Download NAME NOTES: UNIT 8 THE MOLE AND STOICHIOMETRY (2

NAME _________________________________ NOTES: UNIT 8 THE MOLE AND STOICHIOMETRY (2)
APPLYING DIMENSIONAL ANALYSIS TO MOLE THEORY
I-V) Intro to Mole to Percent Composition
VI) Stoichiometry (Greek) * measure of the element (measuring the balance of the element)
stoicheion: element
This connects to the Law of the Conservation of Matter (Big Idea #1) … since
the process is concerned solely with the conversion of one measurement to a
different frame of reference, then the process is really concerned with ensuring
that whatever you “put into the reaction”, you somehow, “get out” with products.
A) How to Organize Your Conversion Factors
1) Every beginning to a stoichiometry problem should look like:
Desiredunit = Given#unit' |
|
unit'
22.4 Liters
1 mole
6.02 x1023species
? grams
22.4 L of gas only at STP
(constant)
1 mole of a substance
use formula &P.T. : changes w/ substance
grams
6.02 x 1023 species
(constant)
2) Calculate the number of grams equal to 0.050 moles of NI3(s).
1 mole of NI3(s)
use the PT
_____grams
3) Calculate the number of molecules equal to 1,245 grams of O2 gas at STP.
1 mole of O2(g)
use the PT
_____grams
4) When do I use just the single mole map? Well, draw a single mole map when you are
*
dealing with only 1 known substance
*
doing most conversion problems when there is no balanced equation
69
DIRECTIONS: Below are a number of simple mole conversion problems. I have provided you with the
conversion factors you require, for many of these problesm. Focus upon becoming comfortable with the way
questions are constructed and the application of your knowledge of unit cancellation.
1) How many grams of I2(s) equal 0.0290 moles of iodine?
What we know…
1 mole
6.02 x1023
molecules
254 g
all are absolute numbers
2) Calculate the mass of 9.73 Liters of H2(g) at STP.
What we know…
22.4 Liters
1 mole
*245 g
6.02 x1023
molecules
all are absolute numbers
3) How many moles of Ca(NO3)2(s) equal 17.0 grams ?
What we know…
1 mole
* 164 g
6.02 x1023
formula units
all are absolute numbers
4) How many atoms of He(g) are equivalent to 40.6 liters of helium gas at STP ?
22.4 Liters
1 mole
* 32g
6.02 x1023
molecules
all are absolute numbers
a) How many molecule of H2(g) are equivalent to 40.6 Liters of hydrogen gas at STP?
Answers: 1) 7.37 grams
2) 0.869 grams
3) 0.104 mol
4) 1.09 x 1024 molecules of H a) 1.09 x 1024 molecules of H2
(notice that it is the same value …hmm! Why?)
70
5) How many grams of Fe(s) equal 0.0750 moles of iron?
1 mole
6.02 x1023
atoms
*56g
all are absolute numbers
6) Calculate the mass, in grams of 1.40 liters of He(g) at STP.
What is the mass, in kilograms?
7) How many moles of Al2(SO4)3(s) equal 24.0 grams ?
8) How many molecules of F2(g) are equivalent to 7.89 Liters of fluorine gas at STP ?
9) Challenge: Calculate the gram equivalent of 0.827 mol NH4Cl(s)
(what do you need to know???)
*
1 mol NH4Cl(s)
* grams NH4Cl = 0.827 mol NH4Cl | 53 grams NH4Cl | =
1 mol NH4Cl
6.02 x 1023
53 g
formula units
10) Challenge: Calculate the mole equivalent of 75.3 grams of Fe2O3(s)
(what do you need to know???)
*
* mol Fe2O3(s) = 75.3 grams | 1 mol Fe2O3(s)
| =
160 grams Fe2O3(s)
Answers:
5) 4.20 gram
6) 0.250 g & 2.50 x10-4 kg 7) 0.0702 mol
8) 2.12 x 1023 molecules
1 mol Fe2O3(s)
160 g
9) 43.8 grams
6.02 x 1023
formula units
10) 0.471 mol
71
11) Super Challenge: Calculate the grams/mol (or the molar mass) of a compound, assuming a 3.41 x 10-6 gram
sample contains 4.67 x 1016 molecules. (Hint: this is just like converting units ….like mg/L to kg/mL …write the “given” as the ratio
between the two values of the problem…as in grams/molecules … No “mole map” is required, for you don’t know the identity of the substance….)
1)
2)
3)
4)
44 grams and the compound is CO2
16 grams and the compound is CH4
17 grams and the compound is NH3
18 grams and the compound is H2O
x grams = * 3.41 x 10-6 grams
|
mol
4.67 x 1016 molecules
6.022 x 1023 molecules
1 mol
12) Avogadro’s number is equal to the number of:
1)
2)
3)
4)
All of these
atoms in 1 mol of Na
molecules in 1 mol of H2O
formula units in 1 mol of KCl
72
NAME _________________________________ NOTES: UNIT 8 THE MOLE AND STOICHIOMETRY (3)
FOCUS ON SOLUTION CONCNETRATION: MOLARITY
VI) Stoichiometry
VII) Molarity
A) You must be able to determine the molar concentration (MOLARITY) of an aqueous solution given
any TWO PIECES of the following information:
a) moles of solute
b) grams of solute
and / or
c) LITERS of solution
2) define the terms : aqueous solution, solute and solvent
There are 2 types of Molarity problems. So, here's what you wish your teacher or your textbook would tell you:
Cutting To The Chase
M O L A R I T Y
There are:
1
s t e p
p r o b l e m s
give you 2 of the 3 variables of the equation
Molarity = moles of solute
Liters of solution
and
There are:
2
s t e p
p r o b l e m s
ask for or give you the number of grams of
solute (rather than the number of moles). Use of
the terms grams & molarity indicates that
it’s 2 steps
If you are given M, Liters &/or moles of solute
then
Begin with the equation
Molarity = moles of solute
Liters of solution
all you need to do is substitute into the equation
the two given values and solve for the third.
then, use unit cancellation to convert
between grams  moles
AND
Much of what you experience deals with "aqueous solutions". Apple juice, milk, soda, lake water, ocean water,
tap water are all examples (complex and simple) of aqueous solutions.
As in all mixtures, an aqueous solution is a physical combination of a SOLVENT and a SOLUTE. The
solvent is the substance which does the "dissolving" or "breaks up the solid". The solute is the stuff that is
dissolved. The solvent must always (powerful word) be in the greater quantity. In the case of aqueous solutions the
solvent is water. Anything water can dissolve may be considered to be a water-soluble solute.
Aqueous solutions (mixtures) are NOT substances. Rather, they are homogeneous mixtures made of 2 or
more substances. For example, you would be unable to differentiate (visually) between a glass of pure water
and (if made correctly) a glass of concentrated salt water (NaCl(aq)). When written, the solute is recorded
first and the water solvent is recorded with the abbreviation (aq). Thus, the formula NaCl(aq) is known
to be a solution of sodium chloride dissolved COMPLETELY in water.
