Download The Mole - Hobbs High School

The Mole
What is a mole?
• Well, yes, but we’re
not discussing
biology or
dermatology now.
• We want the
CHEMIST’S mole.
Relating Mass to Numbers of Atoms
• Recall that we defined the amu in terms of the
carbon-12 atom’s mass. Now we want to
relate this to the number of atoms that
corresponds to this mass.
• Chemists devised a counting unit called the
MOLE. You use counting units all the time—
dozen, pair—to mean a specific number.
• We are going to introduce the MOLE, the SI
unit for amount of substance.
The Mole
A mole is the amount of a substance that
contains as many particles as there are atoms
in exactly 12 g of carbon-12.
A mole is 6.022 x 1023 units of a substance.
That’s 602 000 000 000 000 000 000 000 water
molecules or
602 000 000 000 000 000 000 000 atoms of
copper.
This number is commonly known as Avogadro’s
number, named for Amedeo Avogadro, the 19th
Century Italian chemist.
Molar Mass
• The MOLAR MASS of a PURE substance is the
mass of one mole (6.022 x 1023 units) of that
substance.
• The mass in grams of one mole of an element
is equal to the atomic weight of that element.
What is the molar mass of carbon?
What is the atomic weight of carbon?
The molar mass is 12.01 grams.
1 mol C = 12.01 g C or 12.01 g/mol C
So 12.01 g C contains 1 mole of atoms.
Practice
Give the molar mass of (include units!!):
a) Ca
40.08 g/mol
2 x 19.00g/mol = 38.00 g/mol
b) F2
c) Xe
131.29 g/mol
What is the mass of 2 moles of nitrogen?
2 x 14.01 g/mol = 28.02 g/mol
What is the mass of 0.5 moles of rubidium?
0.5 x 85.47 g/mol = 42.74 g/mol
But what about molecules?
• Well, one mole of water (H2O) has 6.022 x 1023
molecules in it, because the unit here is a
‘molecule’, not an atom.
• To find the molar mass of water, we need the
atomic masses of the elements that make up
water: hydrogen and oxygen.
1 mol H = 1.0079 g H = 1.01 g H
1 mol O = 16.00 g O
Molar Mass of Water
In each mole of water, there are two moles of H
2 mol H x 1.01 g H = 2.02 g H
1 mol H
and one mole of O
1 mol O x 16.00 g O = 16.00 g O
1 mol O
Therefore the molar mass of H2O is the sum
= 2.02 g + 16.00g = 18.02 g
or 18.02 g/mol H2O
Molar Masses of Diatomic Molecules
• What is the molar mass of O2?
What is the atomic weight of oxygen?
16.00 g
The subscript tells us we have two atoms of
oxygen in each molecule, so the molar mass
is
2 mol O x 16.00 g O = 32.00 g O
1 mol O
so 1 mol O2 = 32.00 g O2 or 32.00 g/mol O2
What is the molar mass of carbon dioxide (CO2)?
Start with the molar mass of each element:
1 mol C = 12.01 g C
1 mol O = 16.00 g O
From the compound’s formula, we see that we
need 1 mole of carbon and 2 moles of oxygen:
1 mol C x 12.01 g C = 12.01 g C
1 mol C
2 mol O x 16.00 g O = 32.00 g O
1 mol O
So the molar mass of CO2 = (12.01 g + 32.00g)
= 44.01 g CO2 or 44.01 g/mol CO2
Practice
What is the molar mass of
1. NaCl?
2. CaCl2?
3. Fe2O3?
4. Ca3(PO4)2?
Moles to Gram Conversions
What is the mass in grams of 3.50 mol of the
element copper, Cu?
Given: 3.50 mol Cu
Unknown: Mass of Cu in grams
PLAN:
• We need to convert the amount of Cu in
moles into mass of Cu in grams.
• We multiply the amount of moles by the
molar mass (g/mol) to find the amount in
grams.
