MOLEY, MOLEY, MOLEY, MOLEY, MOLEY, MOLEY, MOLEY, MOLEY, MOLEY
Suppose you were sent into the store to buy 36 eggs. When you picked them up you would get 3 boxes,
each containing 12 eggs. You just used a mathematical device, called DOZEN, to simplify the process.
You might also be asked to obtain a GROSS of some item. This amount is known to contain 144 (12
dozen). Can you think of other terms that are used to help simplify amounts? Yeah, I didn’t think so.
Chemists began to realize that numbers like 12 and 144 were much too small to use when working with
individual parts of matter like atoms and molecules. They chose to use the term MOLE to represent
amounts of matter that were applicable for them.
Keep in mind that this word MOLE is representative of an amount of something. Just as you can have a
dozen eggs you can also have a dozen chairs, a dozen people, a dozen stars, the scientists of the world
can have a mole of water, a mole of sodium chloride, a mole of gold.
Now I need to show you the number associated with the mole:
602 000 000 000 000 000 000 000
Wow! That large a number is difficult to use in calculations so we must use another mathematical
shorthand, scientific notation, to help us.
Lets rewrite that same number as: 6.02 E 23
Another name for this huge amount: Avogadro's Number.
As it turns out this number, named after Amadeo Avogadro, is used as a reference point for most
calculations and equations found in chemistry.
Here are 4 general rules associated with the MOLE concept. These rules allow us to compare mass
with volume, mass with number of particles, and then write balanced chemical equations.
Rules Utilized With MOLES
I. The chemical formula represents a mole of that substance.
II. The formula mass, expressed in grams, represents the mass of one mole of that substance.
III. One mole of any substance contains 6.02 E 23 particles.
IV. One mole of any gas, at STP conditions, occupies 22.4 liters of volume.
Now let’s expand on each of these and include lots of examples.
Rule: The chemical formula represents a mole of that substance.
Remember that any number placed to the left of a chemical symbol or formula is called the
COEFFICIENT. This number (integer, decimal, or in scientific notation) tells us the number of moles of
that substance.
Examples:
1 Pb
 1 mole of lead atoms (understood 1)
3 Pb
4.5 Cl12 CaCl
3.5 E-2 NaOH
 3 moles of lead atoms
 4.5 moles of chloride ions
 2 moles of calcium chloride
 3.5 E-2 moles of sodium hydroxide
Rule: The formula mass, in grams, represents the mass of that substance.
The formula mass of an element is its atomic mass (found on Periodic Table.) The formula mass of a
compound is found by multiplying the number of `moles' of that element (see its subscript in the
formula) by that atom's atomic mass. Then add masses of all elements and record in grams.
The following example is given to demonstrate how to find formula mass.
CaCO3
(calcium carbonate)
Ca
C
O
1 x 40.1 = 40.1
1 x 12.0 = 12.0
3 x 16.0 = 48.0
---100.1 grams = 100. g (3 sig figs)
If you need further help finding formula masses please see your teacher.
Now for some examples involving this rule:
2 Cu
 2 moles of copper atoms  2 x 63.5 = 127
5.00 NaCl  5 moles sodium chloride  5.00 x 58.4 = 292 g
2.5 H2SO4  2.5 moles of sulfuric acid
 2.5 x 98.1 = 245 g
Rule: One mole of any substance contains 6.02 E 23 particles.
Particles here might mean atoms, molecules, ions, electrons, or just about anything you might need to
work with. Remember: just as there are 12 items in a dozen; 6.02 E 23 particles in a mole.
Examples:
HNO3  1 mole of nitric acid, 1.00 x 63.0 = 63.0 grams,
6.02 E 23 molecules of nitric acid
3.00 K  3.00 moles of potassium atoms , 3.00 x 39.1 = 117 grams, 3.00 x 6.02 E 23 =1.81 E 24
potassium atoms
Rule: One mole of any gas, at STP conditions, occupies 22.4 L of volume.
STP is a shorthand way of requiring the temperature to be at 0 degrees C (273 K) and a standard
pressure of 1 atm (101.3 kPa).
This rule is most commonly used when studying gas laws. Suppose you have 4 grams of helium gas.
