Unit 3: Stoichiometry
Note: Page numbers start with the first page of the NMSI Packet, not with these added
instructions/directions. Problems that are not in the packet (specifically the FRQs) are to
be worked out on an individual sheet of paper, one problem per side of paper, and
they will be attached to your packet when you turn it in.
Page 1: Make sure you have a good fundamental understanding of how a mass
spectrometer works.
Page 2: Ms. McCormick gives a dimensional analysis (DA) disclaimer here. Read
through this and be mindful of it as you do the problems.
I tend to naturally resort to DA when I do problems. For me it’s easier to work through
that way and for me to “see” the problem. This comes from many years of habit and at
times I do take shortcuts to the “process”, as would be expected with a certain level of
experience with something.
The bottom line: I don’t care how you do your problems so long as I can make out what
you’re doing, understand how you arrived at an answer, and, really, you get the right
answer.
Page 5: Ms. McCormick’s definition of empirical formula (EF) is fine if you’re mostly
focused on ionic compounds, but in fact when you do problems where you need to find
the actual chemical formula for a compound and the compound is ionic, whatever
answer you get for that compound will be the only formula you come up with, i.e. the EF
IS the formula; this is not always, and it’s frequently not the way it works for molecular
compounds. Ionic compounds don’t have molecular formulas (mostly due to the fact that
they’re not molecular) and so you don’t ever have to divide the molecular mass by the
EF molar mass to give you a number that you then multiply the EF subscripts by to find
the “true” formula for the compound.
Anyway, to avoid confusion I’ve take the definition of EF out of your textbook:
Chemical formulas that indicate the actual number of and types of atoms in a molecular
are called molecular formulas.
Chemical formulas that give only the relative number of atoms of each type in a
molecule are called empirical formulas. The subscripts in an empirical formula are
always the smallest possible while number ratios. The molecular formula for hydrogen
peroxide is H2O2, whereas its EF is HO. The molecular formula for ethylene is C2H4, and
its EF is CH2. For many substances the molecular formula and the EF are identical, as
in the case of water, H2O.
Page 13: The “Dimensional Analysis Disclaimer” presented here is essentially a
duplicate of the one you have on page 2.
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AP* Chemistry
STOICHIOMETRY
3.1 ATOMIC MASSES
•
•
12
C—Carbon 12—In 1961 it was agreed that this would serve as the standard and would be
defined to have a mass of EXACTLY 12 atomic mass units (amu). All other atomic masses
are measured relative to this.
mass spectrometer—a device for measuring the mass of atoms or molecules
o atoms or molecules are passed into a beam of high-speed electrons
o this knocks electrons OFF the atoms or molecules transforming them into cations
o apply an electric field
o this accelerates the cations since they are repelled from the (+) pole and attracted
toward the (−) pole
o send the accelerated cations into a magnetic field
o an accelerated cation creates it’s OWN magnetic field which perturbs the original
magnetic field
o this perturbation changes the path of the cation
o the amount of deflection is proportional to the mass; heavy cations deflect little
o ions hit a detector plate where measurements can be obtained.
o
•
•
Mass13C
= 1.0836129 ∴ Mass13C = (1.0836129)(12amu ) = 13.003355amu
12
Mass C
Exact by definition
average atomic masses—atoms have masses of whole numbers, HOWEVER samples of
quadrillions of atoms have a few that are heavier or lighter [isotopes] due to different
numbers of neutrons present
percent abundance--percentage of atoms in a natural sample of the pure element
represented by a particular isotope
percent abundance =
number of atoms of a given isotope
× 100
Total number of atoms of all isotopes of that element
*AP is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.
© 2008 by René McCormick. All rights reserved.
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•
•
counting by mass—when particles are small this is a matter of convenience. Just as you
buy 5 lbs of sugar rather than a number of sugar crystals, or a pound of peanuts rather than
counting the individual peanuts….this concept works very well if your know an average
mass.
mass spectrometer to determine isotopic composition—load in a pure sample of natural
neon or other substance. The areas of the “peaks” or heights of the bars indicate the relative
20
22
abundances of 10
Ne , 1021 Ne , and 10
Ne
Exercise 3.1
The Average Mass of an Element
When a sample of natural copper is vaporized and injected into a mass spectrometer, the
results shown in the figure are obtained. Use these data to compute the average mass of natural
copper. (The mass values for 63Cu and 65Cu are 62.93 amu and 64.93 amu, respectively.)
63.55 amu/atom
3.2 THE MOLE
•
•
mole—the number of C atoms in exactly 12.0 grams of 12C; also a number, 6.02 × 1023 just
as the word “dozen” means 12 and “couple” means 2.
Avogadro’s number—6.02 × 1023, the number of particles in a mole of anything
DIMENSIONAL ANALYSIS DISCLAIMER: Beginning on page 84 of the Chapter 3 text files
you can find on this CD, you can find all of the remaining exercises worked out with dimensional
analysis. This is most likely the way you were taught in Chemistry I. I will show you some
alternatives to dimensional analysis. WHY? First, some of these techniques are faster and wellStoichiometry
2
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suited to the multi-step problems you will face on the AP Exam. Secondly, these techniques better
prepare you to work the complex equilibrium problems you will face later in this course. The first
problem you must solve in the free response section of the AP Exam will be an equilibrium
problem and you will need to be able to work them quickly. Lastly, I used to teach both methods.
Generations of successful students have encouraged me to share these techniques with as many
students as possible. They did, once they got to college, and made lots of new friends once word
got out they had this “cool way” to solve stoichiometry problems—not to mention their good
grades! Give this a try. It doesn’t matter which method you use, I encourage you to use the method
that works best for you and lets you solve problems accurately and quickly!
