Moles!

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Moles!
Whatcha talkn’ about a mole?
First…a dozen.
12 cookies =
12 bagels =
12 cans =
12 footballs =
12 dance moves =
1 dozen
1 dozen
1 dozen
1 dozen
1 dozen
As scientist we need to determine something to
quantify atoms, molecules, particles, or something very,
very small
As scientist we don't work with dozen....we work with
something called a mole!
Back to original question: "What the heck is a mole?!!"
Same concept as a dozen but larger....much larger.
If I want to have a mole of cookies….
6.022 x 1023 times!!
I mole of cookies = 6.022 x 1023 cookies
What about a mole of dance moves?
6.022 x 1023 times!!
I mole of dance moves= 6.022 x 1023 dance moves
Mole in Chemistry?
Obviously in chemistry you don’t have
cookies and dance moves….we have
 Atoms
 Molecules
 Particles
 Formula units (ionic compounds)
 Electrons
Abbreviation for mole is mol
Moles!
1 mole of atoms =
6.022 x 1023 atoms
1 mole of molecules = 6.022 x 1023 molecules
1 mole of particles =
6.022 x 1023 particles
1 mole of formula units = 6.022 x 1023 formula units
1 mole of electrons =
6.022 x 1023 electrons
6.022 x 1023 is called Avogadro number
(named after Amadeo Avogadro)
So what does this mean?

Converting….DA!!! Yeah!
1. How many moles are in 4.56 x 1024 atoms of hydrogen?
4.56 x 1024 atoms
1 mole
6.022 x 1023 atoms
= 7.57 moles
2. How many formula units are in 8.92 moles of NaCl?
8.92 moles
6.022 x 1023 formula units
1 mole
= 5.37 x 1024 formula units
More moley practice…
1. How many atoms of oxygen are in 5.00
moles of carbon dioxide (CO2)?
5.00 mol CO2
6.022 x 1023 molecules CO2 2 atoms O
1 mole of CO2
= 6.02 x 1024 atoms of O
1 molecule CO2
Why moles, not atoms?
1 atom of Al = 4.48 x 10-23 grams
 Using moles to work with easier numbers
1 mole of Al = 26.98 grams
 Which is called molar mass
◦ Mass (in grams) of a mole
Guess what….another conversion
Yeah!! Boom! No doubt!

Molar Mass
1. Determine the molar mass of Sucrose
(C12H22O11)
C – 12(12.01)
= 342.29 g/mol
H – 22(1.008)
O – 11(15.999)
2. Determine the molar mass of Ca(NO3)2
Ca – 1(40.078)
N – 2(14.0067)
= 164.09 g/mol
O – 6(15.999)
Converting Moles to grams or
grams to moles
All conversions come from molar mass…aka
periodic table
1. How many gramsMolar
of aluminum are in 4.89
mass of Al
moles of Al?
4.89 mol of Al
26.98 grams of Al
1
2.
= 132 g of Al
mol of Al
Molar mass
How many moles
of CaCl2 are in 13.5 g of
of CaCl
CaCl2?
2
13.5 g of CaCl2
1 mol of CaCl2
110.984 g of CaCl2
= 0.122 mol
of CaCl2
Practice
1.
How many moles are in 25.67 grams of
nickel?
0.4374 mol of Ni
2.
How many grams are in 0.234 moles of
lead(II) iodide (PbI2)?
108 grams of PbI2
More Practice
1.
How many atoms are in 14.0 grams of
manganese?
1.53 x 1023 atoms of Mn
2.
How many molecules are in 46.7 grams
of nitrogen gas?
1.00 x 1024 molecules of N2
3.
How many oxygen atoms are in 3.45
grams of CO2?
9.44 x 1022 atoms of O
Molarity

