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Moles
How can we count how many atoms or
molecules are in a piece of matter if we
can’t see them?
How can we count how many atoms or
molecules are in a piece of matter if they
have different masses?
Moles
What can we measure in the laboratory that
will help us?
What is the “common currency”?
MOLES!
Moles Unit
We will incorporate the following skills and
knowledge:
•
•
•
•
metric system (esp. mass)
scientific/exponent notation
conversions and dimensional analysis
sig figs
 the most important mathematical
tool in chemistry – the mole!
Mole
1 mole = the amount of pure substance
that contains as many particles (atoms,
molecules, or fundamental units) as there
are atoms in exactly 12 grams of carbon12
(agreed upon by chemists and physicists
in 1960/61)
Mole
= 6.02 x 1023 particles
= Avogadro’s number
• Mole comes from molekül (German) =
molecule
• Molecule comes from molecula (New Latin),
meaning very small specimen
Background
Lorenzo Romano Amedeo Carlo Avogadro
(1776-1856)
- Italian physics professor
1811 - Avogadro's hypothesis - now a law
"Equal volumes of gases under the same
conditions have equal numbers of molecules.“
 “universal container”
Background
Late 1800s – chemists – developed scale
of relative atomic masses of gases, based
on 1/16 of the average atomic mass of
oxygen
1920s – physicists – developed relative
atomic masses based on 1/16 of the
oxygen-16 atom
Background
1959-1961 – chemists and physicists agreed to switch to carbon-12 as the
standard, setting its atomic mass at 12
(pragmatic reasons – carbon-12 was the
standard in mass spectroscopy, and close
to chemistry’s oxygen standards)
Moles in conversion factors
number of “particles” > moles
6.02 x 1023 particles
1 mole
mass (g) > moles
average atomic mass in g
1 mole
Moles in conversion factors
number of “particles” > mass (g)
both of the above conversion factors, using
moles as the intermediate, or “common
currency”
number of atoms in a compound
# atoms
1 compound
Math and the Mole
(# particles/mole)
1. How many fingers are there on 1 person?
2. How many fingers are there on 1 dozen
people?
3. How many fingers are there on 3 dozen
people?
Math and the Mole
(# particles/mole)
4. How many fingers are there on 1 mole of
people?
5. How many fingers are there on 3.12
moles of people?
Math and the Mole
(# particles/mole)
6. How many F atoms in 3.12 moles of F?
7. How many moles of F do you have if you
have 2.45 x 1022 atoms of F?
Math and the Mole
(continued)
Hint:
If problem includes numbers of atoms or
other representative particles and moles,
use Avogadro’s number:
6.02 x 1023 particles
If problem includes grams and moles use
the periodic table to find molar mass.
Math and the Mole
(continued)
(mass/mole)
8.How many grams of F are in 3.89 moles of
F?
9.How many moles of F atoms in 45.6 g of
F?
Math and the Mole
(continued)
(# particles/mass)
10. How many F atoms are in 65.8 g F?
11. What is the mass, in grams, of 7.62 x
1024 F atoms?
Math and the Mole
(continued)
(# atoms/compound)
12. How many F atoms are in 3.84 moles
of MoF6 molecules?
A: 1.39 x 1025 F atoms
Molar Mass
Molar mass = mass, in grams, of 1 mole of
a substance
(6.02 x 1023 particles)
expressed in g/mol
Molar mass is numerically equal to
average atomic mass in amus (atomic
mass units).
Molar Mass
So, molar mass of oxygen (O)
= mass of 1 mole of O atoms
= mass of 6.02 x 1023 atoms
= 16.00 g/mol
molar mass of lead (Pb)
= mass of 1 mole of Pb atoms
= mass of 6.02 x 1023 atoms
= 207.2 g/mol
How do amu’s compare to grams?
e.g. 1 aluminum atom weighs 26.98 amu.
By definition,
1 amu = 1/12 mass of the nucleus of 1 C-12 atom
= 1.66 x 10-24 g
How do amu’s compare to grams?
\ mass of 1 Al atom in g is:
26.98 amu x 1.66 x 10-24 g = 4.48 x 10-23 g (per atom)
1 amu
How do amu’s compare to grams?
If I have a sample of Al with a mass of
26.98 g, how many atoms do I have?
26.98 g Al x 1 atom Al = 6.02 x 1023 atoms Al
4.48 x 10-23g
How do amu’s compare to grams?
Or how much does 6.02 x 1023 atoms Al
weigh?
4.48 x 10-23 g
x 6.02 x 1023 atoms Al = 27.0 g
Al atom
How do amu’s compare to grams?
\ Mass of 1 mole of any element is the
“molar mass” of that element.
How do we determine the molar
mass of compounds?
Add up the molar masses of all elements
in the compound, taking into account the
number of moles of each element.
How do we determine the molar
mass of compounds?
e.g. What is the molar mass of Na3PO4?