𝑔𝑟𝑎𝑚𝑠 𝐶𝑢
𝑚𝑜𝑙𝑒𝑠 𝐶𝑢 ×
= 𝑔𝑟𝑎𝑚𝑠 𝐶𝑢
𝑚𝑜𝑙𝑒𝑠 𝐶𝑢
Remember, for an element, the molar mass
is just the atomic mass: here 63.55 g for
copper.
molar mass
63.55 𝑔 𝐶𝑢
3.50 𝑚𝑜𝑙 𝐶𝑢 ×
= 222 𝑔 𝐶𝑢
1 𝑚𝑜𝑙 𝐶𝑢
Moles to Grams Practice
Convert moles to grams:
A. 2 moles Co
1 mol Co = 58.93 g Co
58.93 𝑔 𝐶𝑜
2 𝑚𝑜𝑙 𝐶𝑜 𝑥
= 117.86 g Co
1 𝑚𝑜𝑙 𝐶𝑜
B. 5 moles H2
2.016 𝑔 𝐻2
1𝑚𝑜𝑙 𝐻2
2 𝑥 1.008 𝑔 𝐻2
1 𝑚𝑜𝑙 𝐻2
2 𝑥 1.008 𝑔 =
5 𝑚𝑜𝑙 𝐻2 ×
= 10.08 𝑔 𝐻2
C. 2.1 moles H2O
2 x 1.008 + 16.00 = 18.02 g H2O/1 mol H2O
2.1 𝑚𝑜𝑙 𝐻2 O ×
18.02 𝑔 𝐻2 𝑂
1 𝑚𝑜𝑙 𝐻2 𝑂
= 37.84 𝑔 𝐻2 O
Grams to Moles Conversions
A chemist produced 11.9 grams of aluminum.
How many moles of aluminum is this?
Given: 11.9 g Al
Unknown: amount of Al in moles
PLAN: mass of Al in grams → moles of Al
𝑚𝑜𝑙𝑒𝑠 𝐴𝑙
𝑔𝑟𝑎𝑚𝑠 𝐴𝑙 ×
= 𝑚𝑜𝑙𝑒𝑠 𝐴𝑙
𝑔𝑟𝑎𝑚𝑠 𝐴𝑙
1 𝑚𝑜𝑙 𝐴𝑙
11.9 𝑔 𝐴𝑙 ×
= 0.441 𝑚𝑜𝑙 𝐴𝑙
26.98 𝑔 𝐴𝑙
This is just the inverse
of the molar mass!
Grams to Moles Conversions
Another way to view this conversion is as
DIVIDING the mass of aluminum in grams by the
molar mass to get the number of moles of Al:
11.9 𝑔 𝐴𝑙
= 0.441 𝑚𝑜𝑙 𝐴𝑙
26.98 𝑔 𝐴𝑙
1 𝑚𝑜𝑙 𝐴𝑙
Mathematically, it’s the same!
Grams to Moles Practice
Find the number of moles for these masses:
A. 54.2 g N2
14.01 x 2 = 28.02 g N2/1 mol N2
54.2 𝑔 𝑁2 ×
1 𝑚𝑜𝑙 𝑁2
28.02 𝑔 𝑁2
= 1.93 𝑚𝑜𝑙 𝑁2
B. 202.2 g MgO
24.30 g + 16.00 = 40.30 g MgO/1 mol MgO
202.2 𝑔 𝑀𝑔𝑂 ×
1 𝑚𝑜𝑙 𝑀𝑔𝑂
40.3 𝑔 𝑀𝑔𝑂
= 5.02 𝑚𝑜𝑙 𝑀𝑔𝑂
C. 519.3 g FeCl3
55.85 + 3(35.45) = 162.2 g FeCl3/1 mol FeCl3
519.3 𝑔 𝐹𝑒𝐶𝑙3 ×
1 𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3
162.2 𝑔 𝐹𝑒𝐶𝑙3
= 3.20 𝑚𝑜𝑙 𝐹𝑒𝐶𝑙3
What?!? I have to enter Avogadro’s number
EVERY time?!?!!
Want to avoid that?? From the home screen
1) Enter Avogadro’s number into your calculator.
2) Press STO→
3) Press ALPHA and then whatever key
corresponds to the letter you want
4) ENTER
To use this feature, enter the number by which
you’ll be multiplying/dividing Avogadro’s number,
then press ALPHA [your letter] to multiply or ÷
ALPHA [your letter]
Using registers on your calculator
6.022 ×1023 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙𝑒
For instance, 4.32 𝑚𝑜𝑙𝑒𝑠 ×
would be entered like this:
4.32 ALPHA Q ENTER [if I put Avogadro’s
number in register Q]
which equals 2.60 x 1024 [2.6E24 on your
calculator]
Moles to Number of Atoms
Here’s where Avogadro’s number comes in handy:
How many moles of silver (Ag) are in 3.01 x 1023
atoms of silver?