ALTERNATE TECHNIQUE #1—USING THE MOLE MAP:
Simply reproduce this map on your scratch paper until you no longer need to since the image will
be burned into your brain!
MULTIPLY [by the conversion factor on the arrow] when traveling IN THE DIRECTION OF THE ARROW and obviously, divide when “traveling” against an arrow. IF you start with moles, start
here, notice you multiply to “get
away” to any other quantity
FROM moles.
Molarity x Volume (L) = moles also!
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When you draw this it will look more like this:
Liters of gas
@ STP
# particles
6.02 × 1023
22.4 L
# of Moles
Molarity (M)
Volume of solution (liters)
Molar Mass (MM)
Mass in
grams
Exercise 3.2
Determining the Mass of a Sample of Atoms
Americium is an element that does not occur naturally. It can be made in very small amounts in a
device known as a particle accelerator. Compute the mass in grams of a sample of americium
containing six atoms.
[There’s a copy of the AP Chem periodic table at apchemistrynmsi.wikispaces.com PRINT—print it!]
2.42 × 10-21 g
Exercise 3.4
Determining Moles of Atoms
Aluminum (A1) is a metal with a high strength-to-mass ratio and a high resistance to corrosion;
thus it is often used for structural purposes. Compute both the number of moles of atoms and the
number of atoms in a 10.0-g sample of aluminum.
2.23 × 1023 atoms
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Exercise 3.5
Calculating the Number of Moles and Mass
Cobalt (Co) is a metal that is added to steel to improve its resistance to corrosion. Calculate both
the number of moles in a sample of cobalt containing 5.00 × 1020 atoms and the mass of the sample.
8.31 × 10-4 mol Co
4.89 × 10-2 g Co
3.3 MOLAR MASS, MOLECULAR WEIGHT, AND FORMULA WEIGHT
•
•
•
•
•
molar mass, MM--the mass in grams of Avogadro'ss number of molecules; i.e. the mass of a
mole!
molecular weight, MW--sum of all the atomic weights of all the atoms in the formula (it is
essential you have a correct formula as you’ll painfully discover!)
empirical formula--that ratio in the network for an ionic substance.
formula weight--same as molecular weight, just a language problem ) “molecular”
implies covalent bonding while "formula" implies ionic bonding {just consider this to be a
giant conspiracy designed to keep the uneducated from ever understanding chemistry—kind
of like the scoring scheme in tennis}. The AP Exam uses MM for all formula masses.
A WORD ABOUT SIG. FIG.’s—It is correct to “pull” from the periodic table as many
sig. figs for your MM’s as are in your problem—just stick with 2 decimal places for all—
much simpler!
Exercise 3.6
Calculating Molar Mass I
Juglone, a dye known for centuries, is produced from the husks of black walnuts. It is also a natural
herbicide (weed killer) that kills off competitive plants around the black walnut tree but does not
affect grass and other noncompetitive plants [a concept called allelopathy]. The formula for juglone
is C10H6O3.
a. Calculate the molar mass of juglone.
b. A sample of 1.56 × 10-2 g of pure juglone was extracted from black walnut husks. How many
moles of juglone does this sample represent?
a. 174.16 g/mol
b. 8.96 × 10-5 mol juglone
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Exercise 3.7
Calculating Molar Mass II
Calcium carbonate (CaCO3), also called calcite, is the principal mineral found in limestone, marble,
chalk, pearls, and the shells of marine animals such as clams.
a. Calculate the molar mass of calcium carbonate.
b. A certain sample of calcium carbonate contains 4.86 moles. What is the mass in grams of this
sample? What is the mass of the CO32- ions present?
a. 100.09 g/mol
b. 486 g; 292g CO32-
Exercise 3.8
Molar Mass and Numbers of Molecules
Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced
commercially. Interestingly, bees release about 1µg (1 × 10-6 g) of this compound when they sting.
The resulting scent attracts other bees to join the attack. How many molecules of isopentyl acetate
are released in a typical bee sting?
How many atoms of carbon are present?
5 × 1015 molecules
4 × 1016 carbon atoms
ELEMENTS THAT EXIST AS MOLECULES
Pure hydrogen, nitrogen, oxygen and the halogens [I call them the “gens” collectively—easier to
remember!] exist as DIATOMIC molecules under normal conditions. MEMORIZE!!! Be sure
you compute their molar masses as diatomics. Others to be aware of, but not memorize:
• P4--tetratomic form of elemental phosphorous; an allotrope
• S8—sulfur’s elemental form; also an allotrope
• Carbon--diamond and graphite ) covalent networks of atoms
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3.4 PERCENT COMPOSITION OF COMPOUNDS
Two common ways of describing the composition of a compound: in terms of the number of its
constituent atoms and in terms of the percentages (by mass) of its elements.
Percent (by mass) Composition: law of constant composition states that any sample of a pure
compound always consists of the same elements combined in the same proportions by mass.
% comp =
mass of desired element
Total mass of compound
× 100
Consider ethanol, C2H5OH
g
= 24.02 g
mol
g
Mass % of H = 6 mol × 1.01
= 6.06 g
mol
g
Mass % of O = 1 mol × 16.00
= 16.00g
mol
Mass of 1 mol of C2H5OH = 46.08 g/mol
Mass % of C = 2 mol × 12.01
NEXT THE MASS PERCENT CAN BE CALCULATED:
Mass percent of C = 24.02 g C × 100% = 52.13%
46.08 g/mol
Repeat for the H and O present.