This represents 1 mole of helium (see 2nd rule). These 6.02 E 23 atoms of helium would take up 22.4
liters of space. This large volume would be fully occupied if the temperature was 0 degrees Celsius and
the pressure 1 atmosphere. A change in the temperature/pressure would, of course, change the volume
occupied by the gas.
No problems are given for this rule at this time. But there is one more gas law to worry about.
Hooray!
Sample Mole Calculations
The following problems will give you a chance to attempt working mole problems. Use your calculator
and the Periodic Table.
Sample Mole Calculations: Number of Moles to Formula
Given: 3.5 E-2 moles of strontium fluoride, correctly represent the coefficient and formula:
The answer would be 3.5 E-2 mol SrF2 (Strontium has a combining power of 2 and fluoride’s is 1)
There are 3.5 E-2 mol Sr and 7.0 E-2 mol F
Sample Mole Calculations: Moles to Grams
Given: 2.00 moles of Ca(OH)2 would represent ________ grams.
Remember that 1 mole of a compound is represented by the formula mass of that compound. Also, 1
mole of an element equals the atomic mass.
To solve this problem we must first calculate the formula mass and then multiply that number by the
number of moles we have (in this case: 2.00 mol)
To calculate formula mass, first list the elements in the formula along with the number of each (hint:
use the subscripts). Then multiply that number by the atomic mass of that element. Add those masses
and you have the formula mass. Remember to get your final answer you must multiply the formula mass
by the number of moles. Try it on paper.
Calculate formula mass:
Ca
O
H
1 x 40.1 = 40.1
2 x 16.0 = 32.0
2 x 1.01 = 2.02
-----formula mass  74.1 g (rounded)
Calculate mass of 2.00 mol Ca(OH)2
2.00 x 74.1 g = 148 g
Sample Mole Calculations:
Grams to Moles
Given: 48.5 grams of CaCO3 = ___________ moles of calcium carbonate
Remember that you must first find the formula mass of the compound. Then we will use the factor
label method to solve the problem.
Calculate formula mass:
Ca
C
O
1 x 40.1 = 40.1
1 x 12.0 = 12.0
3 x 16.0 = 48.0
-------formula mass  100. g (rounded)
Some of you will readily see that we have less than a full mole and simply divide 48.5/100. to get your
answer. But we should know how to use the factor label method when we encounter more difficult
problems.
Factor label method to solve mole problem:
48.5 g CaCO3
| 1 mol CaCO3
=
---------------------|------------------------|
100. g CaCO3
0.485 mol CaCO3
The gram units will cancel leaving mol as the proper unit
Sample Mole Calculations: Moles to Particles (Atoms, molecules, ions,...)
Given: 4.20 moles of hydrogen fluoride = _______ molecules HF
Remember that 1 mole of any thing has 6.02 E 23 particles.
So to answer this problem we would just multiply 4.20 x 6.02 E 23 and get the answer 2.53 E 24
molecules.
Thr same solution using dimensional analysis
4.20 mole HF | 6.02 E 23 mlcl = 2.53 E 24 mlcl HF
----------------|-------------------|
1 mole HF
The mole units cancel
Sample Mole Calculations: Grams to Particles
Given: 126 g of Lithium Sulfate
= _______________ particles Lithium Sulfate
Remember our motto: "Go To Moles" Since the unit of mole is not used directly in this problem we
must use it indirectly. Dimensional Analysis will do this for us.
We will need to calculate formula mass for lithium sulfate. Do this off to the side of your work
space. Did you get 110. grams? 1 mole = 110 grams. Also we will need to remember that 6.02 E 23 ion
pairs of lithium sulfate equals 1 mole.
126 g Li2SO4 | 6.02 E 23 particles Li2SO4
-------------------|---------------------------| 110. g Li2SO4
= 6.90 E 23 particles Li2SO4
Wow! That last problem was complicated. To work the problem we had to use the transitive property
from dear old math class. Since 6.02 E 23 ion pairs equals 1 mole and 110. g of lithium sulfate equals one
mole they are equal to each other. When we place two items that are equal to each other in a ratio it is
equal to 1.