73
These aqueous solutions may be described in terms of their concentration or their MOLARITY. If you were to
update your Concept Map of Matter you would have the following:
MATTER
may be divided into 2 large categories
HeterogeneousMatter
Homogeneous Matter
exemplified by
Mixtures
like
is divided further into
two different groups
Substances
Mixtures
emulsions &
solid mixtures
that often settle
out into layers
suspensions
or like those
there are 2 types
elements
made of only 1 type of
atom
eg) sand + water
eg) Fe
compounds
made of 2 or more
different species
eg) LiOH
that don't settle out are called
solutions
like
or for our class
alloys &
gas
mixtures
eg) aqueous
solutions
There are many types of homogeneous solutions. The "aqueous solution" is just one type. It is however, the
most important to the understanding of the material contained in this Chemistry syllabus.
SUMMARY: Aqueous solutions are not chemical combinations. Aqueous solutions are ______________
physical, water
solvent
combinations of ______________ and some solute.
Water acts as the dissolving agent.
In its role as the dissolving agent, water is called technically, the ____________________.
homogeneous
While they are not substances, aqueous solutions are ___________________, in that they are
molarity
uniform in their properties. Concentration is expressed as ______________.
74
MOLARITY (symbol M, M or M ): a term of concentration which expresses the number of moles of solute
completely dissolved in 1 Liter of solution .
A) unit:
moles / Liter
(read as moles per Liter)
1) the unit expresses the definition: M, M or M is the number of MOLES per LITER
a) very often when you see M or M, you may "say" it as the word MOLAR
b) so, 0.50 M HCl(aq) is read or spoken as: a 0.50 MOLAR solution of HCl
B) Equation:
M =
# of moles of solute
# of Liters of solution
1) The volume of the solution must be in LITERS
2) If volume is given in milliliters (mL), you must convert from one to the other.
Remember, 1 L = 1,000. mL
_
3) eg) Calculate the molarity of a 2,000 milliliter solution containing 3.0 moles of HCl(g)
M =
M =
# of moles of solute
# of liters of sol'n
3.0 moles HCl____
2.0 Liters of sol'n
M = 1.5 M
4) eg) Calculate the molarity of 2.0 moles of HCl(g) dissolved in a 500. mL solution .
M =
# of moles of solute
# of liters of sol'n
M =
2.0 moles HCl
did you remember to convert the 500. mL of solution
0.500 Liters of sol'n  to 0.500 Liters of solution ???????
M = 4.0 M
The answer has 2 sig figs because 2.0 has the fewer number of sig figs.
75
5) Practice for 1 step Molarity Problems
a) What is the molarity of a 900. mL solution when 1.30 moles of CuSO4 are dissolved?
Equation: M = moles
Liters
Substitution:
Answer: _________ M
ANS: 1.44 M
b) How many moles of NaCl are dissolved in a 0.20 M solution of 500. mL?
Equation:
Substitution:
Ans: _________ moles
Ans: 0.10 moles of NaCl
(note only 2 sig figs)
c) Calculate the volume (Liters) of a 0.500 M solution made with 3.50 moles of NaOH?
Equation:
Substitution:
Ans: 7.00 L
(note 3 sig figs)
Ans: ___________ Liters
76
C) What if you are asked for grams of solute or given, grams of solute?
1) eg) A student made a 5.0 L solution by dissolving 80.0 grams of NaOH(s) into water.
Calculate the molar concentration (molarity) of the solution.
key words for
2-step problems
step 1: calculate moles from grams;
moles of NaOH = 80.0 grams | 1 mole of NaOH|
40 grams ⇚(Na + O + H)
23g + 16g + 1g
therefore: moles of dissolved NaOH = 2.00 moles
(calculating a mole mass)
(note 3 sig figs based on the "given")
step 2: substitute into the M formula:
M =
M =
# of moles of solute
# of liters of solution
2.00 moles NaOH
5.0 L of solution
M = 0.40 mol/L
or
0.40 M
or
4.0 x 10-1 molar solution ….
each is equally correct, with 2 sig figs
3) Practice:
Ans: 1.19 M
N.B. The following problems have been calculated using mole masses rounded to the nearest whole number,
for convenience. In lab, we would use the decimalized values. If you use the decimalized values for theses
problems, your answers would be just slightly different.
a) A 1.50 x 103 mL solution has 100. grams of KOH completely dissolved. Calculate the
molarity of the solution
key words for 2-step problems
77
Ans: 0.85 M
b) Calculate the molar concentration of a 750 mL solution when 54.0 grams of NaNO3(s)
are dissolved completely in water.
clues for a 2-step problem
Ans 0.552 M
c) Calculate the molarity of a 5.00 x 10-1 L solution with 94.5 grams of C12H22O11(s)
completely dissolved in water. clues for a 2-step problem
Ans 2.0 M
d) Calculate the molarity of a 250 mL solution with 53 grams of Na2CO3(s) completely
dissolved
Ans: 1.0 M
e) Calculate the molarity of a 2.0 L solution with 320 grams of CuSO4(s) completely
dissolved.
78
Practice:
1) THIS IS A 1 STEP*: How many moles of NaI are required to make a 750. mL solution , of 0.80 M?
(*You are given M, and the volume in Liters, simply plug into the equation and solve for moles…
Do you need to use the GFM?)
a)
b)
c)
d)
0.29 mole
7.0 moles
0.60 mole
0.10 mole
ans. c
2) THIS IS A 2 STEP**: What is the number of grams of AlCl3 dissolved in 0.500 L of a 0.900 M solution?
**(Note, you're asked for grams, but you must first use the equation to get to moles and then use
unit cancellation to calculate the grams... Will you need the GFM?)
a)
b)
c)
d)
59.4 g
126 g
229 g
74.2 g
ans. a
3) How many grams of KOH are dissolved in a 2.0 x 103 mL solution of 1.5 M KOH(aq)?
How many steps is this? 1 or 2  WHY ??
a)
b)
c)
d)
150 g
140 g
90 g
170 g
ans d
4) A student used 11.0 grams of NaOH to make a 3.0 L solution. Determine the molarity of the solution.
a)
2.5 M
b) 0.092 M
c) 0.50 M
d) 1.8 M
ans b
79
5) How many grams of LiCl are required to prepare 80.0 mL of a 0.150 M solution?
a)
b)
c)
d)
0.504 grams
0.761 grams
1.675 grams
3.19 grams
ans. a
6) Calculate the number of liters of solution made from making a 0.50 M solution by dissolving 2 moles
of KOH into water. (How many steps is this? 1 or 2  WHY ??)
a)
b)
c)
d)
1
2
3
4
ans. d
7) At 25ºC, 0.0018 grams of NaCl dissolves in enough 1-petanol (CH3CH2CH2CH2CH2OH) to give 0.10 L of
solution. What is the molarity of the solution? (What do you have? What do you need?)
a) 0.0031
b) 0.00031
c) 0.00016
d) 0.018
ans. b
8) A sugar-free soft drink contains 7.9 mg of saccharin, C7H5SNO3 in 1.0 ounce (31 mL). What is the molarity
of saccharin in an ounce of the drink? (N.B. 1,000 mg = 1 gram)
a) 4.5 x 10-3
b) 1.4 x 10-1
c) 3.2 x 10-2
d) 1.4 x 10-3
M
M
M
M
ans. d
80
VIII) More Complex Stoichiometry and the Double Mole Map
22.4Liters
22.4 Liters
1 mole
1 mole
Bridging coefficients
____grams
6.02 x 1023
___grams
6.02 x 1023
molecules
formula units
atoms
molecules
formula units
atoms
A) When should I get the “hint” that I need a double mole map?