3 moles of Na @ 22.99 g/mol = 3 mol x 22.99 g
mol
1 mole of P @ 30.97 g/mol
= 1 mol x 30.97 g
mol
4 moles of O @ 16.00 g/mol = 4 mol x 16.00 g
mol
Add these together:
= 163.94 g = molar mass of Na3PO4
How do we use molar mass of a
compound in a conversion
problem?
The same way we use molar mass of an
atom…
How do we use molar mass of a
compound in a conversion
problem?
(#atoms/compound and molar mass of compounds)
13. How many Na atoms are in 252 g of Na3PO4?
(A: 2.78 x 1024 Na atoms)
Where we’ve been
(today’s lab quiz; Quiz 6 – next Th/F)
• # particles-mole and mole-# particles conversions
• mass-mole and mole-mass conversions
• # atoms-molecule/formula unit conversion
• molar mass of compounds
• combination of two or more of above conversions
Where we’ve been
(today’s lab quiz; Quiz 6 – next Th/F)
• Use the sample problems in your notes,
Moles WS #1, your bookwork and the
warmups to help you prepare for Quiz 6.
• Come in for ICP (T/W lunch) for more
practice.
Where we are going next –
applications of mole conversions
• % composition by mass from a chemical formula (today)
(Moles WS #2)
• Find the empirical chemical formula from % composition
data (Moles WS #3 & 4)
•
Find the molecular formula from % composition and
empirical formula (Moles WS #4)
• Practice problems for all moles material (Moles WS #5)
Where we are going next –
applications of mole conversions
• Calculate molarity of a solution (Moles WS #6 & 7)
• Find the mass of a solute or volume of a solution using
molarity (Moles WS #6 &7)
• Find the mass percent of a solution (Moles WS #7)
Calculations relating to Chemical
Formulas
1. Determine the molar mass of a compound:
- Find molar masses of all elements.
- Multiply each mass taking into account the
number of atoms of that element present in the
formula.
- Add all masses together.
e.g. What is the molar mass of Pb3(PO4)2?
Calculations relating to Chemical
Formulas
1. Determine the molar mass of a compound:
- Find molar masses of all elements.
- Multiply each mass taking into account the
number of atoms of that element present in the
formula.
- Add all masses together.
e.g. What is the molar mass of Pb3(PO4)2?
811.5 g/mol
Calculations relating to Chemical
Formulas
2. Determine percent composition of all elements
present in a compound:
- Find the molar masses of each element in the
compound.
- Find the total molar mass.
Calculate:
% composition = molar mass of element x 100%
total molar mass of compound
Calculations relating to Chemical
Formulas
e.g.
What is the % composition of NaCl? CaCl2?
(assume % composition by mass)
Apply % composition:
e.g. What mass of Na is present in 200.0 g NaCl?
Calculations relating to Chemical
Formulas
e.g.
What is the % composition of NaCl? CaCl2?
(assume % composition by mass)
NaCl: 39.34% Na, 60.66% Cl
CaCl2: 36.11% Ca, 63.89% Cl
Apply % composition:
e.g. What mass of Na is present in 200.0 g NaCl?
200.0 g x 0.3934 = 78.68 g Na
3. Convert % composition (mass
percent) empirical formula
a. Assume a 100 g sample – use same
numbers as grams rather than %.
b. Perform mass  mole conversions.
c. Divide each result (mole) by the smallest
result present (mole ratio).
d. Look for whole number ratio.
3. Convert % composition (mass
percent) empirical formula
Rhyme to remember order of steps to convert
% composition  empirical formula:
Percent to mass
Mass to mole
Divide by small
Times till whole
Practice
1) One of the components of fresh
alkaline batteries is a black powdery
compound, of 63% manganese
and 37% oxygen.
What is the compound’s empirical
formula?
Practice
2) While analyzing a dead alkaline battery,
Antonio finds a compound of 70.0%
manganese and 30.0% oxygen.
What is its empirical formula?
Empirical vs. Molecular Formula
Compound
Empirical
formula
Empirical
Molecular
molar mass molar mass
Molecular
formula
Formaldehyde
CH2O
30.03 g
30.03 g
CH2O
Acetic
acid
CH2O
30.03 g
60.06 g
C2H4O2
Glucose
CH2O
30.03 g
180.18 g C6H12O6
To find the molecular formula of glucose,
divide the molecular molar mass by the
empirical molar mass, round to the nearest
whole number.
(molecular) molar mass = 180.18 g = ~ 6
empirical molar mass
30.03 g
then multiply subscripts by 6 => C6H12O6
to get the molecular formula
Practice Problems – Determining
molecular formula
Hydrazine is 87.42% N and 12.58% H.
The (molecular) molar mass of hydrazine is 32.0 g/mol.
a) What is its empirical formula?
b) What is its molecular formula?
(Hint: find the molar mass of the empirical formula)
Practice Problems – Determining
molecular formula
Hydrazine is 87.42% N and 12.58% H.