Given: 3.01 x 1023 atoms
Unknown: amount of Ag in moles
𝑚𝑜𝑙𝑒𝑠 𝐴𝑔
𝑎𝑡𝑜𝑚𝑠 𝐴𝑔 ×
= 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔
𝐴𝑣𝑜𝑔𝑎𝑑𝑟𝑜′𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐴𝑔
3.01𝑥1023
1 𝑚𝑜𝑙𝑒 𝐴𝑔
𝑎𝑡𝑜𝑚𝑠 𝐴𝑔 ×
= 0.500 𝑚𝑜𝑙𝑒𝑠 𝐴𝑔
23
6.022𝑥10 𝑎𝑡𝑜𝑚𝑠 𝐴𝑔
Moles to Number of Atoms Practice
Determine how many atoms/molecules or moles in the
following samples:
A. 1.32 moles Si
6.022 𝑥 1023 𝑎𝑡𝑜𝑚𝑠
1.32 𝑚𝑜𝑙 𝑆𝑖 ×
= 7.95 × 1025 𝑎𝑡𝑜𝑚𝑠
1 𝑚𝑜𝑙
B. 7.29 mol CO2
6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠
7.29 𝑚𝑜𝑙 𝐶𝑂2 ×
= 4.39 × 1024 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠𝐶𝑂2
1 𝑚𝑜𝑙
C. 3.33 x 1028 formula units AuCl
3.33 ×
1028
form. units AuCl ×
= 5.53 × 104 moles
1 𝑚𝑜𝑙
6.022 ×1023 𝑓𝑜𝑟𝑚.𝑢𝑛𝑖𝑡𝑠
Conversions
Mass of
element or
molecule
in grams
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
=
×
1 𝑚𝑜𝑙
1 𝑚𝑜𝑙
×
=
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Amount of
element or
molecule in
moles
=
1 𝑚𝑜𝑙
×
6.022 𝑥 1023 𝑢𝑛𝑖𝑡𝑠
6.022 𝑥 1023 𝑢𝑛𝑖𝑡𝑠
×
=
1 𝑚𝑜𝑙
Number of
atoms of
element or
molecule
How to Calculate Percent Mass
• Let’s use carbon dioxide as an example.
First, we need to know the molar mass of CO2:
1 mol C x12.01g/mol C+ 2 mol O x16.00g/mol O= 44.01 g/mol CO2
Second, now find the percentage of each element in the compound.
Carbon:
Mass of carbon
% carbon =
x 100%
Mass of CO2
12.01
% carbon =
x 100%  27.3%
44.01
Oxygen:
Mass of oxygen
% oxygen =
x 100%
Mass of CO2
32.00
% oxygen =
x 100%  72.7%
44.01
NOTE:
The total
percent
should
equal
100%
Practice
1. PbCl2
2. Ba(NO3)2
3. Ammonia
Determining Chemical Formulae
• When a new substance is discovered or
synthesized, it is analyzed quantitatively to
reveal its percentage compositions.
• From these data, the empirical formula is then
determined.
An empirical formula consists of the symbols for
the elements combined in a compound, with
subscripts showing the smallest whole-number
mole ratio of the different atoms in the compound.
Empirical Formulae
• For an IONIC COMPOUND, the formula unit is
usually the compound’s empirical formula.
• For a MOLECULAR COMPOUND, unfortunately,
the empirical formula doesn’t necessarily
show the actual numbers of atoms present in
each molecule.
For instance, the gas diborane has an empirical
formula of BH3 but the molecular formula is
B2H6. We will only focus on empirical formulas.
Calculation of Empirical Formulae
Starting with a compound’s percentage composition,
you convert this to a mass composition, then a molar
composition.
STEP 1: Using diborane again as an example, the
percentage composition is 78.1% B and 21.9% H. Using
a 100.0g sample as a basis, this would give 78.1 g of B
and 21.9 g of H.