Remember!
I.
The chemical formula represents 1 mole of that substance.
II.
The formula mass (expressed in grams) is the mass of 1 mole of that substance.
III.
1 mole of a substance contains 6.02 E 23 particles (atoms, mlcl, ions, electrons, etc.)
IV.
1 mole of any gas at STP (1 atmosphere of pressure and 0 C) occupies 22.4 liters of volume.
Name
Formula
Formula mass
# of particles
Atomic nitrogen
N
14.0
6.02 E 23 atoms
Nitrogen gas
N2
28.0
6.02 E 23 mlcl (molecules) (1.20 E 24 atoms)
Silver ions
Ag+1
108
6.02 E 23 ions
Sodium chloride
NaCl
58.5
6.02 E 23 ion pairs
Ammonium sulfate
(NH4)2SO4
132
6.02 E 23 mlcl
Can you mole?
1 mole Mo = _________ g
1 mole Mo = _________ atoms
7 moles Mo = _________ g
7 moles Mo = _________ atoms
1 mol Th3(PO4)4 = _________ mlcl = _________ g
0.5 mol CO2 = ________g = ___________ mlcl
1.5 mol MgCl2 = _________g = ___________ ion pairs
1 mol KNO3 = _________ mol K, _________ mol N, ________mol O
1 mol KNO3 = _________ g K, __________g N, ___________ g O
10 g KNO3 = __________ mol KNO3, __________ g K, ___________ atoms O
2.2 mol of H3PO4 = ___________ g H, __________mol P, __________atoms O
I will destroy you!
Molarity - We Use this All the Time with Acids - It’s acid concentration
Molarity: the ratio between the moles of dissolved substance (solute) and the volume of the solution (in
liters or cubic decimeters). Remember that anything dissolved in water is aqueous (aq)
Example:
1 Molar Nitric Acid = 1 M HNO3 = 1 mole of HNO3 in 1 L of solution
0.273 M Ba(NO3)2 contains 0.372 moles of barium nitrate in 1 L of solution
Sample Problem:
What is the molarity of a 250 mL solution containing 9.46 B CsBr?
Solution:
9.46 g CsBr__| 1 mole CsBr__ = 0.0444 mol CsBr
| 213 g CsBr
250 mL__|__1 L__
= 0.250 L
| 1000 mL
molarity = mole =
liter
0.044 4 mol CsBr = 0.178 M CsBr
0.250 L
Problems:
1.
145 g (NH4)2C4H4O6 in 500 mL of solution
2.
13.2 g MnSeO4 in 500 mL of solution
3.
45.1 g cobalt (II) sulfate in 250 mL of solution
4.
41.3 g iron (II) nitrate in 100 mL solution
5.
49.9 g Pb(ClO4)2 in 200 mL of solution
6.
35.0 g MnSiF6 in 50.0 mL of solution
Percentage Composition Calculations
To calculate the correct percentage composition of each element in some compound you must correctly
follow several steps. It is not very difficult, once you get the hang of it. Please take notes on the steps
to follow.
Several problems follow this one for you to work.
Lets look at a sample problem: magnesium chloride  MgCl2
step 1: calculate formula mass
Mg: 1 x 24.3 = 24.3
Cl: 2 x 35.4 = 70.8
---formula mass = 95.1
step 2: divide each component mass by the formula mass and multiply by 100
Mg: 24.3/95.1 x 100 = 25.6%
Cl: 70.8/95.1 x 100 = 74.4%
step 3: make certain the percentages add up to 100 (+/- 0.1)
25.6 + 74.4 = 100
This problem shows all the steps to follow. Be careful to use three significant figures at all times.
Remember to first find the correct formula, then find formula mass, then calculate the percentage
composition.
Practice Problem #1
Find the percentage composition for each element in: iron (III) silicate
iron (III) silicate
Fe: 32.9%
--> Fe2(SiO3)3
Si: 24.8%
O: 42.4%
Remember that with significant figures your answer can be one above or below the final digit.
Example: iron could have been reported as 32.8%, 32.9%, or even 33.0%.