Exercise 3.9
Calculating Mass Percent I
Carvone is a substance that occurs in two forms having different arrangements of the atoms but the
same molecular formula (C10H14O) and mass. One type of carvone gives caraway seeds their
characteristic smell, and the other type is responsible for the smell of spearmint oil. Compute the
mass percent of each element in carvone.
C = 79.94%
H = 9.41%
O = 10.65%
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Exercise 3.10
Calculating Mass Percent II
Penicillin, the first of a now large number of antibiotics (antibacterial agents), was discovered
accidentally by the Scottish bacteriologist Alexander Fleming in 1928, but he was never able to
isolate it as a pure compound. This and similar antibiotics have saved millions of lives that might
have been lost to infections. Penicillin F has the formula C14H20N2SO4. Compute the mass percent
of each element.
C = 53.82%
H = 6.466%
N = 8.969%
S = 10.27%
O = 20.49%
3.5 DETERMINING THE FORMULA OF A COMPOUND
When faced with a compound of “unknown” formula, one of the most common techniques is to
combust it with oxygen to produce CO2, H2O, and N2 which are then collected and weighed.
•
empirical and molecular formulas: assume a 100 gram sample if given %
o
empirical gives smallest ratio
•
need to know molar mass to establish molecular formula which is (empirical
formula)n, where n is an integer
•
determining empirical and molecular formulas
•
hydrates—“dot waters” used to cement crystal structures.
•
anhydrous--without water
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Example: A compound is composed of carbon, nitrogen and hydrogen. When 0.1156 g of this
compound is reacted with oxygen [burned, combusted], 0.1638 g or carbon dioxide and 0.1676 g of
water are collected. What is the empirical formula of the compound?
Compound + O2 Æ CO2 + H2O + N2
but NOT balanced!!
You can see that all of the carbon ended up in CO2 so…when in doubt, FIND THE NUMBER OF
MOLES!!
0.1638 g ÷ 44.01 g/mol = 0.003781 moles of CO2 = 0.003781 moles of C
Next, you can see that all of the hydrogen ended up in H2O, so….FIND THE NUMBER OF
MOLES!!
0.1676 ÷ 18.02 g/mol = 0 .009301 moles of H2O, BUT there are 2 moles of H for each mole of
water [ think “organ bank” one heart per body, one C per molecule of carbon dioxide—2 lungs per
body, 2 atoms H in water and so on…] so DOUBLE THE NUMBER OF MOLES TO GET THE
NUMBER OF MOLES OF HYDROGEN!! moles H = 0.01860 moles of H
The rest must be nitrogen, BUT we only have mass data for the sample so convert your moles of C
and H to grams:
g C = 0.003781 moles C × 12.01 = 0.04540 grams C
g H = 0.01860 moles H × 1.01
= 0.01879 grams H
0.06419 grams total thus far
SUBTRACT!
0.1156 g sample – 0.06419 g thus far = grams N left = 0.05141 g N so….
0.05141 g N ÷ 14.01 = 0.003670 moles N
Chemical formulas represent mole to mole ratios, so…divide the number of moles of each by the
smallest # of moles of any one of them to get a guaranteed ONE in your ratios…multiply by 2, then
3, etc to get to a ratio of small whole numbers!!
Element
# moles
C
H
N
0.003781
0.01860
0.003670
ALL Divided by
0.003670
1
5
1
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Therefore the correct EMPIRICAL formula is CH5N.
Next, if we are told that the MM is 31.06 g/mol, then simply use this relationship:
(Empirical mass)
×
n
=
MM
(12.01 + 5.05 + 14.01)
×
n
=
31.06
Solve for n
n = 0.999678… or essentially one, so the empirical formula and the molecular formula are one in
the same.
One last trick of the trade: When you don’t know the mass of your sample, assume 100 grams so
that any percents become grams….proceed by finding the number of moles!
Exercise 3.11
Determining Empirical and Molecular Formulas I
Determine the empirical and molecular formulas for a compound that gives the following analysis
(in mass percents):
71.65% C1
24.27% C
4.07% H
The molar mass is known to be 98.96 g/mol.
Empirical formula = CH2Cl
Molecular formula = C2H4C12
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Exercise 3.12
Determining Empirical and Molecular Formulas II
A white powder is analyzed and found to contain 43.64% phosphorus and 56.36% oxygen by mass.
The compound has a molar mass of 283.88 g/mol. What are the compound’s empirical and
molecular formulas?
Empirical formula = P2O5
Molecular formula = (P2O5)2 or P4O10
Exercise 3.13
Determining a Molecular Formula
Caffeine, a stimulant found in coffee, tea, and chocolate, contains 49.48% carbon, 5.15% hydrogen,
28.87% nitrogen, and 16.49% oxygen by mass and has a molar mass of
194.2 g/mol. Determine the molecular formula of caffeine.
Molecular formula = C8H10N4O2
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3.6 & 3.7 BALANCING CHEMICAL EQUATIONS
CHEMICAL REACTIONS
Chemical reactions are the result of a chemical change where atoms are reorganized into one or
more new arrangements. Bonds are broken [requires energy] and new ones are formed [releases
energy].
CHEMICAL EQUATIONS
chemical reaction--transforms elements and compounds into new substances
balanced chemical equation--shows the relative amounts of reactants [left] and products [right] by
molecule or by mole.
•
s, l, g, aq--solid, liquid, gas, aqueous solution
•
NO ENERGY or TIME is alluded to
•
Antoine Lavoisier (1743-1794)--law of conservation
of matter: matter can neither be created nor destroyed
) this means “balancing equations” is all his fault!!