When given a balanced chemical reaction equation
When asked to deal with two different substances, in a single problem
B) You'll notice that the "double mole map" requires the use of a balanced equation.
We've introduced the issues of balancing in light of the Law of the Conservation of Matter.
Now, very briefly, I would like to introduce the role of the balanced equation in terms of
stoichiometry.
I am quite aware that you may not yet know how to balance an equation ... but I wish to discuss
what a balanced equation is.
1) First, the coefficients of the balanced equation represent the mole ratios between each of the
reactants, each of the products and each reactant to each product.
2) The coefficients tend to be in the simplest whole number ratio
3) Many chemistry teachers approach the balanced equation using a recipe...
For instance ... take the basic recipe for brownies:
For 1 batch of brownies:
2 sticks of butter (B)
4 eggs (E)
2 cups sugar (S)
2 cups flour (F)
5 squares unsweetened chocolate (C)
... and assorted flavorings...
So the recipe goes like this:
2 B + 4 E + 2 S + 2 F + 5 C +....  1 batch of brownies
When I have only 2 eggs ...how much butter is required:
how many cups of sugar :
how many cups of flour:
how many squares of chocolate
How many batches of brownies can I make
*1
*1
*1
*2.5
*1/2 batch
81
Given a chemical reaction:
2 C2H6(g) + 7 O2(g) 
4 CO2(g) + 6 H2O(l) + 3170 kJ
Meaning: For every 2 mol of ethane, 7 mol of O2 molecules are consumed in the combustion
THUS, producing 4 mol of carbon dioxide and 6 moles of water with the release of
3,170 kJ of thermal energy into the environment.
This set of relationships can indicate, at a glance that:
Twice as many moles of CO2 are produced as moles of ethane consumed
Three times as many moles of CO2 are produced as moles of ethane used
3.5 times more moles of dioxygen are required than moles of ethane used
For every 2 mol of ethane combusted, 3,170 kJ of energy are produced
So ... Here's that balanced equation again .... Try to answer the following quesitons:
2 C2H6(g) + 7 O2(g) 
4 CO2(g) + 6 H2O(l) + 3,170 kJ
example 1) When 4 mol of ethane are combusted completely,
example 2)
* 14
mol of dioxygen are required.
* 8
mol of carbon dioxide are produced.
* 12
mol of water are produced.
* 6,340
kJ energy are released to the environment.
Imagine a container, that has ONLY 1 mole of ethane molecules but there are still
7 moles of dioxygen present. Assume that all of the ethane molecules are combusted.
* 3.5
mol of dioxygen are required.
and * 3.5
mole of dioxygen molecules are left in the container, unreacted
* 2
mol of carbon dioxide are produced.
* 3
mol of water are produced.
* 1,585
kJ energy are released to the environment.
82
NAME _________________________________ NOTES: UNIT 8 THE MOLE AND STOICHIOMETRY (4)
VIII) Dimensional Analysis and Mole
IX) Advanced Stoichiometry
Sometimes, a chemist just needs to consult a map....
22.4Liters
22.4 Liters
bridging coefficients
1 mole
____grams
1 mole
6.02 x 1023
___grams
molecules
formula units
atoms
6.02 x 1023
molecules
formula units
atoms
The above mole map can represent the numeric relationships between any two substances of a
balanced chemical reaction.
**************************************************
A) The key to knowing when you need to use the double mole map to organize your conversion factors
rests in whether you are given * a balanced reaction equation (one with coefficients)
and told to work between *two different substances
(as in one reactant & one product OR 2 reactants or 2 products)
B) Process
STEP 1 Read the problem & circle the important words and
STEP 2 Write your "desired" = givenunit
measurements
___
unit
STEP 3 Draw a mole map and fill-in each variable with the correct conversion factor& label start/finish
STEP 4
Solve the problem by manipulating conversion factors so that units cancel out.
83
1) Given:
C5H12(g) + 8 O2(g)  5 CO2(g) +
6 H2O(g)
Given the balanced equation, calculate the number of grams of carbon dioxide produced when
89.6 L of O2(g) are consumed completely, in an excess of C . (The term in “excess” is used to
ensure that the reader understands that there is so much of a reactant, that the other reactant will be consumed completely
and be depleted first….leaving some of the “excess” reactant, unreacted, in the vessel)
2) Given: 2 Al(s) + 3 H2SO4(aq)  Al2(SO4)3(aq) + 3 H2(g)
Given the balanced equation, calculate the grams of dihydrogen gas produced, when
189 grams of aluminum are consumed completely in an ◊excess of sulfuric acid (H2SO4(aq))
3) Given:
O2(g) + 4 K(s)  2 K2O(s)
Calculate the liters of dioxygen gas at STP required to oxidize 150.0 grams of potassium
metal completely.
84
MOLE TO MOLE CONVERSIONS
(Farily basic, fairly useless … it just gives molar relationships … it doesn’t deal with mass … This just helps to
understand the use of the coefficients as “bridging” two different substances.)
1) Given the balanced equation:
2 C2H6(g) +
7 O2(g) 
4 CO2(g) + 6 H2O(l)
How many moles of dioxygen (O2(g)) must be consumed in order to produce 2.0 moles of water?
1 Mole O2
1 Mole H2O
bridging coefficients
moles of O2(g) = 2.0 moles of H2O |
|
moles of H2O
2) Given the balanced equation:
2 C7H14(g) + 21 O2(g)
 14 CO2(g) + 14 H2O(g)
How many moles of carbon dioxide are produced when 10.5 moles of dioxygen are consumed in an excess
of heptene (C7H14)?
1 Mole
1 Mole
bridging coefficients
moles of CO2 = 10.5 moles O2 |
|
moles O2
3) Given the balanced equation:
2 C8H14(g)
+ 23 O2(g)  14 H2O(g) + 16 CO2(g)
How many moles of C8H14 must be combusted in an excess of oxygen to produce 28.00 moles of water?
