NAME_______________________________ INDEPENDENT PACKET: PERCENT COMPOSITION
GOAL :
Upon completion of this independent packet, you will be able to determine the percent, by
mass, of any given element found in a compound.
The equation for determining the percent by mass is on Table T of your Reference Charts.
In general, to determine percent, you must be concerned with the
Mass of Part x100
Mass of the Whole
With this packet, you will learn how to determine the percent composition of a compound, USING MOLE
MASS. You know how to determine the gram atomic mass of any element (by looking it up on the periodic
table) and you know how to determine the gram formula mass of any compound.
But if you need a visualization, try the following.
PRETEND: 1 giraffe weighs 500 lbs and 1 balancing elephant weighs 2,500 lbs.
Effectively, we could say the formula of the above grouping is
3
2
To determine the percent composition by mass of due to the giraffe is something like:
3(mass of a single giraffe) x 100
mass of the total group
OR
3(500 lbs) x 100
6500 lbs
With this visualization, I am trying to reinforce the idea that the "part" you want (e.g. mass of an element) is
divided by the entire sample's mass.
Given the following sample,
What per cent of the sample is
?
Part you want x 100
Whole
therefore:
or you could say:
3 (mass
)
x 100
x 100
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Process for determing percent composition using Mole Mass: (use the E.S.A. Method)
Step 1) Determine 1 mole mass of the compound. (any mass will work ...but 1 mol mass is such a nice
beginning...)
Step 2) Multiply the GAM of the element in question by its subscript to find the element's total mass
Step 3) Divide the element's total mass of the element by the GFM of the compound from step 1.
(This fulfills the equation of part/whole)
Step 4) Multiply this quotient (from step 3) by 100 so that you may obtain a percent or percentage.
Most answers will be recorded in two or three sig figs… I will not hold you accountable though
for sig figs with this topic. With all the rounding ...it is too difficult.
Use the following as an example of a percent composition problem.
Calculate the percent composition, by mass of potassium ion potassium sulfide (K2S)? (You can assume 1 mol)
a) Equation: % K = Total Mass of K
GFM of K2S
b) Substitution:
(2) (39.1) g
110.3 g
which equals:
c) Answer :
x 100
Taken from the periodic table
by looking up potassium (#19).
x
100
78.2 g x 100
110.3 g
In AP Chem and in lab, we should use the
entire value of the relative atomic mass.
70.89 % or 70.9 % (sig figs are a bit iffy …because of the rounding.
Thus, for this topic, I focus on process
& have been limiting answers to 3 sig figs …based upon the rounded atomic mass value.)
1) Calculate the percent composition of chloride ion, by mass in 1 mol mass of lithium chloride (LiCl).
Use E.S.A.
Equation:
% = Part x 100
Whole
Substitution:
Answer: __________% chloride ion
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2) Assuming 1 mol of substance, determine the percent composition of iodide ion, by mass in BaI2
Equation:
Substitution:
Answer: ___________% iodide ion
3) Find the percent composition of chloride ion in CrCl3
4) Find the percent composition of cadmium ion in CdCO3
5) Find the percent composition of barium ion in BaF2
6) What is the percent composition of hydrogen in CH3COOH? (Hint: There are multiple C's, H's,& O's. What
do you think you need to do to find the GFM?....Yep, group the C’s with each other & group the hydrogen ... and oxygen….)
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7) Find the percent composition of carbon in CH3CH2CH2CH2OH
8) Find the percent composition of nitrogen in
ANSWERS: 1) 83.3 % Cl ion 2) 64.9 % I ion 3) 66.6 % Cl ion
N
(NH4)3PO4
4) 65.2 % Cd ion
5) 78.3 % Ba ion
6) 6.67 % H 7) 64.9 % C
8) 28.1 %
NOTE: At this point, to make things a bit simpler, I am just going to use atomic mass values, rounded to the
nearest whole number … In AP or in lab use the entire decimalized values. But for the learning process, we’ll
round to the nearest whole number value. So 1 mol of Cl atoms = 35 grams, 1mol of O atoms = 16 grams etc…
Now try something … a bit more complex ...
Percent Composition and Hydrates
Hydrates are crystals that have a specific number of water molecules trapped in the crystals. This translates nicely
to mole theory. For every 1 mole of formula units of an ionic compound, a specific number of moles of water are trapped
inside the crystals.
Many ionic compounds when crystallized from aqueous solutions retain a definite proportion of water as part of their
crystalline structure. This water is referred to as the water of hydration. These ionic crystals appear to be perfectly dry.
When an ionic crystal is in possession of this "trapped water", the crystal is called a HYDRATE. Upon heating a sample
of the ionic crystal, this water may be driven off. As the water is driven off, the crystalline structure is shattered and all
that remains is a dry powder. This dry powder is said to be the ANHYDROUS salt. (The prefix "AN-" simply means
"WITHOUT" and "HYDROUS" refers to water, so, an "anhydrous salt" refers to an ionic crystal WITHOUT WATER.)