BALANCING CHEMICAL EQUATIONS
•
Begin with the most complicated-looking thing (often the scariest, too).
•
Save the elemental thing for last.
•
If you get stuck, double the most complicated-looking thing.
•
MEMORIZE THE FOLLOWING:
•
metals + halogens Æ MaXb
•
CH and/or O + O2 Æ CO2(g) + H2O(g)
•
H2CO3 [any time formed!] Æ CO2 + H2O; in other words, never write carbonic acid
as a product, it spontaneously decomposes [in an open container] to become carbon
dioxide and water.
•
metal carbonates Æ metal OXIDES + CO2
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Exercise 3.14
Balancing a Chemical Equation I
Chromium compounds exhibit a variety of bright colors. When solid ammonium dichromate,
(NH4)2Cr2O7, a vivid orange compound, is ignited, a spectacular reaction occurs, as shown in the
two photographs on page 105. Although the reaction is actually somewhat more complex, let’s
assume here that the products are solid chromium(III) oxide, nitrogen gas (consisting of N2
molecules), and water vapor. Balance the equation for this reaction.
(NH4)2Cr2O7(s) → Cr2O3(s) + N2(g) + 4H2O(g)
(4 × 2) H
(4 × 2) H
Exercise 3.15
Balancing a Chemical Equation II
At 1000ºC, ammonia gas, NH3(g), reacts with oxygen gas to form gaseous nitric oxide, NO(g), and
water vapor. This reaction is the first step in the commercial production of nitric acid by the
Ostwald process. Balance the equation for this reaction.
4 NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
3.8 STOICHIOMETRIC CALCULATIONS: AMOUNTS OF REACTANTS AND
PRODUCTS
Stoichiometry is the most important thing you can learn as you embark upon AP Chemistry!
Get good at this and you will do well all year. This NEVER goes away!
It’s time to repeat my dimensional analysis disclaimer.
DIMENSIONAL ANALYSIS DISCLAIMER: Beginning on page 84 of the Chapter 3 text files you can find on this
CD, you can find all of the remaining exercises worked out with dimensional analysis. This is most likely the way you
were taught in Chemistry I. I will show you some alternatives to dimensional analysis. WHY? First, some of these
techniques are faster and well-suited to the multi-step problems you will face on the AP Exam. Secondly, these
techniques better prepare you to work the complex equilibrium problems you will face later in this course. The first
problem you must solve in the free response section of the AP Exam will be an equilibrium problem and you will need
to be able to work them quickly. Lastly, I used to teach both methods. Generations of successful students have
encouraged me to share these techniques with as many students as possible. They did, once they got to college, and
made lots of new friends once word got out they had this “cool way” to solve stoichiometry problems—not to mention
their good grades! Give this a try. It doesn’t matter which method you use, I encourage you to use the method that
works best for you and lets you solve problems accurately and quickly!
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First you have to be proficient at the following no matter which method you choose!:
• Writing CORRECT formulas—this requires knowledge of your polyatomic ions and being
able to use the periodic table to deduce what you have not had to memorize. Review section
2.8 in your Chapter 2 notes or your text.
• Calculate CORRECT molar masses from a correctly written formula
• Balance a chemical equation
• Use the mole map to calculate the number of moles or anything else!
Remember the mole map? It will come in mighty handy as well!
# particles
Liters of gas
@ STP
6.02 × 1023
22.4 L
# of Moles
Molar Mass (FW)
Mass in
grams
Here’s the “template” for solving the problems…you’ll create a chart. Here’s a typical example:
What mass of oxygen will react with 96.1 grams of propane?
[notice all words—you supply chemical formulas!]
Molar Mass:
Balanced
Eq’n
mole:mole
# moles
amount
(44.11)
C3H8
(32.00)
+ 5 O2
1
5
(44.01)
(18.02)
Æ 3 CO2
+ 4 H2 O
3
4
1. Write a chemical equation paying special attention to writing correct chemical formulas!
2. Calculate the molar masses and put in parentheses above the formulas—soon you’ll figure
out you don’t have to do this for every reactant and product, just those you’re interested in.
3. Look at the coefficients on the balanced equation, they ARE the mole:mole ratios!
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4. Next, re-read the problem and put in an amount—in this example it’s 96.1 g of propane.
Molar Mass:
Balanced
Eq’n
mole:mole
# moles
amount
(44.11)
C3H8
(32.00)
+ 5 O2
(44.01)
(18.02)
Æ 3 CO2
+ 4 H2 O
1
5
3
4
2.18
96.1 grams
10.9
6.53
8.71
5. Find the number of moles of something, anything! Use the mole map. Start at 96.1 grams,
divide [against the arrow] by molar mass to get the # moles of propane.
6. USE the mole: mole to find moles of EVERYTHING! If 1 = 2.18 then oxygen is 5(2.18)
etc…. [IF the first you find is not a “1”, just divide to make it “1” and then it’s easy greasy!]
Leave everything in your calculator—I only rounded to save space!
7. Re-read the problem to determine which amount was asked for…here’s the payoff….AP
problems ask for several amounts! First, we’ll find the mass of oxygen required since that’s
what the problem asked. 10.9 moles × 32.00 g/mol = 349 g of oxygen
Molar Mass:
Balanced
Eq’n
mole:mole
# moles
amount
(44.11)
C3H8
(32.00)
+ 5 O2
(44.01)
(18.02)
Æ 3 CO2
+ 4 H2 O
1
5
3
4
2.18
96.1 grams
10.9
349 g
6.53
146 L
8.71
Now, humor me…What if part b asked for liters of CO2 at STP [1 atm, 273K]? Use the mole map.