1 Mole
1 Mole
bridging coefficients
4) Given the balanced equation : 11 O2(g) + 2 C4H6(g)  8 CO2(g) + 6 H2O(l)
Calculate the moles of CO2 produced when 0.500 moles of C4H6 are combusted completely in an excess of
dioxygen.
85
3 STEP CONVERSION PROBLEMS
___1. Given: 4 Al(s) + 3 O2(g)  2 Al2O3(s)
How many grams of O2(g) (GFM=32 g), at STP are consumed completely in the production of 43.012 grams
of Al2O3(s) (GFM=102 g)
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a) 20.241 grams
b) 9.4460 grams
L
|
1 Mole
/
\
g. 6.02 x 1023
c) 22.400 grams
d) 0.85203 grams
___2. Given: 4 BCl3(s) + P4(s) + 6 H2(g)  4 BP(s) + 12 HCl(g)
Calculate the number of grams of BCl3 (GFM=116 g) required to react completely with 4.00 grams of
H2 gas (GFM= 2g) at STP, in an excess of tetraphosphorus (P4 or white phosphorus).
L
L
|
|
1 Mole
1 Mole
bridging coefficients
/ \
/
\
g. 6.02 x 1023
g. 6.02 x 1023
a) 91.5 g
b) 13.2 g
c) 155 g
d) 268 g
___ 3. Given: 2 Mg(s) + O2(g)  2MgO(s)
Calculate the number of grams of Mg (GAM=24 g) required to produce 100. grams of MgO (GFM = 40 g), in
an excess of dioxygen gas.
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a) 75.0 g
b) 781 g
L
|
1 Mole
/
\
g. 6.02 x 1023
c) 9.00 g
d) 60.0 g
86
___4. Given: C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(l)
Assuming an excess of propane, calculate the volume of oxygen gas (GFM= 32g) required to produce 88.0 grams of
CO2 gas (GFM= 44 g) at STP. (Think...you want volume …What UNIT will you need to use?))
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a) 74.7 L
b) 62.8 L
L
|
1 Mole
/
\
g. 6.02 x 1023
c) 19.5 L
d) 21.2 L
___5. Given: F2(g) + 2 KCl(aq)  2 KF(aq) + Cl2(g)
Assuming an excess of KCl, calculate the number of grams of fluorine gas (GFM =38 g) required to produce
103.67 grams of dichlorine gas (GFM=70 g)
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a) 64.115 g
b) 58.285 g
L
|
1 Mole
/
\
g. 6.02 x 1023
c) 56.278 g
d) 38.027 g
___ 6. Given: 2 KClO3(s) 
2 KCl(s) + 3 O2(g)
How many grams of KCl (GFM = 74 g) must also be produced, when 134.0 L of O2 (GFM=32 g) are produced, at STP?
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a) 43.01 g
b) 295.1 g
L
|
1 Mole
/
\
g. 6.02 x 1023
c) 84.30 g
d) 110.2 g
87
___ 7. Given: Fe(s) + CuSO4(aq)  Cu(s) + FeSO4(aq)
Calculate the mass, in grams, of copper metal (GAM= 64 g) produced when 112.0 g of Fe (GAM = 56 g)
are reacted completely.
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a) 57.10 g
b) 108.5 g
___8. Given:
L
|
1 Mole
/
\
g. 6.02 x 1023
c) 52.28 g
d) 128.0 g
2 Ca3(PO4)2(s) + 6 SiO2(s) + 10 C(s)  P4(s) + 6 CaSiO3(s) + 10 CO(g)
Calculate the number of molecules of tetraphosphorus (P4) produced when 892.43 grams of Ca3(PO4)2 are
consumed completely in excesses of the other reactants.
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
a)
b)
c)
d)
L
|
1 Mole
/
\
g. 6.02 x 1023
4.6620 x 1024 moleucles
8.6652 x 1023 moleules
9.4016 x 1023 moleucles
1.0262 x 1015 molecules
____9. Given
2 C8H18 + 15 O2  16 CO2 + 18 H2O
Calculate the mass, in kilograms, of CO2(g) produced when 114.0 grams of octane are combusted completely in
an excess of dioxygen gas. Recall : 1,000 grams = 1 kilogram
L
|
1 Mole
bridging coefficients
/ \
g. 6.02 x 1023
ANS. 1. a
L
|
1 Mole
/
\
g. 6.02 x 1023
a) 3.520 x 10-1 Kg
c) 2.861 x 10-1 Kg
b) 4.511 x 10-2 Kg
d) 7.205 x 10-2 Kg
2. c
3. d
4. a
5. c
6. b
7. d
8 b
9. a
88
MIX IT ON UP!!!
1. GIVEN: 4 Al(s) + 3 O2(g)  2 Al2O3(s)
AT STP,how many mol of O2(g) (GFM=32g) are required to produce 0.500 mol of Al2O3(s) (GFM=102 g) ?
(Ans= 0.750 moles)
Draw a mole map
2. GIVEN: Fe3(PO4)2(aq)
+ 6 NaOH(aq)  2 Na3PO4(aq)
+
3 Fe(OH)2(s)
How many grams of NaOH (GFM=40) are required to produce 450. grams of Fe(OH)2 (GFM= 90 g)
(Ans = 400. g)
Draw a mole map
3. GIVEN:
2 Na(s) + Cl2(g)  2 NaCl(s)
How many moles of NaCl are produced when 45.0 grams of Cl2 are reacted completely?
(Ans=1.29 moles)
Draw a mole map
89
4. GIVEN:
2 C8H18(g) + 15 O2(g)  16 CO2(g) + 18 H2O(l)
At STP how many liters of CO2 are produced when 342 grams of octane are combusted completely
in an excess of oxygen ?
(Ans= 538 liters)
Draw a mole map
5. GIVEN:
3 H2(g) + N2(g)  2 NH3(g)
How many moles of hydrogen gas are required to react completely with 6.50 moles of nitrogen gas?
(Ans=19.5 moles)
Draw a mole map
6. GIVEN: 4 Fe(s) + 3 O2(g) + 6 H2O(l)  4 Fe(OH)3(s)
At STP, how many liters of O2 gas are required to produce 233.9 grams of Fe(OH)3 in an excess of iron?
(ans=36.72 L)
Draw a mole map
90
7. GIVEN: 2 PbO(s) + 2 SO2(g)  2 PbS(s) + 3 O2(g)
In an excess of sulfur dioxide, calculate the moles of PbO required to produce, 1.55 mol of dioxygen gas at STP
(ans=1.03 moles)
Draw a mole map
8. GIVEN:
2 Fe2O3(s) + 3 C(s) 
4 Fe(s) + 3 CO2(g)
How many grams of Fe are produced when 49.3 grams of carbon dioxide gas are produced as well?
(ans= 83.7 grams)
Draw a mole map
91
NAME _________________________________ NOTES: UNIT 8 THE MOLE AND STOICHIOMETRY (4)
FOCUS ON LIMITING REAGENTS
First, a little vocabulary. You might have noticed that in class I have used the terms: reactant & reagent
interchangeably. Well, the terms are synonyms of each other.