What are these bottles we keep working
with numbers and a big M?
Well, Molarity is how scientist describe concentration of a
solution
Molarity = how many moles per 1 liter of solution
Molarity
What does this mean?
For example: 2.00 M HCl
2.00 moles of HCl per 1 liter OR
2.00 moles of HCl = 1 L
That’s right another conversion factor!! Yeah!
Other examples
1.
0.500 M of Pb(NO3)2
0.500 moles of Pb(NO3)2 = 1 L
2.
1.56 M of NaCl
1.56 moles of NaCl = 1 L
Using molarity…All about the DA
1.
A 250. mL container has a solution of NaCl with a
molarity of 1.54 M. How many moles of NaCl is
present?
250. mL
1L
1.54 mol
1000 mL
2.
1
How many grams
L
= 0.385 mol of NaCl
Molar mass
ofNaCl
NaCl would
of
58.443 g
1
mol
that contain?
= 22.5 g of NaCl
Why finding mass?
You need the mass…
- how much you need to measure in the
lab (because you can’t measure moles on
a scale)
- make accurate concentrations of your
solutions
Practice
1.
How many moles are present in 500.0 mL of
0.800 M solution of HCl?
500. 0 mL
1L
1000 mL
2.
0.800 mol
= 0.400 mol of HCl
1L
mass mL of
How many grams are present Molar
in 250.
of Pb(NO3)2
0.500 M solution of Pb(NO3)2?
250. 0 mL
1L
1000 mL
0.500 mol
1L
331.21 g Pb(NO3)2
1 mol
= 41.4 g of Pb(NO3)2
More Practice…
1.
How many grams are needed to make 4.00 L
of 0.250 M solution of NaCl?
= 58.4 g of NaCl
2.
How many grams are needed to make 2.50 L
of 0.500 M solution of Na2CO3?
= 132 g of Na2CO3
More Practice…
Complete the table:
Formula
Grams
Dissolved
NaCl
4.00 grams
KOH
?
Moles
Dissolved
?
2.56 x 10-3
Volume of Molarity
Solution
1.50 L
?
1.80 L
?
Moles NaCl = 0.0684 mol
Grams KOH = 0.144 g
Molarity of NaCl = 0.0456 M
Molarity of KOH= 0.00142 M
Making Solutions
1.
Obtain correct volumetric flask
Determine the correct number of grams
needed to measure
3. Add the massed out compound to flask
4. Rinse weigh boat with DI into flask
5. Add DI to etched line on flask
2.
Example
1.
How many grams are needed Molar
to make
mass 250 mL
of 0.250 M solution of CuSO4of
? CuSO4
250. 0 mL
1L
1000 mL
0.250 mol
1L
159.608 g CuSO4
1 mol
= 9.98 g of CuSO4
-Mass out 9.98 grams of copper(II) sulfate
-Hot dog weigh boat and use DI water to add CuSO4 to 250
mL volumetric flask
-Add DI water to 250 mL mark
-Add stir bar and stir till all dissolves
-Add solution to appropriate containers
Percent Mass and Determining
Compound Formulas

Find percent of each element in the
compound

Define Molecular and Empirical Formulas
Mass Percent
What percent of each element in a compound?
Example:
Find the percent of each element in ethanol (C2H5OH).
C – (2)(12.01) = 24.02
H – (6)(1.008) = 6.048
O – (1)(15.999) = 15.999

46.07 g/mol
%C=
24.02 g/mol
46.07 g/mol
x 100 = 52.14% of C
Mass Percent
What percent of each element in a compound?
Example:
Find the percent of each element in ethanol (C2H5OH).
C – (2)(12.01) = 24.02
H – (6)(1.008) = 6.048
O – (1)(15.999) = 15.999

46.07 g/mol
%H=
6.048 g/mol
46.07 g/mol
x 100 = 13.13% of H
Mass Percent
What percent of each element in a compound?
Example:
Find the percent of each element in ethanol (C2H5OH).
C – (2)(12.01) = 24.02
H – (6)(1.008) = 6.048
O – (1)(15.999) = 15.999