For instance, MgSO4•7H2O (called magnesium sulfate heptahydrate) has 7 moles of water molecules per 1 mole of
magnesium sulfate formula units. The prefix hepta means 7, and hydrate refers to the water.
In the example of BaCl2•10 H2O (barium chloride decahydrate), there are 10 moles of water trapped within 1 mole of the
barium chloride. This trapped water is insufficient to dissolve away the crystal. The prefix deca means 10.
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The calculation of the gram formula mass is pursued along the same lines as any other compound. However, the
coefficient for the moles of water matters tremendously.
Can you follow how the gram formula mass of MgSO4•7H2O is calculated? Analyze the following…
mass of 1 mol Mg + mass of 1 mol S + mass of 4 mols O + mass of 14 mols H + mass of 7 mols O
(1)24 g
+
(1) 32 g
+
4(16 g)
+
14(1 g)
+ 7 (16 g) = 246 g/mol
Now, once you have the GFM of the hydrate, determine the percent composition follow the same procedure you
have been practicing right along.
1) Calculate the percent composition, by mass, of water, trapped in 1.00 mole of MgSO4•7H2O
2) Calculate the percent composition of water, by mass, in 1.00 mole of BaCl2•10 H2O
3) Calculate the percent composition of water, by mass, in 1.0 mole of CuSO4•5 H2O
ans: 1) 51.2%
2) 46.5%
3) 36%
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Now try something a bit more complex ....Yes, AGAIN!!
Percent Composition of a Mixture
Hey, what if you did not have a "pure" sample of a compound? What if only 38.2% of the mixture were
actually made of the compound, in question? How would you calculate the amount of the specific compound?
Well, it would involve taking a percentage of another percentage value.
Well, how about treating it as if the whole sample were the compound ... and then multiplying by the actual
percentage once you have that value?
Look, try this on for size....
Consider the issue of sandstone mixtures. They were important to the glass industry, but today, sandstone
mixtures are very important in hydraulic fracturing (fracking)…Fracking is that process of using highly
pressurized fluids injected into the ground to create small fissures (fractures) allowing trapped methane and
other fossil fuels to move up the fissures to be collected. The sandstone mixture is used as a proppant. A
proppant is material that is used to “prop” open or keep the small fissures that are formed in the rock open for
more effective fuel recovery. http://www.gp.com/chemicals/Proppant_Resins_Product_Category )
Now, consider you have found a sandstone deposit, made primarily of silicon dioxide SiO2 (silica). The only source
of silicon in the sandstone is the silica. An assay of the mixture shows that 81.5% of the sandstone mixture is SiO 2
Question: What is the total mass of silicon, in grams in a 15.0 gram sample of a sandstone mixture of
which 81.5% of the mass is due to SiO2?
Solution:
1) find the percent composition of silicon in SiO2. This will give you the maximum mass ... as if the entire
sample were silicon dioxide.
Equation: % Silicon = Total Mass of Silicon x100
GFM of SiO2
Substitution:
%Si = 28 grams x 100
60 grams
Answer: 46.7 % silicon in SiO2 by mass (I rounded to 3 sig
figs because the values in the original question had 3)
Understand this point: If the entire sample were silicon dioxide, then 46.7% of 15.0 grams would be about
7.01 grams of Si.
But Remember (!) Only 81.5% of the 15.0 gram mixture is SiO2
2) Determine how much of the mixture's mass is the desired compound: (0.815)(15.0 grams) = 12.2 grams SiO2
3) Determine the grams of the element, in the mass due to the compound, using the percent composition from
step 1 and the mass of step 2:
total mass of element = (grams of compound in mixture) (percent element by mass)
(12.23 grams of SiO2) (0.467) = 5.71 grams of silicon (Answer)
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There are all sorts of variations on that last problem. You'll see a question like this on your quiz.... There is
more than one way to solve the problems … I offer only one of those ways, but TRY and THINK!
1) A 455 grams mixture of rock is 20.5% Fe2O3 by mass, and it is the only source of iron in the mixture.
Calculate the number of grams of iron in the mixture.
Here's the answer: The mass of Fe in the original mixture is 65.3 grams.
For the solution, get this page off the website. Highlight this section of the * and turn the font to black....
*a) Find the % composition of iron in Fe2O3.
% Fe = 112 x 100
160
= 70.0%
b) Find the mass of the Fe2O3 in the 455 gram sample: mass Fe2O3 = (455)(0.205) = 93.3 grams Fe2O3
c) You know that you have only 93.3 grams of Fe2O3 and you know that 70% of the compound is Fe.