Secondly, have you noticed that the stoichiometry practice problems often contain the phrase “in an excess of...”?
For instance:
Given: 2 Ag2S(s) + 3 O2(g)  2 SO2(g) + 2 Ag2O(s)
Calculate the mass, in grams of Ag2O(s) produced when 7.85 grams of Ag2S(s) are consumed
completely, in an excess of dioxygen.
This phrase tells the reader that there is so much oxygen available that the silver sulfide will be consumed completely
and the reaction ends with left-over dioxygen in the vessel.
Thus the masses of SO2(g) and Ag2O(s) that can be produced are limited by the limited mass of Ag2S(s)
Think about it this way: You’re making peanut butter and jelly sandwiches (mmmmm!)
Assume the term “sandwich” means 2 pieces of bread, two dips of the knife into the peanut butter
and 1 dip into the jelly.
You have a 1 lb jar of PB, a 1 lb jar of “J” with four pieces of bread. What is the limiting reagent?
Well, the reactant (reagent) of a chemical reaction which gets used up first is called the limiting reagent. This has big
implications, for it is the concentration of the limiting reagent which determines the amount of product(s) produced.
METAPHOR!! A chain is only as strong as its weakest link. … You can only produce as much
product as you have reactants in the stoichiometric amounts. When one or more of the
reactant concentrations is/are limited, either due to mass or large requirement (as given
by the molar ratio of the coefficients of the balanced equation), then a reaction may not
“go to completion”. The amount of product may be limited by the reactant.
Given:
2 Na(s) + Cl2(g)  2 NaCl(s)
How many moles of of sodium chloride that can be prepared by the reaction of:
i) 0 mol of of Na and 1 mol of Cl2
* no NaCl can be made
Thougth Experiment
ii) 2 mol of Na and 1 mol of Cl2
* 2 mol NaCl per bal. equation
iii) 2 mol Na and 2 mol of Cl2
* 2 mol of NaCl ... all the Na would be
consumed and 1 mol of Cl2 is left
over, bringing us back to the
92
X) Limiting Reagent Problems:
A) Recognizing limiting reagent problems: The problem will have quanitites for 2 or more different
reactants ... The reactansts will not "completely react".

Recognition: Given: Al4C3(s) + 12 H2O(l)  4 Al(OH)3(s) + 3 CH4(g)
Determine the limiting reagent, when 156 grams of Al4C3(s) are allowed to react with
192 grams of water.
(note, you may solve for the mass of either product)

Recognition:
Given: Al4C3(s) + 12 H2O(l)  4 Al(OH)3(s) + 3 CH4(g)
Determine the mass, in grams of methane, when 156 grams of Al4C3(s) are allowed to
react with 192 grams of water.
(note, you must solve for the mass of methane)

Recognition:
Given: Al4C3(s) + 12 H2O(l)  4 Al(OH)3(s) + 3 CH4(g)
Determine the mass, in grams of the reactant which is in excess, when 156 grams of
Al4C3(s) are allowed to react with 192 grams of water.
B) Strategy: Perform a double mole map stoichiometry problem using the quantity of each reactant, as
a "given"... solving for an amount of only 1 product, or the named product.
Whichever calculation gives the smallest amount of product, is the limiting reagent.
To determine how much of a reactant is left over, you could determine how much of it
was used to make the maximum amount of the product, and subtract the mass of the
reactant consumed, from the "given"quantity.
Example: Given: 4 FeS2 + 11 O2  2 Fe2O3 + 8 SO2
When given 951.2 grams of FeS2 and 704.0 grams of O2, determine which reactant
is the limiting reagent.
Run the stoichiometries. Use a reactant value as the "given". You can use either
product as the "desired" ...but use it for both stoichiometric calculations.
calculation 1: g SO2 = 951.2 g FeS2 |1 mol FeS2 | 8 mol SO2 | 64 g SO2 | = 1,015 g SO2
120 g FeS2 4 mol FeS2 1 mol SO2
calculation 2: g SO2 = 704.0 g O2 |1 mol O2 | 8 mol SO2 | 64 g SO2 | = 1,024 g SO2
32 g O2 11 mol O2 1 mol SO2
Since the calculation, using the mass of FeS2 results in the smaller mass of sulfur dioxide, then
FeS2 is the limiting reagent. This would hold true, even if you were to run the stoichiometric calculations
solving for a mass of produced Fe2O3.
93
C) Practice (Basic)
1) Given: 2 Ag2S(s) + 3 O2(g)  2 SO2(g) + 2 Ag2O(s)
Determine the limiting reagent when the above reaction is initiated between 8.0 moles of
silver sulfide and 9.0 moles of dioxygen gas. (notice, the phrase "reacted completely" is missing and you already
have “moles”)
Think about it: What do you know and What can you
figure out?
 You know mole values and the balanced equation
 You can run 2 stoichiometry calculations.
Set the desired as either SO2 or Ag2O [it doesn't really
matter ..but use the same desired for both calculations]
Use the reactant values (8.0 mol of SO2 and then use the
9.0 mol value of O2 as the "given" values.) Whichever
reactant yields the smaller mass is the limiting reagent.
2) Given:
C3H8(g) + 5 O2(g) 
3 CO2(g) + 4 H2O(l)
Determine the limiting reagent when 0.75 mol of propane and 0.75 mol oxygen gas are
mixed in a reaction vessel and allowed to react, according to the ratios of the above balanced
reaction equation.
* mol CO2 = 0.75 mol propane | 3 mol CO2| = 2.25 mol
1 mole propane
* mol CO2 = 0.75 mol oxygen | 3 mol CO2 | = 0.45 mol
5 mole oxygen
(dioxygen gas is the limiting reagent)
….I want to introduce a type of problem, we’ll see later …How to calculate the amount of a reactant remaining after a reaction is completed.
3) Given: 3F2 + S  SF6
A mixture containing 9 mol of F2 and 4 mol of S is allowed to
react. How many moles of F2 remain after 2 mol of S have reacted?
Strategy: First, rephrase the question: How many moles of F2 react with 2 mol of S?
Second: Use that value for moles of F2 and subtract it from the original 9 mol.
mol F2 react = 2 mol S| 3 mol F2| = 6 mol of F2 reacted
1 mol S
Thus: amount of unreacted or remaining reactant = (original amount – reacted amount)
3 =
9
6
94
D) Practice (Going Further)
1) Given: Cu(s) + 2 AgNO3(aq)  Cu(NO3)2(aq) + 2 Ag(s)
Calculate the maxium mass of silver metal produced, when 58.3 grams of copper and
85.0 g of silver nitrate are allowed to react according to the above balanced reaction
* g Ag = 58.3 g Cu | 1 mol Cu | 2 mol Ag | 108 g Ag | = 197 g Ag
64 g
1 mol Cu
1 mol
* g Ag = 85.0 gAgNO3 | 1 mol AgNO3 | 2 mol Ag
| 108 g Ag | = 54.0 g Ag
170 g
2 molAgNO3
1 mol
2) Given: 2 N2(g) + 3 O2(g)  2 N2O3(g)
a) Determine maximum grams of dinitrogen trioxide produced, when 14.0 grams of
dinitrogen gas and 96.0 grams of dioxgyen gas are allowed to react.