Work: Fe: 2 x 55.8 = 112
Si: 3 x 28.1 = 84.3
O: 9 x 16.0 = 144
-----340.
112/340. x 100 = 32.9%
84.3/340. x 100 = 24.8%
144/340. x 100 = 42.4%
----100.1 (ok)
Practice Problem #2
Find the percentage composition for each element in: tin (IV) arsenate.
tin (IV) arsenate
Sn: 39.1%
--> Sn3(AsO4)4
As: 32.8%
O: 28.0%
Work: Sn: 3 x 119 = 357
As: 4 x 74.9 = 300.
O: 16 x 16.0 = 256
----913
357/913 x 100 = 39.1%
300./913 x 100 = 32.8%
256/913 x 100 = 28.0%
----99.9 (ok)
Empirical Formula Calculations (Empirical = smallest whole number ratio)
Finding the empirical formula is somewhat the reverse of finding percentage composition. First you
will be given the percentages of each element in a particular compound.
Assuming that you will always be given a 100 gram sample we will convert these percentages to grams.
The next step is to find the ratio of moles of each element and then calculate the simplest ratio of
subscripts that maintains that ratio.
It is much easier done than said! Take notes as we work through this sample problem.
Lets start with this data: 36.5% Na, 25.4% S, and 38.1% O
First, convert each percentage to grams: 36.5 g Na, 25.4 g S, 38.1 g O.
Next, divide each by the grams/mole of that element:
Na:
36.5 g
----23.0 g/mol
= 1.58 mol Na
S:
25.4 g
-----32.1 g/mol
= 0.791 mol S
0:
38.1 g
-----16.0 g/mol
= 2.38 mol O
Now we can set up the ratio of moles of each element:
Na1.58S0.791O2.38
Na1.58S0.791O2.38
This is the mole ratio, but it cannot be an atomic ratio. You simply cannot have 1.58 sodium atoms or
2.38 oxygen atoms. To convert these decimal numbers into whole numbers and maintain the same ratio
between them, just divide each by the smallest of the subscripts. Choose 0.791.
Na
S
O
1.58 0.791 2.38
---- --------0.791 0.791 0.791
Our final formula would look like: Na2SO3  sodium sulfite
Following will be two problems for you to work.
Practice Problem #1
Given the following data, find the correct empirical formula:
49.0% C, 2.70% H, 48.2% Cl
The correct formula would be: CH3Cl2
To solve:
C: 49.0g
----- =
4.08 mol
12.0 g/mol
C
4.08
---1.36
H: 2.70g
----- = 2.67 mol
1.01 g/mol
H
Cl
2.57
1.36
------1.36
1.36
Cl:
48.2g
----=
35.5 g/mol
gives the final formula: CH 3Cl2
Practice Problem #2
Given the following data find the empirical formula:
N = 26.2%,
H = 7.50%,
Cl = 66.4%
The correct formula would be: NH4Cl
(ammonium chloride)
The work:
N: 26.2g
----- =
1.87 mol
14.0 g/mol
H: 7.50g
----- =
1.01 g/mol
Cl: 66.4g
7.50 mol
----- =
1.87 mol
35.5 g/mol
1.36 mol
N
H
Cl
1.87 7.50
1.87
---------1.87
1.87
1.87
 NH4Cl
To find the molecular formula, first find the empirical formula for the data you have been given. Then
compare the formula mass for your empirical formula with the formula mass for the molecular formula.
You may have to multiply the subscripts in your empirical formula by some factor.
Sample Problem: Molecular Formula Calculation
Given: 38.7% C, 9.70% H, 51.6% O and a molecular formula mass of 62.0 u find the true molecular
formula:
First, find the empirical formula:  CH3O
Next find the formula mass:
C
H
O
1 x 12.0 = 12.0
3 x 1.01 = 3.03
1 x 16.0 = 16.0
-------formula mass = 31.0 u
Now divide the molecular mass by this formula mass:
62.0/31.0 = 2
Multiply each subscript in your formula by this factor (2):
Your final molecular formula: C2H6O2
To check your answer, find the formula mass of your final formula. It should be the same as the
molecular formula mass.