So, to find the total mass due to just Fe in the mixuture: (0.700)(93.3) = 65.3 grams Fe
2) All of the zinc ion in a 1.22 gram mixture of zinc-containing material is reacted to produce 1.6374 grams of
a precipitate, Zn2P2O7. Calculate the percent by mass of zinc ion in the original mixture.
Here's the answer: The % composition of Zn (as zinc ion) in the originall mixture is 57.5%.
For the solution, get this page off the website. Highlight this section of the * and turn the font to black....
* a) Since all the zinc is trapped in the precipitate, find the % composition of zinc in the precipiatate.
% Zn = mass of Zn
x 100
GFM of Zn2P2O7
=
% = (65)(2) x 100
304
= 130 x100 = 42.8%
304
b) Now that you know the % by mass of zinc in the precipitate, you can find the mass of Zn in that
precipitate of Zn2P2O7 .... (0.428)(1.6374 grams) = 0.701 grams of zinc ion
c) You can now infer that 0.701 grams of zinc ion is the total mass of the original mixture (since that
is the mass of zinc in the precipitate and all of the zinc ended up in the precipitate).
d) use this mass to find the % composition of zinc in the mixture ...and be sure to use the mixture's mass
% composition Zn = 0.701 grams x 100
1.22 grams
percent comp of Zn in the original mixuture = 57.5%
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3) A 500. gram mixture of soluble salts contains sodium chloride (NaCl). The NaCl is the only source of
chloride ion in the mixture.
When the mixture is dissolved in water and reacted with an aqueous solution of silver nitrate, 42.5 grams of
AgCl are precipitated out of solution.
How many grams of NaCl are in the mixture, and what is the percent composition by mass of NaCl in the
mixture?
Here are the answers: The mass of NaCl in the original mixture = 17.2 grams and the % composition of
NaCl in the mixture is 3.44%.
For the solution, get this page off the website. Highlight this section of the * and turn the font to black....
* a) find the percent composition of Cl in AgCl ...
% = 35 x 100
143
= 24.5% Cl
b) You now know the percent by mass of Cl in AgCl. You can use this value to find the mass due to
Cl in the 42.5 grams of AgCl. This will give you the total amount of Cl, in grams of the original
mixture as well.
24.5% of 42.5 grams of AgCl is due to the mass of Cl
OR
(0.245) (42.5) = 10.4 grams Cl
c) Since Cl is the only common species, between AgCl and NaCl , it would be helpful to find the
percent composition of Cl in NaCl.
%Cl = mass Cl x100 =
% Cl = 35 x 100 = 60.3 %
GFM NaCl
58
Now you know that 60.3% by mass of NaCl is due to Cl. AND! You know that you have a mass
of 10.4 grams of Cl, in the original mixture.. .
So, 10.4 grams is 60.3% of what total mass? OR
10.4 grams = (0.603)(x) = 17.2 grams NaCl
d) Thus the mass of NaCl in the mixture = 17.2 grams
and the % by mass of NaCl in the mixture must be % = 17.2 x 100 = 3.44% NaCl
500.
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NAME ____________________________________ PREQUIZ 1: PERCENT COMPOSITION
DIRECTIONS: Complete each question by selecting the most correct answer. You do not need to show your
work - but following the ESA process limits the "human" error dramatically.
___ 1) What is the percent composition of iron, ion, by mass in the compound, Fe2O3(s)?
a) 75 %
c) 70 %
b) 23 %
d) 67 %
___2) Which of the following compounds is mostly chloride ion, according to its percent composition by mass?
a) MnCl2
c) ICl
b) NaClO4
d) LiCl
___ 3) Which of the following compounds has the greatest percent composition by mass, of oxygen?
a) BaO
c) Al2O3
b) CO
d) NaClO
4) Calculate the percent composition by mass of water in 1 mol of the hydrate LiCl•H2O
Answers: PQ #1: 1) c 112/160 x 100
2) d
3) b
4) 30% … and, don’t forget the more complex percentage of mixtures will be on the quiz.
239
NAME ____________________________________
PREQUIZ 2: PERCENT COMPOSITION
DIRECTIONS: Complete each question by selecting the most correct answer. You do not need to show your
work - but following the ESA process limits the "human" error dramatically.