* moles N2O3(g) = 14.0 grams | 1mol N2
28 grams
| 2 mol N2O3
2 mol N2
* moles N2O3(g) = 96.0 grams| 1 mol O2 | 2 mol N2O3
32 grams 3 mol O2
| 76 grams N2O3 | = 38.0 grams N2O3
1 mol N2O3
| 76 grams N2O3 | = 152 grams N2O3
1 mol N2O3
b) So, how many grams of the limiting reagent will react? * all 14.0 grams
c) How many grams of the other reactant will react? *24.0 g
are left over? * 72.0 g
* Think about re-writing the question such as: Since I now know 14.0 grams of N2 will react, how
many grams of the dioxygen will react?....And then set up a stoichiometric solution ….
*grams O2 react = 14.0 g of N2 | 1 mol N 2 | 3 mol O2 | 32 grams O2| = 24.0 grams O2 react,
28 grams
2 mol N2 1 mol O2
leaving 72g of the original
96.0 grams unreacted....
95
3) Given: 4 BCl3 + P4 + 6 H2  4 BP + 12 HCl
232 grams of BCl3 were reacted with 155 grams of phosphorus and 10.0 grams of dihydrogen
gas. What’s the maximum mass of BP produced?
* gBP = 232 gBCl3 | 1 mole BCl3 | 4 mol BP | 42 grams BP | = 84.0 grams
116 g BCl3
4 mol BCl
1 mol BP
* g BP = 155g P4 | 1 mole P4 | 4 mol BP
124 g
1 mol P4
| 42 grams BP | = 210. grams
1 mol BP
* gBP = 10.0 g H2| 1 mole H2 | 4 mol BP
2g
6 mol H2
| 42 grams BP | = 140. grams
1 mol BP
*Because it produces the fewest moles of BP…. BCl3 is the limiting reagent, thus, only 84.0 grams of
BP may be produced …regardless of the fact that there is left over H2 and P4
4) Calculate the maximum number of grams of aluminum iodide produced when120.5 grams
of aluminum metal are reacted with 543.1 grams of iodine according to the balanced equation:
2 Al(s) + 3 I2(s)

2 AlI3(s)
NB: The problem does not say “are reacted completely”. This is one clue as to approaching the problem as a limiting reagent probelm
* grams of AlI3 = 120.5 g Al | 1 mol Al | 2 mol AlI3| 408mol AlI3| =
27 grams
2 mol Al
1 mol AlI3
*
1,821 g AlI3
grams of AlI3 = 543.1 g I2 | 1 mol I2 | 2 mol AlI3| 408mol AlI3| = 581.6 g AlI3
254 g I2 3 mol I2
1 mol AlI3
96
5) The complete combustion of ethanol (C2H5OH) is represented by the following reaction.
C2H5OH(l) + 3 O2(l)  2 CO2(g) + 3 H2O(g)+ 1,236 kJ
a) What is the limiting reagent in the reaction when 77.0 grams of ethanol and
19.5 grams of O2 are allowed to react. NB: The problem does not say “are reacted completely”.
* grams CO2 = 19.5 gO2 | 1mole O2| 2 mol CO2 | 44 grams CO2| = 17.9 grams
32 g
3 mol O2
1 mole CO2
* grams CO2 = 77.0 g eth | 1mole eth | 2 mol CO2 | 44 grams CO2| = 147 grams
46 g eth
1 mol eth
1 mole CO2
b) What reagent is in excess? * ethanol or C2H5OH
c) How many grams of the excess reagent should be left in the reaction vessel, once the
limiting reagent is consumed and the reaction has ended?
Think about it: What do you know and What can you figure out?
You know which reactant is used up and goes to 0 grams
Thus, you know how many grams of that reactant are consumed
You can use this measurement to calculate how many grams of
the other reactant get used up (stoichiometry with the limiting
reagent as the given, and the grams of the other reactant that
get consumed, as the desired).
*g eth used = 19.5 g O2 | 1 mol O2 | 1mol eth | 46 g eth| = 9.34 g eth used
32 grams 3 mol O2 1 mol eth
*Since the vessel contained 77.0 grams of ethanol, but ONLY 9.34 grams
were consumed then: 77.0 - 9.34 = 67.7 grams are left over unreacted...
Once you know how many grams of the excess reagent are
consumed, you can subtract that value from the calculation
from the original amount of the reactant. This difference
is what is left over in the beaker of the excess reactant.
Ans: * 67.5 grams ethanol left in the vessel
97
6) When 35.0 grams of C6H10 and 45.0 grams of O2, are allowed to react, based upon the
following balanced chemical reaction, how many grams of the excess reagent will remain
after the reaction ceases?
6 C6H10 + 17 O2  12 CO2 + 10 H2O
Think about it: What do you know and What can you figure out?
You know the masses of the reactants and the ratios by which
the molar equivalents of those masses, react.
Thus, by finding the limiting reagent ... you also know the
reagent that is in excess. Once you know the excess reagent's
identity, you can use the mass of the limiting reagent to
calculate how many grams of the excess are consumed (just
as in the last problem)
Once you know how many grams of the excess reagent are
consumed, you can subtract that value from the calculation
from the original amount of the reactant. This difference
is what is left over in the beaker of the excess reactant.
* C6H10 is the limiting reagent
* g O2 = 35.0 g C6H10 | 1 mol C6H10| 17 mol O2
82 grams
| 32 g O2| = 38.7 g O2 used
6 mol C6H10 1mol O2
* Only 38.7 g of dioxygen of the original 45.0 grams get consumed.
So, 45.0 -38.7 = 6.30 grams left over.
*Ans: 6.30 grams of O2 are left over
7) Given:
6 I2O5 + 20 BrF3  12 IF5 + 15 O2 + 10 Br2
a) Calculate the maximum mass, in grams of IF5 would be produced using 44.01 grams
of I2O5 and 107.0 grams of BrF3.
* g IF5 = 44.01 g I2O5 | 1 mol I O | 12 mol IF | 222 g IF | = 58.50 grams IF5
334 g I O 6 mol I O 1 mol IF
2
5
2
5
5
2
5
5
5
* g IF5 = 107.0 g BrF3 | 1 mol BrF3 | 12 mol IF | 222 g IF | =
137g BrF3 20 mol BrF3 1 mol IF
5
5
104.0 grams IF5
5
b) Calculate the number of grams of the excess reactant left in the reaction vessel,
once the reaction ended.