Start in the middle with 6.53 moles × [in direction of arrow] 22.4 L/mol = 146 L
AND…how many water molecules are produced? Use the mole map , start in the middle with 8.71
mol (6.02 × 1023) = 5.24 × 1024 molecules of water.
Try these two exercises with whichever method you like best!
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Exercise 3.16
Chemical Stoichiometry I
Solid lithium hydroxide is used in space vehicles to remove exhaled carbon dioxide from the living
environment by forming solid lithium carbonate and liquid water. What mass of gaseous carbon
dioxide can be absorbed by 1.00 kg of lithium hydroxide?
919 g
Exercise 3.17
Chemical Stoichiometry II
Baking soda (NaHCO3) is often used as an antacid. It neutralizes excess hydrochloric acid secreted
by the stomach:
NaHCO3(s) + HC1(aq) → NaC1(aq) + H2O(l) + CO2(aq)
Milk of magnesia, which is an aqueous suspension of magnesium hydroxide, is also used as an
antacid:
Mg(OH)2(s) + 2HC1(aq) → 2H2O(l) + MgC12(aq)
Which is the more effective antacid per gram, NaHCO3 or Mg(OH)2 ?
Mg(OH)2
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3.9 CALCULATIONS INVOLVING A LIMITING REACTANT
Ever notice how hot dogs are sold in packages of 10 while the buns come in packages of 8?? The
bun is the limiting reactant and limits the hot dog production to 8 as well! The limiting reactant [or
reagent] is the one consumed most entirely in the chemical reaction.
Plan of attack: First, you’ll know you need a plan if you are given TWO amounts of matter that
react.
Next, when in doubt…find the number of moles. Set up your table like before, only now you’ll
have TWO amounts and thus TWO # ‘s of moles to get you started. I cover one up and “What if?”
More to follow! It doesn’t matter where you start the “What if?” game….you get there either way.
Let’s use a famous process [meaning one the AP exam likes to ask questions about!], the Haber
process. This is basically making ammonia for fertilizer production from the nitrogen in the air
reacted with hydrogen gas. The hydrogen gas is obtained from the reaction of methane with water
vapor. This process has saved millions from starvation!! The reaction is below:
Molar Mass:
Balanced Eq’n
(28.04)
mole:mole
1
N2
(2.02)
+
3 H2
(17.04)
Æ
2 NH3
3
2
# moles
amount
Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. What mass of
ammonia can be produced? Which reactant is the limiting reactant? What is the mass of the
reactant that is in excess?
Insert the masses in the amount row and find the number of moles of BOTH!
Molar Mass:
Balanced
Eq’n
mole:mole
# moles
amount
(28.02)
N2
+
(2.02)
(17.04)
Æ 2 NH3
3 H2
1
3
892 moles
25,000 g
2,475 moles
5,000 g
2
WHAT IF I used up all the moles of hydrogen? I’d need 1/3 × 2,475 moles = 825 moles of
nitrogen. Clearly I have EXCESS moles of nitrogen!! Therefore, hydrogen limits me.
OR
WHAT IF I used up all the moles of nitrogen? I’d need 3 × 892 moles = 2,676 moles of hydrogen.
Clearly I don’t have enough hydrogen, so it limits me!! Therefore nitrogen is in excess.
Either way, I’ve established that hydrogen is the limiting reactant so I modify the table:
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That means I’ll use up all the hydrogen but not all the nitrogen!
Molar Mass:
(28.02)
(2.02)
(17.04)
Balanced Eq’n
N2
+
3 H2 Æ 2 NH3
mole:mole
1
3
2
1650 mol
# moles
825 mol used
produced
892 moles
2,475 moles
1650 mol (17.04)
amount
825 mol (28.02) =
= 28,116 g
23,116 g used
produced
5,000 g
25,000 g
1,884 g excess!!
Here’s the question again, let’s clean up any sig.fig issues:
Suppose 25.0 kg of nitrogen reacts with 5.00 kg of hydrogen to form ammonia. (3 sig. fig. limit)
What mass of ammonia can be produced? 23,100 g produced = 23.1 kg (always polite to respond in
the unit given).
Which reactant is the limiting reactant? hydrogen—once that’s established, chunk the nitrogen
amounts and let hydrogen be your guide!
What is the mass of the reactant that is in excess? 1,884 g = 1.88 kg excess nitrogen!!
Exercise 3.18
Stoichiometry: Limiting Reactant
Nitrogen gas can be prepared by passing gaseous ammonia over solid copper(II) oxide at high
temperatures. The other products of the reaction are solid copper and water vapor. If a sample
containing 18.1 g of NH3 is reacted with 90.4 g of CuO, which is the limiting reactant ? How many
grams of N2 will be formed?
CuO is limiting; 10.6 g N2
Stoichiometry
18
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Theoretical Yield: The amount of product formed when a limiting reactant is completely
consumed. This assumes perfect conditions and gives a maximum amount!! Not likely!
Actual yield: That which is realistic!
Percent yield: The ratio of actual to theoretical yield.
Actual Yield
× 100% = Percent yield
Theoretical Yield
Exercise 3.19
Calculating Percent Yield
Methanol (CH3OH), also called methyl alcohol, is the simplest alcohol. It is used as a fuel in race
cars and is a potential replacement for gasoline. Methanol can be manufactured by combination of
gaseous carbon monoxide and hydrogen. Suppose 68.5 kg CO(g) is reacted with 8.60 kg H2(g).
Calculate the theoretical yield of methanol. If 3.57 × 104 g CH3OH is actually produced, what is the
percent yield of methanol ?