___ 1) What is the percent composition, by mass, of manganese in the compound, MnSO4(s)?
a) 55.7 %
c) 10.8 %
b) 36.4 %
d) 27.1 %
___ 2) In which of the following compounds is mostly fluoride ion, according to its percent composition
by mass?
a) HF
c) NaF
b) CCl3F
d) CrF2
___ 3) Which of the following compounds has the greatest percent composition by mass, of nickel?
a) NiClO2
c) NiCl3
b) Ni2S3
d) NiO
4) Calculate the percent composition by mass of water in 1mol of the hydrate BaCl2•2 H2O
PQ # 2
1) b
2) a
3) d
4) 14.8% … and don’t forget the more complex percentage of mixtures will be on the quiz
240
NAME _____________________________ NOTES: UNIT 4 (2): THE MOLE & STOICHIMETRY
APPLYING DIMENSIONAL ANALYSIS TO MOLE THEORY
I-II) Mole and Mole Mass
III) Stoichiometry (Greek) * measure of the element (measuring the balance of the element)
stoicheion: element
This connects to the Law of the Conservation of Matter (MR #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) RECAP: The basic Mole Map and Simple Conversions
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
1 mole of a substance
use formula &P.T. : changes w/ substance
grams
(constant)
(constant)
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 substance
*
doing any conversion problem when there is no balanced equation
241
DIRECTIONS: Below are a number of simple mole conversion problems. I have provided you with the
conversion factors you require. Focus upon becoming comfortable with the way questions are constructed
and the application of your knowledge of unit cancellation. All answers should have 3 sig figs.
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
2 g
6.02 x1023
molecules
all are absolute numbers
What we know…
3) How many moles of Ca(NO3)2(s) equal 17.0 grams ?
1 mole
164 g
6.02 x1023
formula units
all are absolute numbers
4) How many molecules of O2(g) are equivalent to 40.6 liters of oxygen gas at STP ?
22.4 Liters
1 mole
32g
6.02 x1023
molecules
all are absolute numbers
Answers: 1) 7.37 grams
2) 0.869 grams
3) 0.104 mol
4) 1.09 x 1024 molecules of O2
242
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
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NAME _____________________________
STOICHIOMETRY: EXTRA 1 TO 2 STEP MOLE PRACTICE
These are just a few extra practice problems. I urge you to keep to the technique …. of
Desired unit = Given unit’|
| and then designing a mole map, to keep you organized.
unit’
1) Calculate the volume, in liters equivalent to 71.92 grams of H2(g) at STP
2) Calculate the number of molecules equivalent to 38.9 grams of NH3(g).
Did you draw a mole map?
3) Calculate the mass, in grams of 1.20 x 1024 molecules of O2(g)
Did you draw a mole map?
4) Calculate the mole equivalent of 454 grams of CO2(g)
5) How many moles are equivalent to in 13.00 grams of NaHCO3(s)?
244
6) Calculate the number of formula units equivalent to 327.89 grams of MgSO4
Did you draw a mole map?
7) Calculate the number of moles equivalent to 0.9201 Liters of He(g) at STP.
8) Which has a greater number of molecules A) 0.507 L of CO2(g) at STP OR B) 0.987 grams of C6H12O6?
Did you draw 2 mole maps?
Answers:
22.4 L at STP
1) LitersH = 71.92 grams H | 22.4 L | = 805.8 L
2 grams
2(g)
1 mol
2(g)
2 grams
1 mole
2)moleculesNH = 38.9grams NH | 6.02 x 1023 molecules| = 1.38x 1024 molecules
17 grams NH3
3(g)
3(g)
3) gramsO2 = 1.20 x 1024 molecules O2 |
32 grams
| = 63.8 grams
6.02 x 1023 molecules
6.02 x1023molecules
17 grams
1 mol
32 grams
6.02 x 1023molecules
245
1 mol
4) mol = 454 grams CO2 | 1 mol
| = 10.3 mol
44 grams
44 grams
5) mol = 13.00 grams NaHCO | 1mol | = 0.1548
84 grams
1 mol
3
84 grams
6) formula units = 327.89 grams MgSO4 | 6.02 x 1023 formula units | 1.6449 x 1024 formula units
120 grams
1 mol
7) moles = 0.9201 L He | 1 mol | = 0.04108 mol
22.4 L
22.4 L at STP
120 grams
6.02 x 1023formula units
1 mol
2 grams
22.4 L
8) to solve, run both conversions…whichever gives the larger value is the answer
44 g
1 mol
6.02 x 1023molecules
A) molecules = 0.507L CO2 | 6.02 x 1023 molecules| =1.36 x 1022
22.4 L
1 mol
B) molecules = 0.987 grams C6H12O6 |6.02 x 1023 molecules| = 3.30 x 1021
180 grams
180 g
6.02 x 1023molecules
ANS: Thus, the 0.507L of CO2 has the greater number of molecules.
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IV) More Complex Stoichiometry and the Double Mole Map
A) When should you get the “hint” that you 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?
*1
how many cups of sugar ?
*1
how many cups of flour?
*1
how many squares of chocolate? *2.5
How many batches of brownies can I make? *1/2 batch
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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 dioxygen molecules are consumed in the
combustion (a ratio of 2 to 7) This produces 4 moles of carbon dioxide, 6 moles of
water and releases 3,170 kJ of energy from the bonding chemicals to 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 H2O 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 questions:
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.
248