The Berkeley Review MCAT General Chemistry Part 1

356 Pages·2011·52.8 MB·English

by  The Berkeley Review

General
Chemistry
Parti
Sections I-V
Section I
Stoichiometry
Section II
Atomic Theory
Section III
Equilibrium
Section IV
Acids & Bases
Section V
Buffers & Titrations
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BERKELEY
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Specializing in MCAT Preparation
ERRELEY
REV- I^E • W*
P.O. Box 40140, Berkeley, California 94704-0140
Phone: [phone] (800) N CAT-TB R
Internet:
Unit Conversion
a) Dimensional Analysis
b) Density Determination
c) Typical Conversions
Section I
Elemental Analysis
a) Mass Percent
b) Empirical Formulas
Stoichiometry c) Molecular Formulas
by Todd Bennett d) Combustion Analysis
Solution Concentration
O.lOMMX(aq) a) Units and Terminology
i. Molarity
ii. Molality
iii. Mass Percent (in Solution)
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iv. Density
b) Dilution
c) Beer's Law
Balancing Reactions
a) Standard Balancing
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b) Limiting Reagents
Reaction Types
a) Common Reactions
b) Oxidation States
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Test-Taking Tips
a) General Advice
b) Mathematical Tricks
i. Addition and Subtraction
ii. Averaging
iii. Multiplication
iv. Division
f?EBKELEY
Specializing in MCAT Preparation
Stoichiometry Section Goals
Know how to convert one kind of concentration unit into another.
The concentration of a solution can be measured in terms of molarity, molality, and density. You
mustknow thedefinitions ofeach unitandhow they differ from one another. Although the test
does not feature a great deal of math, you should have an idea of how to convert between units.
Understand the difference between empirical and molecular formulas.
Know the difference between the molecular formula (actual ratio of atoms in a molecule) and the
empiricalformula (simplestwhole number ratio of the atoms in a molecule). Befamiliarwith the
experimentsand informationneeded to determineboth of the formulas.
Know the effect of standard conditions.
Standard conditions are defined as 1 atm. and 298 K for thermodynamics, but STP (standard
temperature andpressure) isdefined as1atm. and273 K. Many calculations ofgasvolume usethe
ideal gas assumption that at STP, one mole of gas occupies22.4liters.
Understand dilution and its effect on concentration of a solute.
Dilution involves a reduction in the concentration of a solute in solution by the addition of solvent
to the mixture. The addition of solvent therefore dilutes the concentration, but does not change the
moles of solute. The equation that you must recallis based on the constant number of solute moles:
Minitial-Vinitial = Mfinal-Vfinal.
Recognize standard reactions from general chemistry.
The most commonly recurring reactions in general chemistry that you are expected to know include
combustion, single replacement, double displacement, and proton transfer, to name just a few. You
must recognize these reactionsand have a basicunderstanding of them.
Recognize the limiting reagent in a reaction and know its effect on the reaction.
The limiting reagent dictates theamount ofproduct thatcanbeformed andconsequently thepercent
yield for a reaction. Using only starting values andthestoichiometric equation, you mustbe able
to determine which reactant is the limiting reagent in the reaction.
Understand the stoichiometric ratios in combustion reactions.
In the combustion of both hydrocarbons and carbohydrates, there is a consistent relation between
the number of oxygen molecules on the reactant side, and the number of water and carbon dioxide
molecules that form on the product side. Know each reaction so that you may easily balance the
coefficients.