Theoretical yield is 6.86 × 104 g
Percent yield is 52.3%
Stoichiometry
19
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Unit 3 – Stoichiometry FRQs
The following applies to the NMSI stoichiometry FRQs (Free Response Questions) that
come with the NMSI Stoichiometry packet. This is intended to clarify some of the
problems and provide you some general guidance in how to do the problems and to
direct your attention to things you want to pay attention to.
Abbreviations:
CB: College board.
EF: Empirical formula
LR: Limiting reagent.
MM: Molecular mass or molar mass.
PSF: Pretty Straight forward (i.e. you should (thought admittedly this isn’t always true)
have the prior knowledge to solve this.)
Strategy for these packets and FRQs: I expect you to work through every packet,
which isn’t to say that you should have to listen to every word uttered by Ms.
McCormick. If you can work through the packet sans listening to Ms. McCormick, do the
problems in the packets (the problems inside the packets, NOT the FRQs, a different
story there), and you understand what’s given to you there, then all you need do is
confirm your answers with Ms. McCormick. This means you can simply fast forward the
NMSI videos until you get to the answers, check them if they’re good keep going, if
they’re not sit through that section of the video that deals with what you got wrong and
work through it until you get it.
This strategy works fine for the FRQ problems (I’m sure you don’t need me to tell you
this, so I’m just sort of letting you know that I know.) There’s no need to sit through
every single minute of the NMSI videos. This isn’t a chore put upon you to make your
life harder, but rather an alternate means by which to learn the material, providing you a
means to check your work as you’re doing it.
FRQs are to be done one problem per side of a piece of paper, so you can have a max
of 2 problems per page. Units are to be cancelled out, answers are to be circled.
As of this writing (July 2013) trying to manipulate the NMSI videos online was difficult,
i.e. it was hard to move the video to where you wanted to by interfacing with the video
through a browser. It’s MUCH easier to do what you want with the videos if you
download them, and I’ve provided you a means by which to do this. The videos for this
course can be found at the following link:
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Questions
#1 (1982)
a)
b)
c)
d)
Focus on the gas.
PSF
PSF
PSF
#2 (1986)
a) Remember the ideal gas equation, PV = nRT, and recall that this can be
manipulated to give you molar mass:
MM = (gRT)/PV = (DRT)/P [g/V = density]
Remember this as you’re sure to see it again.
b) PSF
c) The easiest way to do this problem is to look for the smallest common multiple of
the 3 values that you obtained in b.
d) PSF, and it may pay to put in your head or your calculator what the percentage of
C in CO2 and H in H2O is – you can be sure that you’ll have a reason to use
these numbers again.
#3 (1991)
a) See # 2 part d above, it’s déjà vu!
b) There are 2 ways to solve this problem. One of which, and it may not be so
evident, is to burn the hydrocarbon using the EF value that you determined in a)
above, i.e. put together a combustion equation using the EF. Since you know
from the problem how much of the product is formed you don’t need the
molecular formula to get the right answer, the EF will work just fine. If you use
this method, when you balance your equation one of the coefficients will be 4.5 –
use this, don’t multiply to get a whole #, using the fractional value will work just
fine.
c) Freezing point depression and boiling point elevation, along with colligative
properties in general will be covered in a bit more depth later. For now recall from
honors chem the following equation:
∆Tf = mKfi
This reads: delta, i.e. the change in the freezing point temperature, Tf = molality
times the freezing point depression constant (you have to be given the K value
for whatever solvent you’re given to work with in these sorts of problems) times i,
the Van Hoft factor.
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Remember that molality is
Molality = moles of solute/kilograms of solvent
The Van Hoft factor, i, is the number of particles created when the solute is
added to the solvent. So in the case of NaCl, when you place this in water the
Van Hoft factor or i value is 2 as two separate ions are created when the salt
dissolves. Glucose, C6H12O6 has an i value of 1, it does not break down into
particles when it dissolves in water.
To find the MM in this problem you need to perform a similar trick as you did with
the ideal gas law equation in problem 2, though in this case it should be a
somewhat easier so long as you’re on top of what molality is and how to
manipulate what you have to give you the MM.
d) For an AP problem this was not as “clean” as I’m accustomed to. Normally when
you divide the MM found by the MM of the EF you get a pretty clear cut whole
number or very near to a whole number, e.g.:
MM/EF = 3.08 or MM/EF = 4.98
No question that you would round down the first number to 3, the second you’d
round up to 5.
If you do this part of the problem correctly you’ll find that the multiple you get
from MM/EF is not as neatly “clear cut”. I’m guessing the CB did this deliberately
to cause you to doubt your answer – don’t doubt your answer.
#4 (1993)
a) The way that (a) presents itself may be oft-putting to some of you, especially
since you’ve not done Reduction-Oxidation reactions, i.e. ReDox, before. To do
this particular problem you don’t need to know anything about ReDox reactions,
though you will soon enough know plenty about them.
This is 3-step ReDox reaction. So it looks like three different reactions, but
actually what you have here is one reaction that has been broken down into three
separate steps. The easiest way to do this (the way I see it anyway, Ms.
McCormick does this somewhat differently) is to combine all three reactions into
1 reaction. Once it’s all in one reaction you cancel out whatever is the same on
both sides of the reaction symbol () and combine any OH- with H+ if they’re on
the same side of the . Here’s an example of what of a reaction combined
together:
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This is what you would get in the initial bringing together of the multiple steps,
however many of them there may be (to show what’s going on here I’m
stretching this out a bit, you’re not expected to do it exactly as you see this here)
A + B + C + 5OH- + 4H+  2A +BC + 2H2O
This becomes
B + C + 4H2O + OH+  A + BC + 2H2O
Which becomes
B + C + 2H2O + OH-  A + BC
Dealing with this sort of thing will become more familiar to you as we do ReDox
problems.