* g BrF3 consumed = 44.01 g I2O5| 1 mol I O | 20 mol BF | 137 g BrF3 | = 60.16 grams consumed
2
334 g I2O5
5
3
6 mol I2O5 1 mol BrF3
* 107.0 - 60.16 = 46.8 grams left over
98
XI) Percent Yield = actual yield
x 100
theoretical yield
A) If you have ever ended up with a percent error other than 0%, you understand that chemical
reactions can go wrong. There is a difference between the amount you produced and the amount you
should have (theoretically) produced.
1) Procedure:
a) determine the limiting reagent (sometimes it is given ...& sometimes you need to find it)
b) determine the theoretical yield assuming a perfect reaction using the limiting reagent
value ...but, realize that the theoretical answer of the calculation, is different than the
experimental value (amount of the of the product), given in the problem.
c) determine the actual yield. Often, this is given in the problem.
d) calculate the % yield, using the above equation and two yield values.
2) Practice: (answers are on the next page)
When 84.8 grams of iron (III) oxide (GFM = 160 g) react with an excess of carbon
monoxide, 52.1 grams of iron (GFM = 56 g) were experimentally produced.
(NB the CO is in excess, thus Fe2O3 is the limiting reagent, automatically)
Fe2O3 + 3 CO  2 Fe + 3 CO2
a) Calculate the theoretical yield (...just do the stoch problem...)
b) Calculate the % yield
3) When 50.0 g of silicon dioxide (GFM = 60 g) are heated with an excess of carbon, a student
found that 27.9 g of silicon carbide (GFM = 40 g) are produced.
SiO2 + 3 C  SiC + 2 CO
What is the percent yield of this reaction?
you need the actual yield (from the problem)
& the theoretical yeild from the stoich (math)
99
4) The following reaction has a per cent yield of 96.8%. How many grams of
calcium sulfate (GFM = 100 g) are formed when 53.4 grams of sulfur dioxide (GFM = 64 g)
reacts with an excess of calcium carbonate and dioxygen gas?
2 CaCO3(s) + 2 SO2(g) + O2(g)  2 CaSO4(s) + 2 CO2(g)
Answers for percent yield
2a g Fe = 84.8 gFe2O3| 1 mole Fe2O3| 2 mol Fe
| 56 g Fe| = 14.8 g
160 grams
1 mol Fe2O3 1 mol Fe
b) % yield = actual yield
x 100
theoretical yield
% yield = 52.1 x 100
59.4
87.7% yield
3) First calculate the theoretical grams, using silicon dioxide as the limiting reagent.
g SiC = 50.0 g SiO2 | 1mol SiO2| 1mol | 40 grams|
60 grams 1 mol 1 mol SiC
= 33.3 g SiC
thus:
% yield = actual yield
x 100
theoretical yield
% yield = 27.9 x 100
33.3
83.8% yield
4) g CaSO4 = 53.4 g SO2 | 1mol | 2 mol | 100 g CaSO 4| = 81.9 (theoretically)
64 g
2 mol
1 mol
With a % yield of 96.8%, only 79.3 grams are produced really.
100
NAME ________________________________ NOTES: UNIT 8 MOLE AND STOICHIOMETRY (6)
FOCUS ON: EMPIRICAL FORMULA USING PERCENT COMPOSITION
XII) Empirical Formula
Point1 : Definition: The empirical formula is the simplest mole ratio between the elements of a compound.
It is the chemical formula, written with the smallest, whole number subscripts,
thus giving the mole ratio between the elements: e.g. CaCl2 has 1:2 ratio or rather, in
1 mol of CaCl2, there is ration of 1 mol of Ca2+ ions to 2 mol of Cl1- ions.
The formula C6H12O6 is NOT an empirical formula. The empirical formula is CH2O, but no
chemical is thought to exist with that formulation (it is just the mole ratio of elements)
To find the simplest mole ratio:
Point2 : Calculate the number of moles of each element of a compound
Determine the ratio between each of the mole numbers.
Point3 : Find the ratio by dividing each mole measurement by the smallest of the mole values.
Point4 : When a quotient value is within ONE TENTH of a whole number, you may round UP (or
down) to that whole number. IF NOT, you must multiply all quotient mole values by a common
factor to get whole numbers. The mole ratio (empirical formula) must represent whole numbers
in the simplest ratio.
A) PROCESS FOR DETERMINING THE EMPIRICAL FORMULA: THERE ARE 3 STEPS
1) Assume 100g of the compound and convert the percentage compositions to grams
2) Use unit cancellation to convert the grams of each element to moles of each element
3) Divide each mole number by the smallest of the mole numbers to get the ratio between the elements.
TRY THIS: The percent composition of a compound is 11.1% hydrogen and 88.9% oxygen.
(So, assuming 100 grams, you have: 11.1 grams of hydrogen (H) and 88.9 grams of oxygen (O)
moles hydrogen = 11.1 grams  1 mole of H atom  = 11.1 moles
1 gram  Did you note the GAM
of 1 and NOT the
GFM of 2 ??? You’re dealing with moles of H atom,
NOT moles of H2 molecule, the diatomic element
moles of oxygen = 88.9 grams
 1 mole of O atom| = 5.56 moles
16 grams  See that use of a GAM of 16 and not 32???
You’re dealing with moles of O atom, NOT moles
of O2 molecule, the diatomic element
Thus, the number of moles of hydrogen and oxygen is 11.1 : 5.56
or:
H11.1O5.56
….but we’re not done … we need the ratio between these mol values.
101
H = 11.1 mole
5.56 mole
O = 5.56 mole
5.56
= 1.996 which rounds UP to
2
= 1
Thus, the simplest ratio is 2 : 1 or rather, the EMPIRICAL FORMULA = H2O1 or simply H2O
 NOTE:
You may round (as in the above case) values such as 1.9. You could NOT round were it 1.8 You can only round
when the value is 0.1 off from a whole number.
GUIDED PRACTICE PROBLEM:
Determine the empirical formula for the compound which when analyzed showed 70.9% potassium by mass and 29.1%
sulfur by mass.
STEP 1: Assume 100g. Thus, you have
_____________g of K and
_____________g of S
Review: This value is a mole mass
(a GFM) from the periodic table
STEP 2: Determine the mole ratio, by converting grams to moles
K
x moles K =
___________________|
39 grams
=
S
x moles S =
__________________|
32 grams
=
STEP 3: Divide each by the smallest mole number to determine the simplest whole number ratio
K
=
S
=
The empirical formula therefore is:
Ans. = K2S
102
COMPLETE THE FOLLOWING PRACTICE PROBLEMS
1) Determine the empirical formula of a compound analyzed to be 75.0% carbon and 25.0% hydrogen.
(Remember … you really need to determine the number of moles and then the mole ratio….)
Step 1)
Step 2)
Step 3)
Ans:=CH4
2) Determine the empirical formula of a compound analyzed to be 85.0% silver, and 15.0% fluorine.
(Remember … determine the number of moles and then the ratio….)