General Chemistry Stoichiometry Introduction
Stoichiometry
The perfect spot to start any review of general chemistry is the basics, which
traditionally include stoichiometry and chemical equations. The most
fundamental perspective of a chemical reaction, where bonds are broken so that
new bonds can be formed, is at the atomic and molecular levels. Due to the
minute size of atoms, we can never actually view a chemical reaction (so states
Heisenberg's uncertainty principle). We must therefore rely upon developing
models that can account for changes in all of the atoms and molecules involved
in a chemical reaction or physical process. At the molecular level, we consider
molecules. At the macroscopic level, we consider moles. Stoichiometry allows
us to convert one into the other and to shift between these two perspectives. The
number of molecules is converted into the number of moles using Avogadro's
number (6.022 x 1023). The concept of a mole is based upon the amount of
carbon-12 that is contained in exactly 12.0 grams of carbon, a quantity
determined by knowing the volume of a 12.0-g carbon sample, the type of
molecular packing in it, and the dimensions of the carbon atom. This task of
quantifying atoms in a mole is similar to guessing the number of peas that are
contained in an aquarium. It is important that you utilize the mole concept to
understand, and later to balance and manipulate, chemical equations.
In the stoichiometry section, we focus on those skills needed to solve ratio
questions. Stoichiometry is most commonly thought of as the mathematical
portion of general chemistry. The MCAT, however, has relatively few
calculations. It is a conceptual test, emphasizing logical thought process rather
than calculations. For some of you, this is great news. But before celebrating too
much, consider where the mathematical aspects of general chemistry fit into a
conceptual exam. The MCAT does involve some math, but it is not too
complicated. Math-related calculations required for MCAT questions involve
making approximations, determining ratios, setting up calculations, and
estimating the effect of errors on results. The initial problems presented in this
section involve slightly more calculations than you should expect to see on the
MCAT. Some of them may look familiar to you from your general chemistry
courses and should stimulate your recall. As the section proceeds, less emphasis
is placed on calculating and estimating, and more emphasis is placed on the art
of quickly determining ratios and approximating values.
The focus of the stoichiometry section is problem-solving, with special attention
to the idiosyncrasies of each type of problem. Definitions of important terms are
presented with sample questions and their solutions. Answer solutions discuss
test strategy and the information needed to obtain the correct answer. Each
problem in the stoichiometry section represents what we might call the "book
keeping" of reactions in general chemistry, and it offers an ideal opportunity to
begin work on fast math skills as well. As you do each of the questions, learn the
definitions and develop an approach that works well for you. You may want to
consider multiple pathways to arrive at the correct answer. It is important that
you be able to solve questions in several different ways and to get into the
mindset of the test writers. As you read a passage, think about the questions that
could be asked about it. If a passage gives values for various masses and
volumes, there will probably be a question about density. If it gives values for
moles and solution volume, there will probably be questions on concentration
and dilution. Use your intuition and common sense as much as you can, and
make every effort to develop your test-taking logic.
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General Chemistiy Stoichiometry Unit Conversion
Unit Conversion
Dimensional Analysis
Dimensional analysis is a mathematical conversion from one set of units into
another. It involves multiplyinga given value by a conversion factor or a series
of conversion factors until the value is finally expressed in the desired units.
Converting from one set ofunits to another is a critical skill needed to eliminate
incorrect answer choices in the physical sciences section. Be systematic when
converting between units. The standard measurements that can be expressed in
a variety ofunits are distance (1 m = 1.094 yd, 2.54 cm = 1.00 in,and 1.609 km=
1.00 mile), mass (1.00 kg = 2.205 lb and 453.6 g = 1.00 lb), volume (3.79 L = 1.00
gal and 1.00 L = 1.06 qt), and time (3600 s = 1.00 hr). Always convert units as
they appear in a problem into the units indicated in the answer choices (the so-
called "target units").
Example 1.1
Sprinters can run 100 meters injustunder 10 seconds. Atwhat average speed in
miles per hourmusta runner travel to cover 100 meters in 10.0 seconds?
A. 3.7 miles/hour
B. 11.2 miles/hour
C. 22.4 miles/hour
D. 36.0 miles/hour
Solution
The first task is to determine the given units and the target units. From there,
convert the given units into the target units. We are given 100 meters in 10
seconds, but the answer choices are expressed in miles per hour. Use the correct
conversion factors, as follows:
10.0 s hour
Conversion ofdistance: 10°mxmiles - miles
10.0 s rn s
Conversion of time: 10° m x-§- =-^3_
10.0 s hr hour
mnm.. 1km x 1 mile y 3600s _ 100x3600 miles
10.0 s 1000 m 1.609 km 1.00 hr 10 x 1000 x 1.609 hour
100x3600 _ 3600 _ 36 miles
10x1000x1.609 10x10x1.609 1.609 hour
36< 36 ,<36 where 36 _ 18and 36 = 36. So18<^6_ < 36
2 1.609 1 2 1 1.609
Only choice C falls within the range of 18 to 36.
