound
Ex: H2O is always 88.9% O and 11.1% H by mass
Only the simplest formula (ratio) can be found… in
other words, you can only find empirical formulas
Using % composition to calculate the
formula
Process is as follows:
1.Calculate % by mass of each element
2.Determine mass of each element

Easy if you use 100 g of the chemical
3.Use mass to find the # of moles of each element
4.Find the smallest ratio of the atoms


÷ the number of moles of each element by the element
with the smallest # of moles
Round to the nearest whole #
Example
 A molecule is 75% C & 25% H. Calculate the empirical
formula.
 Using 100g total = 75g C and 25 g H
 Calculate moles of each =
H
 1 mol 
 1 mol 
  6.2 mol
  25mol
75g C 
25 g H 
 12.01g 
 1.01g 
 Ratio = 6.2C : 25H, simplify by ÷ each by 6.2. Whole
number only!!
C
6.2
C

6.2
1
25
H

6.2
 Final ratio ≈ 1C : 4H so CH4
4
C
H
Percent
75%
25%
100 g total
Moles
75g
25g
Ratio
75g C  1 mol

 
 12.01g 
 1.01g 
6.2 mol
25 mol
6.2
6.2
25
6.2
=
1
Formula
25 g H  1 mol  
=
4
CH4
Find the molecular formula
 Ex: C3H6O2 is an empirical formula for a
chemical. The molar mass of the compound is
148 g/mol.
 What is the molecular formula of the
compound??
 Point: The ratio of C:H:O will always be what the
empirical formula shows
 Steps
1. Calculate the empirical formula mass
2. Calculate molar mass/empirical formula mass
3. Multiply your subscripts by that #.
Steps
 1. Calculate the empirical formula mass:
C3H6O2 mass = (3)(12.01) + (6)(1.01) + (2)(16.00)
= 148.09 g/mol
 2. Calculate (molar mass)/(empirical formula mass)
Round to a WHOLE number.
 148 

2
 74 
 3. Multiply the subscripts of the empirical formula by
that number.
C3x2H6x2O2x2 = C6H12O4