Step 1)
Step 2)
Step 3)
Ans:=AgF
3) A sample of a compound was analyzed to be 58.8% barium, 13.75% sulfur and 27.45% oxygen. Determine the
empirical formula.
Step 1)
Step 2)
Ans:=BaSO4
103
4) By chemical analysis, a compound was found to be 80.0% carbon and 20.0% hydrogen. What is the
compound’s empirical formula?
Ans:=CH3 ..note that no stable compound exists with this formula ... it is just an empirical formula, illustrating the basic mole ratio...
5) A sample of a substance was found to contain 26.7% phosphorous, 12.1% nitrogen and 61.2% chlorine. What is the
empirical formula of the compound?
Ans:=PNCl2
TRY THIS:
___1. A compound contains 16% carbon and 84% sulfur by mass. What is the empirical formula of this
compound?
1) CS2
3) CS
2) C2S2
4) C2S
ans.1
___2. A sample of a compound contains 24 grams of carbon and 64 grams of oxygen. What is the empirical
formula of the compound?
1) CO
3) C2O2
2) CO2
4) C2O4
ans.2
___3. Using the definition and your understanding of chemistry, which is an empirical formula?
1) C2H2
3) N2O4
2) H2O
4) C6H12O6
ans.2
104
MORE PRACTICE FOR DETERMINING AN EMPIRICAL FORMULA
1. Determine the empirical formula of a substance which is 85.6% carbon and 14.4% hydrogen.
(Remember … determine the number of moles and then the ratio….)
a)
b)
c)
d)
CH2
C2 H
CH
C2H3
2. Which of the following substances is written in an EMPIRICAL formula?
a) Ni4S6
b) P2O5
c) C6H6
d) C2H6
3. Given the following per cent composition, by mass, determine the empirical formula. A substance was
40.0% calcium, 12.0% carbon, and 48.0% oxygen. (Remember … determine the number of moles and then the ratio….)
a)
b)
c)
d)
CaCO3
CaCO
Ca2CO3
Ca2C2O3
5. A compound contains 0.5 mole of sodium, 0.5 mole of nitrogen, and 1.0 mole of hydrogen.
The empirical formula of the compound is:
a) NaNH
b) Na2NH
c) NaNH2
d) Na(NH)2
5. Define the term, empirical formula ____________________________________________________
6. Assuming each of the following exist as the formula in which it is written, which formula represents BOTH
the a empirical formula and molecular formula of the substance?
a) P2O4
b) NO2
c) C2H2
d) C2O2H6
Ans.: 1) a 2) b (simplest whole # mole ratio) 3) a 4) c 5) The system of symbols and subscripts representing the simplest whole number mole
ratio between the elements (atoms, ions) of a compound. The empirical formula may or may not represent how the substance exists in nature. 7) b
105
NAME ________________________________ NOTES: UNIT 8 MOLE AND STOICHIOMETRY (7)
FOCUS ON: GAS LAWS
XII) The Mole and Gas Laws
A) There are a variety of gas laws: Ideal, Combined, Gay-Lussac’s, Boyle’s, Charles’s, Dalton’s Law of
Partial Pressures, Graham’s Law of Diffusion, Henry’s Law …. At a glance, the gas laws address
the changes experienced in the physical properties of a gas.
1) These physical properties of a gas surround, mass, the Kelvin temperature, pressure, volume
and the number of moles.
2) The gas phase is the phase of matter, most susceptible to alterations in temperature and
pressure.
3) Essentially, most gas laws connect somewhere with the Kinetic Molecular Theory
Idea1: The Kinetic Energy of an ideal gas is directly proportional to the absolute (Kelvin)
temperature
Double the Kelvin temperature … you double the average kinetic energy
of the gas molecules
Idea2: When several different gases are present at a given temperature, all the gases
will have the same average kinetic energy … for the average kinetic energy
depends upon the absolute (Kelvin) temperature, and NOT the identity of the
gas.
Idea3: The volume and the mass of an individual ideal gas particle is insignificant in
comparison to the volume and the mass of the ideal gas sample
Idea4: There are no forces of attraction between the gas molecules of an ideal gas
Idea5: Gas molecules are in constant motion, colliding with each other, and with the
walls of their container …Pressure, exerted by a gas, is essentially the measure
of those collisions.
2) An ideal gas really is just a mathematical construct … no such gas exists … thus real gases,
like the noble gases, carbon dioxide, dioxgyen, ozone etc… exhibit. Real gases do
have a measureable volume
exert intermolecular forces of attraction upon other molecules
have a measureable mass
106
B) Ideal Gas Law:
PV = nRT
Where:
P = the pressure of the gas (atmospheres or kPa)
V = the volume of the gas (Liters)
n = the number of moles of gas
T = the absolute temperature (Kelvin)
R= the universal gas constant, 0.0821 L•atm/mol• K
1) Some relationships derived from the ideal gas law
a) At constant temperature: P is inversely proportional to V
or you could say that
Pressure is proportional to 1
V
e.g. as pressure doubles, volume is reduced to ½
Take Home Message: Pressure and volume operate oppositely…
b) At constant volume: P is directly proportional to TK
e.g. imagine a gas with a pressure of 2 atm and 300K … when the temperature is
raised to 600K, the pressure becomes 4 atm.
c) Pressure is proportional to n
e.g. add more moles of gas to a sample, and the pressure increases proportionately
Hence: when we summarize the 3 rules:
P is proportional to 1
V
P is proportional to T
P is proportional to n
P is proportional to nT
V
The universal gas constant was introduced to make the proportionality into an equality
Hence:
P = R (nT)
V
OR
P = nRT and this gets us:
V
PV=nRT
107
With all of that done (PHEW!) … there are some further generalities that can be observed, before moving on!
2) According to the ideal gas law, what happens to the pressure of a gas in a container when:
So, let’s solve the ideal gas law PV=nRT for Pressure:
P = nRT
V
you double the Kelvin temperature of the gas? * the pressure doubles, since T is in the numerator
you double the volume of the container?
* the pressure is cut by ½
you double the number of moles of gas?
* the pressure is doubled.
TRY THIS:
You work in a factory filling balloons with helium gas. One day your boss accuses you of over-filling
balloons, with helium… thus costing him money. Your boss demands to know the mass, in grams, of
helium in a balloon, at 0.984 atm and 22.0 ºC. Each balloon is to have an average volume of 240. mL.
The Ideal Gas Law comes to your rescue …. It does not have “grams” as a variable … but it does have
moles … and if you solve for moles … you can of course, convert to grams.
Hence a quick rearrangement: * n= PV
RT
Substitution:
*n = (0.984 atm) (0.240L)
(0.0821 L•atm)(295K)
K•mol
n = 0.00975 mol He.
Now, use unit cancellation:
*grams He = 0.00975 mol | 4 grams | = 0.0390 grams
1 mol
You are in good shape… as long as the company policy agrees with your value!
108