STEP 2: Convert these masses into moles:
1 𝑚𝑜𝑙 𝐵
78.1𝑔 𝐵 ×
= 7.22 𝑚𝑜𝑙 𝐵
10.81 𝑔 𝐵
1 𝑚𝑜𝑙 𝐻
21.9𝑔 𝐻 ×
= 21.7 𝑚𝑜𝑙 𝐻
1.01 𝑔 𝐻
Calculation of Empirical Formulae
STEP 3: These values give a mole ratio of
7.22 𝑚𝑜𝑙 𝐵 ∶ 21.7 𝑚𝑜𝑙 𝐻
but this is not a ratio of smallest whole numbers. To
find such a ratio, divide each number of moles by the
smaller/smallest number in the existing ratios:
7.22 𝑚𝑜𝑙 𝐵 21.7 𝑚𝑜𝑙 𝐻
:
= 1 𝑚𝑜𝑙 𝐵: 3.01 𝑚𝑜𝑙 𝐻
7.22
7.22
The molar ratios may not be exact whole numbers due to
rounding or experimental error. You get to ignore these
differences and assume they are exact whole numbers!
Here “3.01” is assumed to be exactly 3!
Calculation of Empirical Formulae
STEP 4: Convert the mole ratio into the
empirical formula
1 𝑚𝑜𝑙 𝐵: 3 𝑚𝑜𝑙 𝐻
Which means that in this compound, for every 1
mole of boron, there are three moles of
hydrogen, so the empirical formula is
𝐵𝐻3
Calculation of Empirical Formulae
Example 1: Quantitative analysis shows that a
compound contains 32.38% sodium, 22.65%
sulfur, and 44.99% oxygen. Find the empirical
formula of this compound.
Plan: Convert the percentage composition into
mass composition (based on a 100 g sample),
then convert into moles, and then find the
smallest whole-number mole ratio of atoms.
STEP 1: Convert % to mass:
For a 100 g sample, we would have:
32.38 g Na, 22.65 g S, and 44.99 g O.
For a 100 g sample, we would have:
32.38 g Na, 22.65 g S, and 44.99 g O.
STEP 2: Convert these masses into moles:
1 𝑚𝑜𝑙 𝑁𝑎
32.38 𝑔 𝑁𝑎 ×
= 1.408 𝑚𝑜𝑙 𝑁𝑎
22.99 𝑔 𝑁𝑎
1 𝑚𝑜𝑙 𝑆
22.65 𝑔 𝑆 ×
= 0.7063 𝑚𝑜𝑙 𝑆
32.07 𝑔 𝑆
1 𝑚𝑜𝑙 𝑂
44.99 𝑔 𝑂 ×
= 2.812 𝑚𝑜𝑙 𝑂
16.00 𝑔 𝑂
STEP 3: Find the smallest whole-number ratio of atoms
by dividing all by the smallest number of moles:
1.408 𝑚𝑜𝑙 𝑁𝑎 0.7063 𝑚𝑜𝑙 𝑆 2.812 𝑚𝑜𝑙 𝑂
:
:
0.7063
0.7063
0.7063
1.408 𝑚𝑜𝑙 𝑁𝑎 0.7063 𝑚𝑜𝑙 𝑆 2.812 𝑚𝑜𝑙 𝑂
:
:
0.7063
0.7063
0.7063
This reduces to
1.993 mol Na : 1 mol S : 3.981 mol O
Rounding each number to the nearest whole
number gives us
2 mol Na : 1 mol S : 4 mol O
STEP 4: Write the formula.
So the empirical formula of the compound is
therefore
Na2SO4
Calculation of Empirical Formulae
Example 2: Analysis of a 10.150 g sample of a
compound known to contain only phosphorus and
oxygen gives a phosphorus content of 4.433 g.
What is the empirical formula of this compound?
Plan: Use the masses of the sample and the
phosphorus content to find the mass of the oxygen
content. Then convert these masses to moles, and
then find the smallest whole-number mole ratio of
atoms.
Givens:
10.150 g sample
4.433 g phosphorus
Subtracting gives an oxygen mass of 5.717 g
Now convert mass to moles:
1 𝑚𝑜𝑙 𝑃
4.433 𝑔 𝑃 ×
= 0.1431 𝑚𝑜𝑙 𝑃
30.97 𝑔 𝑃
1 𝑚𝑜𝑙 𝑂
5.717 𝑔 𝑂 ×
= 0.3573 𝑚𝑜𝑙 𝑂
16.00 𝑔 𝑂
Find the smallest whole-number ratio of atoms:
0.1431 𝑚𝑜𝑙 𝑃 0.3573 𝑚𝑜𝑙 𝑂
:
0.1431
0.1431
1 𝑚𝑜𝑙 𝑃: 2.497 𝑚𝑜𝑙 𝑂
1 𝑚𝑜𝑙 𝑃: 2.497 𝑚𝑜𝑙 𝑂
Hmm, not as neat as before. We need the smallest
WHOLE-number ratio here. We can round this to
1 𝑚𝑜𝑙 𝑃: 2. 5 𝑚𝑜𝑙 𝑂
but certainly not up to 3 or down to 2. Now what?