So for this part of the problem combine the 3 steps to this ReDox reaction into
one, do what I show you to do above, and then balance the final equation that
remains and you’ll get the answer to this problem.
b) This is a titration problem that doesn’t involve hydrogen or hydroxide ino. If you
recall (though in all fairness you likely don’t) at the equivalence point in a titration
you have equal quantities of titrate and titrant.
c) PSF
d) Note that they want you to come up with the volume of O2 needed to get the
same concentration of O2 as in part b. So you’ll need to figure out the
concentration of O2 in part b to get a handle on this part of the problem.
e) The indicator you would use would have to be sensitive to I2 concentration. This
is one of those things that you either know of don’t; if you took bio you should
know what you use to show iodine in solution of something – starch.
#5 (1998)
a) PSF
b) This brings us back to question #3, part c. After this, PSF
c) PSF – don’t worry that the value for this is radically different than that in part b,
but do take notice of how much of a difference there is as this will be important.
d) The explanation for this can be a bit complicated, addressing dimerization and
other things that are beyond the scope of what you now know or will need to
know. That said, the explanation is sort of self-apparent if you look at the values
obtained for the MM in both cases. Essentially the MM obtained using the ideal
gas law is ½ that obtained using the freezing point depression equation. Ok, that
says something. The argument you can make here, without necessarily using the
term dimerization, which effectively is what’s going on, (Definition: A dimer is a
chemical entity consisting of two structurally similar monomers joined by bonds
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that can be either strong or weak, covalent or intermolecular) is that it appears
that when a molecule of the compound in question hits the water it combines with
another molecule of the compound, i.e. you go from having two individual
molecules to having one molecule made up of two of the original molecules. So
effectively the i (Van Hoft ) value for the freezing point depression equation
becomes ½. Sort of unusual, but it can happen. If you use i = ½ in the freezing
point depression equation you’ll get the same MM as was arrived at using the
ideal gas law equation.
#6 (2000)
a) PSF
b) PSF. There are two ways of doing this, my way is to find out how much water
there is in the hydrate and go from there.
c) i. You’re not familiar with oxidizing and reducing agents from honors chemistry so
this is not something you would normally be able to answer based on prior
knowledge.
An oxidizing agent will take away from something else. For example, Fluorine is
a very strong oxidizing agent as is, surprise of surprises, oxygen. When the
oxidizing agent takes away electrons it itself is reduced, i.e. it gains electrons. So
in a situation where Na is put next to F2, F would take an electron from Na, being
reduced to F-, and Na would be oxidized to Na+.
To make this a bit easier I’ll go through what’s going on in this problem:
16H+ + 2MnO4- + 5C2O42-  2Mn2+ + 10CO2 + 8H2O
This isn’t quite as straightforward as the Na and F situation, but it’s not that hard
either. Here Mn goes from a +7 charge in MnO4- to a +2 charge (actually it’s not
entirely accurate to use the term “charge”, oxidation number would be more
accurate), so clearly it’s been reduced. That said the Mn is in MnO4- so we don’t
say that Mn has been reduced but rather that permanganate (MnO4-) is reduced.
The reducing agent, therefore, must be oxalate (C2O42-) where C goes from a +3
in oxalate to a +4 charge in CO2.
Listening to Ms. McCormick’s explanation for this part of the problem would likely
be helpful.
For the most part the rest of this problem is PSF, you just need to carefully and
methodically work through what you’re being asked for.
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#7 (2003B)
a) i. PSF
ii. LR problem.
iii. PSF, if you did ii above correctly.
b) Again you’re dealing with a LR.
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AP* Stoichiometry Free Response Questions
page 1
1982
Water is added to 4.267 grams of UF6. The only products are 3.730 grams of a solid containing only
uranium, oxygen and fluorine and 0.970 gram of a gas. The gas is 95.0% fluorine, and the remainder is
hydrogen.
(a) From these data, determine the empirical formula of the gas.
(b) What fraction of the fluorine of the original compound is in the solid and what fraction in the gas after
the reaction?
(c) What is the formula of the solid product?
(d) Write a balanced equation for the reaction between UF6 and H2O. Assume that the empirical formula
of the gas is the true formula.
1986
Three volatile compounds X, Y, and Z each contain element Q. The percent by weight of element Q in each
compound was determined. Some of the data obtained are given below.
Compound
Percent by weight
of Element Q
Molecular
Weight
_______________________________________ X
Y
Z
64.8%
73.0%
59.3%
?
104.
64.0
(a) The vapor density of compound X at 27°C and 750. mm Hg was determined to be 3.53 grams per liter.
Calculate the molecular weight of compound X.
(b) Determine the mass of element Q contained in 1.00 mole of each of the three compounds.
(c) Calculate the most probable value of the atomic weight of element Q.
(d) Compound Z contains carbon, hydrogen, and element Q. When 1.00 gram of compound Z is
oxidized and all of the carbon and hydrogen are converted to oxides, 1.37 grams of CO2 and 0.281
gram of water are produced. Determine the most probable molecular formula of compound Z.
1991
The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and
investigating its colligative properties.
(a) The hydrocarbon burns completely, producing 7.2 grams of water and 7.2 liters of CO2 at standard
conditions. What is the empirical formula of the hydrocarbon?
(b) Calculate the mass in grams of O2 required for the complete combustion of the sample of the
hydrocarbon described in (a).