A frequent task in chemistry is the conversion between various types of
temperature units, volume units, pressure units, and concentration units.
Chemists, like most scientists, employ the MKS system, so the conversion from
conventional units less commonly used in science to MKS units is routine.
Example 1.2 demonstrates the interconversion between the conventional
Fahrenheit unit and the scientific Celsius unit of temperature.
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General ChemiStiy Stoichiometry Unit Conversion
Example 1.2
At what temperature is the numerical value the same, whether the units are in
Celsius or Fahrenheit?
A. 32°
B. 0°
C. -40°
D. -273°
Solution
The formulas for conversion between Celsius and Fahrenheit are as follows:
T.F =£t.c +32 .-. T-c = §• (T-f- 32)
5 9
Answering the question requires setting T>f= T>(\
T-F = £ T-c +32 becomes: T=£T+32, soT=1.8 T+32
5 5
T = 1.8T + 32=> -0.8T = 32.\ T = -40°
Density Determination
The density of a material or solution is the mass of the sample divided by the
volume of the sample. Density is a measured quantity, determined
experimentally. Understand the techniques used to measure density. The term
specific gravityrefers to the density of a material relative to the density of water,
and may be used in a question in lieu of density. For our purposes, specific
gravity means the same thing as density, but it has no expressed units.
Determining density is a typical example of dimensional analysis.
Example 1.3
Exactly 10.07 mL of an unknown non-volatile liquid is poured into an empty
25.41-gram open flask. The combined mass of the unknown non-volatile liquid
and the flask is 34.12 grams. What is the density of the unknown liquid?
A 34.13 g n 8.71 g
B.
10.07 mL 10.07 mL
c 10-07 g D 8.71 mL
8.71 mL * 10.07 g
Solution
The density of the liquid is found by dividing the mass of the liquid by the
volume of the liquid. This results in units of grams per milliliter, which
eliminates choice D. The volume of the liquid is 10.07 mL, so 10.07 should be in
the denominator. This eliminates choice C. The mass of the liquid is the
difference between the final mass of the flask and liquid combined, and the mass
of the flask (34.12 - 25.41), which is equal to 8.71 grams. This means that the
numerator should be 8.71. The correct answer is choice B. incidentally, the
question did not state the reason for using a non-volatile liquid. The liquid must
be non-volatile, to prevent any loss due to evaporation from the open flask. In
the event the liquid evaporates away, then the mass you determine is too small,
due to the loss of vapor molecules.
Copyright © by The Berkeley Review 5 Exclusive MCAT Preparation
General Chemistry Stoichiometry Unit Conversion
Density questions may on occasion involve a more intricate conversion ofunits.
Forany density question, keep in mind that the targetunits are mass solution
divided by volume solution. The mass percent ofa solute is mass solute divided
by mass solution. The product of the density and mass percent is mass solute
dividedby volume solution. Converting the massof solute into moles of solute
yields molarity. Example 1.4shows this.
Example 1.4
Whatis themolarity ofa 3% NaClsolution with a densityof1.05 grams/mL?
A. 0.497 M NaCl
B. 0.504 M NaCl
C 0.539 M NaCl
D. 0.724 M NaCl
Solution
Thefirststep is to determine the units you are looking for, whichin this example
is moles solute per liter solution. You must find both moles solute and liters
solution. The density of the solution is 1.05 grams/mL which means that one
liter of the solutionweighs1050 g. Threepercent (3%) of the solution is sodium
chloride, so the mass of sodium chloride is 0.03 x 1050g. This is the same as 3%
of 1000 g + 3% of50g, which is 30g + 1.5 g = 31.5 g of NaCl per liter solution.