46.07 g/mol
%O=
15.999 g/mol
46.07 g/mol
x 100 = 37.73% of O
Mass Percent Practice
1.
Find the percent of each element in CuSO4.
Cu = (1)63.546 = 63.546
S = (1)32.065 = 32.065
O = (4)15.999 = 63.996
%Cu = 39.81%
%S = 20.09%
%O = 40.10%
159.607 g/mol
2.
Find the percent of each element in C2H2.
H = (2)1.008 = 2.016
C = (2)12.01 = 24.02
26.04 g/mol
%H = 7.743%
%C = 92.26%
Empirical and Molecular Formulas
Look at C2H2 (Acetylene gas)
Acetylene + O2 + pumpkin = awesome!
What is the ratio between hydrogen and carbon atoms?
2 atoms of H
= 1 H :1 C
2 atoms of C
What about the mole ratio between hydrogen and carbon?
Same: 1 H : 1 C
Why?
2 atoms O 1 mol
2 atoms H 1 mol
6.022 x
3.32 x
10-24
1023
moles H
atoms
6.022 x 1023 atoms
3.32 x 10-24 moles O
What does this tell us?
A chemical compound’s formula can be determined by
finding the moles of each element.
Yeah!! Moles!
I
Moles
This formula is called Empirical (strikes back) formula
The empirical formula only tells you the ratio between
elements…
For example:
C2H2 and C6H6 both have the same empirical
formula… CH
To find the correct formula (molecular formula)…you need
the molar mass of the compound to be given.
Example:
Given molar mass of 78.1 g/mol and empirical formula of
CH…
Molar mass of CH = 13.02 g/mol (Periodic table)
It takes 6 times of 13.02 to get 78.1… therefore the
empirical formula (CH) gets multiplied by 6 and thus
C 6 H6
Molecular Formula
The exact formula for the
compound.
Ex. C4H10 or P4O8
Empirical Formula
The reduced formula for the
compound.
Ex. C2H5 or PO2
The empirical can be the molecular just as MgCl2 or Fe2O3
Determine Empirical Formula
Must find moles of each element (molar mass)
2. Find ratios between each element (large over
smallest)
Problem: A compound is made up of sulfur and
oxygen. 25.0 grams each of sulfur and oxygen are
present. Find the empirical formula.
1.
25.0 g S
1 mol of S
32.06 g S
25.0 g O
= 0.780 mol of S
1.56 mol of O
= 2 O :1 S
0.780 mol of S
= 1.56 mol of O
15.999 g O
Formula: SO2
1 mol of O
Determine Molecular Formula
Must find molar mass of empirical formula (periodic
table)
2. Find ratio between given molecular molar mass and
empirical molar mass (large over small)
Problem: A compound is made up of sulfur and oxygen.
25.0 grams each of sulfur and oxygen are present.
The molar mass of the compound is 128.12 g/mol.
Determine the molecular formula.
1.
Empirical Formula: SO2
Molar Mass of SO2 = 64.06 g/mol
128.12 g/mol
= 2 :1
64.06 g/mol
Molecular Formula: S2O4
Determine Empirical/Molecular
Formula
1.
2.
3.
4.
5.
Must find moles of each element (molar mass)
Find ratios between each element (large over
smallest) to get empirical formula
Find molar mass of the empirical formula
Find the ratio between molar mass
Multiply the ratio to each element in the
empirical formula
Example 1
An unknown compound contains both hydrogen and carbon
has a molecular molar mass of 26.04 g/mol. The compound
contains 1.85 grams of hydrogen and 22.15 grams of
carbon. Determine the empirical and molecular formulas.
1.85 g H 1 mol of H
1.008 g H
22.15 g C 1 mol of C
12.01 g C
= 1.84 mol of H
1.84 mol of H
= 1 H :1 C
1.84 mol of C
= 1.84 mol of C
Emp Formula: CH
Molar mass of CH = 13.02 g/mol
26.04 g/mol
13.02 g/mol
= 2:1
Mol. Formula: C2H2
Example 2
An unknown compound contains both sulfur and oxygen has
a molecular molar mass of 160.1 g/mol. The compound
contains 40.0%
sulfur andAlert!!*****
60.0% oxygen Determine the
*****Math
empirical and molecular formulas.
40.0 g S
1 mol
of S
Attention:
This is a math ALERT!!!
= 1.25 mol of S
32.06 g S
3.75 mol of O
= 3 O :1 S
Treat percentage just like grams….it doesn’t
1.25 mol of S
1 mol of O
60.0 g O matter
how= 3.75
much
sample you have!!
mol of O
15.999 g O
Emp. Formula: SO3
Therefore
have
%...make
Molar
mass of SOif3 you
= 80.06
g/mol
160.1 g/mol
80.06 g/mol
= 2:1
it grams!
Mol. Formula: S2O6
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