Practice Problem #2
Given the following data, find the correct molecular formula:
24.3% C, 4.1% H, 71.6% Cl molecular formula mass = 99.0
To see the correct answer and the method to solve the problem, please continue.
The correct formula would be: C2H4Cl2
The empirical formula would be CH2Cl with a formula mass of 49.5
Dividing 99.0 by 49.5 gives a factor of 2. Each subscript is doubled.
Again, check the formula mass of your final answer to confirm your answer.
Practice Problem #3
Given the following data, find the correct molecular formula:
54.6% C, 9.00% H, 36.4% O and a molecular mass of 176
The correct answer would be: C8H16O4
The empirical formula would be: C2H4O with a formula mass of 44
The multiplication factor would be 4 (176/44.0)
WORKSHEET -- Alpha
1. 1 mole of iron weighs ____________________ grams.
2. 1 mole of iron contains __________________ atoms.
3. 7.4 moles of iron weighs _________________ grams.
4. 7.4 moles of iron contains _________________ atoms.
5. 1 gram of iron contains __________________ atoms.
6. 1 atom of iron weighs ____________________ grams.
7. 1.2 E 24 atoms of iron would be _________________ moles of
iron.
8. 6.4 E 31 atoms of iron would be __________________ moles of
iron.
9. 6.4 E 31 atoms of iron weighs ____________________ grams.
10. 140 grams of iron would be _______________________ atoms.
11. 1 mole of Ca(OH)2 weighs _______________________ grams.
12. In one mole of Ca(OH)2 there are __________________ ion
pairs.
13. 4 moles of Ca(OH)2 weighs ____________________ grams.
14. In 2.2 E - 3 moles of Ca(OH)2 there are _________________
ion pairs.
15. The total number of atoms in one mole of Ca(OH)2 would be
_____________.
16. 2.2 E - 3 moles of Ca(OH)2 weighs ________________ grams.
17. 1.6 E 4 grams of Ca(OH)2 would be ____________________
particles.
18. A compound contains 29.1% sodium, 40.5% sulfur, and 30.4%
oxygen.
What would be the empirical formula for this compound?
_______________
19. 3 grams of hydrogen combines with 42 grams of nitrogen and 144
grams of oxygen to form a new compound.
What would be the empirical formula of this compound?
___________
How many moles of this compound would be created? _______.
WORKSHEET -- Omega
1.
2.
3.
4.
5.
6.
1 mole of CO2 weighs _____________________ grams.
3 moles of CO2 contains __________________ molecules of CO2.
3 molecules of CO2 weighs ________________ grams.
3 moles of CO2 contains __________________ atoms of carbon.
3.1 moles of CO2 contains ________________ molecules of CO2.
3.1 moles of CO2 contains ________________ grams of CO2.
7. 1 mole of KClO3 weighs ___________________ grams.
8. 6 grams of KClO3 would be ________________ moles.
9. 6 grams of KClO3 if split up will produce ___________ grams of K,
____________ grams of Cl, and ___________ grams of O.
10. Find the percentage composition of each element in KNO3.
______________% K, _____________% N, _____________% O
11. 1 mole of KNO3 weighs ________ grams and contains ___________ ion pairs.
12. In 1 mole of KNO3 there are ____________ moles of K,
____________ moles of N, and 3 moles of _________________.
13. In 0.25 moles of KNO3 there are ___________ grams of K,
___________ grams of N, and ____________ grams of O.
14. 5 E 23 ion pairs of KNO3 is ___________ moles and weighs _________
grams.
15. 10 grams of KNO3 would be ____________ moles of KNO3 and would be
________________ ion pairs.
16. In 2.2 moles of H3PO4 there are __________ grams of H,
___________ grams of P, and _____________ grams of O.
17. 2.1 E 24 molecules of H3PO4 would be ______________ moles and would
weigh ____________ grams.
18. In 2.2 moles of H3PO4 there are _________________ molecules of H3PO4
and ________ atoms of H, _________ atoms of P, and _________ atoms of
O
19. In 100 grams of H3PO4 there would be __________ grams of H,
___________ grams of P, and ___________ grams of O.
20. Find the percentage composition of each element in H3PO4
_____________% H, ______________% P, _______________% O