The (molecular) molar mass of hydrazine is 32.0 g/mol.
a) What is its empirical formula? (NH2)
b) What is its molecular formula? (N2H4)
(Hint: find the molar mass of the empirical formula)
Describing solution concentration –
Percent by Mass
Mass % of component
= mass of component in solution x 100%
total mass of solution
Describing solution concentration –
Percent by Mass
e.g.
In order to maintain a sodium chloride
(NaCl) concentration similar to ocean
water, an aquarium must contain 3.6 g
NaCl per 100.0 g of water. What is the
percent by mass of NaCl in the solution?
Describing solution concentration –
Percent by Mass
3.6 g NaCl
x 100% = 3.6 g NaCl x 100%
100.0 g H2O + 3.6 g NaCl
= 3.5%
103.6 g total
Describing solution concentration –
Percent by Mass
e.g.
A solution contains 2.7 g of CuSO4 in 75
mL of solution. Assume the density of the
solution is 1.0 g/mL.
What is the mass percent of the solution?
2.7 g (1 mL) x 100% = 3.6%
75 mL 1.0 g
Molarity
(Moles WS #6)
• One way to measure concentration
= the # moles solute dissolved in 1 L of
solution
Molarity ( M ) = moles of solute
1 liter of solution
Molarity
•
Divide # moles by # liters.
•
If the problem gives you grams,
first convert to moles by dividing by
molar mass.
•
Also, don't forget to convert mL to L.
Example
A saline solution contains 0.90 g of NaCl
in 100.0 mL. What is the molarity?
(Molar mass of NaCl is 58.5 g.)
•
Begin by converting the solute and
solution into the correct units.
Example
Begin by converting the solute and solution into
the correct units.
Solute:
0.90 g (1 mol NaCl) = 0.015 mol NaCl
58.5 g NaCl
Solution:
100.0 mL (1 L) = 0.1000 L
103 mL
Example
Next, divide the number of moles by the
volume of the solution:
• Molarity = 0.015 mol NaCl = 0.15 M NaCl
0.1000 L
Example
OR, in one step:
0.90 g (1 mol NaCl) (103 mL) = 0.015 M NaCl
100.0 mL 58.5 g NaCl
1L
Molarity
• You can also use the molarity equation to
calculate moles (and grams) or volumes
(measured in L or mL).
If M = mol
L
then M x L = # moles
and L = mol
M
Molarity
• Notice that all three of the equations
above are just different algebraic versions
of each other.
• Use them to solve the next few problems,
or use standard conversion techniques:
Example of # moles = M x L
How many moles of CaCl2 are in 250. mL
of 2.0 M CaCl2?
How many grams is this?
Begin with the value you are given in the
problem (250. mL), then use molarity of
CaCl2 (2.0 M) as a conversion factor.
Example of # moles = M x L
250. mL (1 L) (2.0 mol CaCl2)
103 mL
1L
molarity as conversion factor
= 0.50 mol CaCl2
then
0.50 mol CaCl2 (110.08 g CaCl2) = 55.0 g CaCl2
1 mol CaCl2
Example of # moles = M x L
OR, all in one step:
250. mL (1 L) (2.0 mol CaCl2) (110.08 g CaCl2)
103 mL
1L
1 mol CaCl2
molarity as conversion factor
= 55.0 g CaCl2
Example of : # L (or mL) = mol
M
• How many mL contain 6.25 g of 2.00 M
CaCl2?
• Begin with the value you are given in the
problem (6.25 g), then use molarity of
CaCl2 (2.00 M) as a conversion factor.
Example of
# L (or mL) = mol
M
6.25 g CaCl2 (1 mol CaCl2)
= 0.0568 mol CaCl2
110.08 g CaCl2
then
0.0568 mol CaCl2 (1 L
) (103 mL) = 28.4 mL
2.00 mol
1L
Note upside down molarity
Making Dilutions (Honors only)
• Chemists often use concentrated (or
“stock”) solutions to make dilute solutions.
**The number of moles of solute does
not change when a solution is diluted.**
Number of moles before dilution =
Number of moles after dilution
Making Dilutions
• Since moles = Molarity (M) x liters ( V ),
then: Mc x Vc = Md x Vd
e.g.
How would you prepare 100. mL of 0.40 M
MgSO4 from a solution of 2.0 M MgSO4 ?
Mc=2.0 M
Vc= ????
Md = 0.40 M Vd = 100. mL
Making Dilutions
Rearrange the equation:
Vc = Md x Vd
Mc
Vc = 0.40 M x 100. mL
2.0 M
= 20. mL of the concentrated solution
Making Dilutions
So, you would measure out 20. mL of the
original solution, then add enough water to
it to bring the volume to 100. mL.
Important: You can do these dilution
problems in either mL or L, but the V1 and
V2 must both be in the same units (either
both mL or both L; don't mix them up).