(c) The hydrocarbon dissolves readily in CHCl3. The freezing point of a solution prepared by mixing
100. grams of CHCl3 and 0.600 gram of the hydrocarbon is −64.0°C. The molal freezing-point
depression constant of CHCl3 is 4.68°C/molal and its normal freezing point is −63.5°C. Calculate
the molecular weight of the hydrocarbon.
(d) What is the molecular formula of the hydrocarbon?
(1) AP® is a registered trademark of the College Board. The College Board was not involved in the production of and does not endorse this product. (2) Test
Questions are Copyright © 1984-2008 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes,
classroom teachers are permitted to reproduce the questions. Web or Mass distribution prohibited.
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1993
AP* Stoichiometry Free Response Questions
page 2
I. 2 Mn2+ + 4 OH− + O2(g) → 2 MnO2(s) + 2 H2O
II. MnO2(s) + 2 I− + 4 H+ → Mn2+ + I2(aq) + 2 H2O
III. 2 S2O32− + I2(aq) → S4O62− + 2 I−
The amount of oxygen, O2, dissolved in water can be determined by titration. First, MnSO4 and NaOH
are added to a sample of water to convert all of the dissolved O2 to MnO2, as shown in equation I
above. Then H2SO4 and KI are added and the reaction represented by equation II proceeds. Finally, the
I2 that is formed is titrated with standard sodium thiosulfate, Na2S2O3, according to equation III.
(a) According to the equation above, how many moles of S2O32− are required for analyzing
1.00 mole of O2 dissolved in water?
(b) A student found that a 50.0-milliliter sample of water required 4.86 milliliters of 0.0112-molar
Na2S2O3 to reach the equivalence point. Calculate the number of moles
of O2 dissolved in this sample.
(c) How would the results in (b) be affected if some I2 were lost before the S2O32− was added? Explain.
(d) What volume of dry O2 measured at 25°C and 1.00 atmosphere of pressure would have to be
dissolved in 1.00 liter of pure water in order to prepare a solution of the same concentration as that
obtained in (b)? (cont.)
(e) Name an appropriate indicator for the reaction shown in equation III and describe the change you
would observe at the end point of the titration.
1998
An unknown compound contains only the three elements C,H, and O. A pure sample of the compound
is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.
(a) Determine the empirical formula of the compound.
(b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze
at a temperature 15.2 Celsius degrees below the normal freezing point of pure camphor.
Determine the molar mass and apparent molecular formula of the compound. (The molal
freezing-point depression constant, Kf, for camphor
is 40.0 kg⋅K⋅mol−1.)
(c) When 1.570 grams of the compound is vaporized at 300 °C and 1.00 atmosphere, the gas
occupies a volume of 577 milliliters. What is the molar mass of the compound based on this
result?
(d) Briefly describe what occurs in solution that accounts for the difference between the results
obtained in parts (b) and (c).
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AP* Stoichiometry Free Response Questions
page 3
2000
Answer the following questions about BeC2O4(s) and its hydrate.
(a) Calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC2O4 • 3H2O
(b) When heated to 220.°C, BeC2O4 • 3 H2O(s) dehydrates completely as represented below.
BeC2O4 • 3 H2O(s) → BeC2O4(s) + 3 H2O(g)
If 3.21 g of BeC2O4 • 3 H2O(s) is heated to 220.°C, calculate
(i) the mass of BeC2O4(s) formed, and,
(ii) the volume of the H2O(g) released, measured at 220.°C and 735 mm Hg.
(c) A 0.345 g sample of anhydrous BeC2O4 , which contains an inert impurity, was dissolved in sufficient
water to produce 100. mL of solution. A 20.0 mL portion of the solution was titrated with KMnO4(aq).
The balanced equation for the reaction that occurred is as follows.
−
+
2−
2+
16 H (aq) + 2 MnO4 (aq) + 5 C2O4 (aq) → 2 Mn (aq) + 10 CO2(g) + 8 H2O(l).
The volume of 0.0150 M KMnO4(aq) required to reach the equivalence point was 17.80 mL.
(i) Identify the reducing agent in the titration reaction.
(ii) For the titration at the equivalence point, calculate the number of moles of each of the
following that reacted.
•
−
MnO4 (aq)
2−
•
C2O4 (aq)
(iii) Calculate the total number of moles of C2O42−(aq) that were present in the 100. mL of
prepared solution.
(iv) Calculate the mass percent of BeC2O4(s) in the impure 0.345 g sample.
2003B
Answer the following questions that relate to chemical reactions.
(a) Iron(III) oxide can be reduced with carbon monoxide according to the following equation.
Fe2O3(s) + 3 CO(g) → 2 Fe(s) + 3 CO2(g)
A 16.2 L sample of CO(g) at 1.50 atm and 200.°C is combined with 15.39 g of Fe2O3(s).
(i) How many moles of CO(g) are available for the reaction?
(ii) What is the limiting reactant for the reaction? Justify your answer with calculations.
(iii) How many moles of Fe(s) are formed in the reaction?
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AP* Stoichiometry Free Response Questions
page 4
(b) In a reaction vessel, 0.600 mol of Ba(NO3)2(s) and 0.300 mol of H3PO4(aq) are combined with
deionized water to a final volume of 2.00 L. The reaction represented below occurs.
3 Ba(NO3)2(aq) + 2 H3PO4(aq) → Ba3(PO4)2(s) + 6 HNO3(aq)
(i) Calculate the mass of Ba3(PO4)2(s) formed.
(ii) Calculate the pH of the resulting solution.
–
–1
(iii) What is the concentration, in mol L , of the nitrate ion, NO3 (aq), after the reaction
reaches completion?
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