Aha, if we multiply both numbers by 2, we get
2 𝑚𝑜𝑙 𝑃: 5 𝑚𝑜𝑙 𝑂
which is a whole-number ratio. The empirical
formula is therefore
𝑃2 𝑂5
Consider the following …
• What if you get a whole number ratio of
2.33 𝑚𝑜𝑙 𝐶: 2 𝑚𝑜𝑙 𝐻 ?
You should recognize ‘.33’ as the decimal equivalent
of one-third (1/3). Multiply through by 3 to get
7 𝑚𝑜𝑙 𝐶: 6 𝑚𝑜𝑙 𝐻
• What if you get a whole number ratio of
1.75 𝑚𝑜𝑙 𝐾: 3 𝑚𝑜𝑙 𝑆 ?
You should recognize ‘.75’ as the decimal equivalent
of three-fourths (3/4). Multiply through by 4 to get
7 𝑚𝑜𝑙 𝐾: 12 𝑚𝑜𝑙 𝑆
Molecular Formulae
• We have just practiced finding empirical formulas
that contain the smallest possible whole-number
ratios of atoms.
• The molecular formula is the actual formula of a
molecular compound. The empirical formula
may or may not be a correct molecular formula.
• We used diborane earlier as an example of this
(empirical formula of BH3 but the molecular
formula is B2H6). Ethene, C2H4, and
cyclopropane, C3H6, have the same atomic ratio
(1C:2H) yet they are very different substances.
Relationship between Empirical
and Molecular Formulas
𝑥 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
The number represented by x is a whole number
indicating the factor by which subscripts in the
empirical formula are multiplied to get the molecular
formula.
So we can find the factor x by comparing the formula
masses (same as molar mass for the compound).
𝑥 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎r 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑥=
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
Calculation of Molecular Formula
Example 3: In the last problem, the empirical
formula of a compound of phosphorus and
oxygen was found to be 𝑃2 𝑂5 . Experimentation
shows that the molar mass of this compound is
283.89 g/mol. What is the compound’s
molecular formula?
Plan: Using the empirical formula, find the
empirical molar mass and compare it to the
molecular formula mass to find the multiplying
factor.
Empirical molar mass (based on 𝑃2 𝑂5 ):
P has a molar mass of 30.97 g
O has a molar mass of 16.00 g
30.97g/mol x 2 + 16.00 g/mol x 5 = 141.94 g/mol
𝑥 𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 = 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
𝑥=
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
283.89 𝑔/𝑚𝑜𝑙
𝑥=
= 2.0001
141.94 𝑔/𝑚𝑜𝑙
Therefore the compound’s molecular formula is
2 × 𝑃2 𝑂5 = 𝑃4 𝑂10
Example 4: If 4.04 g of N combine with 11.46 g
of O to produce a compound with the formula
mass of 108.0 amu, what is the molecular
formula of the compound?
1 𝑚𝑜𝑙 𝑁
4.04 𝑔 𝑁 ×
= 0.2884 𝑚𝑜𝑙 𝑁
14.01 𝑔 𝑁
1 𝑚𝑜𝑙 𝑂
11.46 𝑔 𝑂 ×
= 0.71625 𝑚𝑜𝑙 𝑂
16.00 𝑔 𝑂
0.2884 𝑚𝑜𝑙 𝑁 0.71625 𝑚𝑜𝑙 𝑂
:
=
0.2884
0.2884
1 mol N: 2.4835 mol O
The empirical formula is thus N2O5.
The empirical formula is thus N2O5.
The empirical molar mass would be:
Nitrogen 14.01 g/mol
Oxygen 16.00 g/mol
2 x 14.01 g/mol + 5 x 16.00 = 108.02 g/mol
The empirical formula mass is 108.02 amu.
𝑥=
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
=
108.0 𝑎𝑚𝑢
108.02 𝑎𝑚𝑢
= 0.9998
Therefore, the molecular formula is the same as
the empirical formula, N2O5.