The grams of sodium chloride are converted to moles by dividing by the
molecular weightofNaCl(58.6 grams/mole). Theunit factor method is shown
below:
1.05 gsolution x1000 mLx 3gNaCl x1mole NaCl
mL solution L 100gsolution 58.6g NaCl
_ 1.05 x 1000 x 3 moles NaCl - 3.15 x lOmoles NaCl
100 x 58.6 L solution 58.6 L solution
On the MCAT, you will not have time to solve for values precisely, so you must
makean approximation. Select the answer that is closest to that approximation.
3L5. > 30 _ 1_ so mevalue is greaterthan 0.500 M
58.6 60 2
315.<33 _11 __55./ so mevalue is lessthan 0.550 M
58.6 60 20 100
The value falls between 0.500 M and 0.550 M, so choices A and D are eliminated.
Next you must choose between 0.504 M and 0.539 M. The value is not close
enough to 0.504 M, so you should choose C, and be a wise student! Wise
students are a good thing. Some questions on the MCAT may present
mathematical set-ups,without solvingfor an exactnumber.
The physical sciences section of the exam incorporates physics and general
chemistry, so from the beginning of your review, make a conscious effort to
consider physics when working on general chemistry and to consider general
chemistry whenworking on physics. Determining density is a problem common
to both disciplines. Example 1.5 shows an approach to the concept of density
that is more typical of what is found in a physicsproblem.
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General Chemistry Stoichiometry Unit Conversion
Example 1.5
What can be concluded about the density of a metal object which, when placed in
a beaker of water at room temperature, sinks to the bottom?
A. The density of the metal is less than the density of either water or ice.
B. The density of the metal is less than the density of water, but greater than the
density of ice.
C. The density of the metal is greater than the density of water, but less than the
density of ice.
D. The density of the metal is greater than the density of either water or ice.
Solution
When an object floats in a liquid medium, its density is less than that of the
medium surrounding it. The fact that it floats means the buoyant force pushing
upward against it (pmedium'Vobjecfg) Is greater than gravitation force pushing
downward (weight = mg = PobjecfVobject'g)- Thus, an object floats when
Pmedium > Pobject- Because the metal object sinks in water, it must be denser
than water. Ice floats in water, meaning that ice is less dense than water and thus
less dense than the metal object. The density of the metal must be greater than
the density of either water or ice. The correct answer is therefore choice D.
Typical Conversions
In chemistry, conversions between products and reactants are common, so the
mole concept is frequently employed. The mole concept is pertinent in the
interconversion between moles and mass, using either atomic mass (for
elements) or molecular mass (for compounds). These calculations involve using
the unit factor method (also known as dimensional analysis.)
Example 1.6
How many moles of NaHCC>3 are contained in 33.6grams NaHCC>3?
A. 0.20 moles NaHC03
B. 0.40 moles NaHC03
C. 0.50 moles NaHC03
D. 0.60 moles NaHC03
Solution
The first step in determining the number of moles is to determine the molecular
mass of NaHC03. The mass is 23 + 1 + 12 + 48 = 84 grams. The number of moles
of NaHC03 is found by dividing 33.6 by 84, which is less than 0.50. This
eliminates choices C and D. The number is greater than 0.25 (21 over 84) and
thus greater than 0.20, so choice A is eliminated. The only value left is choice B,
0.40 moles.
Beyond deteirnining the moles from grams for the same compound are questions
where the moles of products are determined from the grams of reactants. These
questions require converting from grams of a given substance to moles of the
given substance, and then expressing the quantity of a final substance in terms of
moles, grams, or liters. By balancing the reaction, the mass of a selected product
that is formed in the reaction can be calculated based on the mass of a selected
reactant (which must be the limiting reagent). Examples 1.7 and 1.8 involve
determining moles, mass, and volume from the given values.
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General Chemistry Stoichiometry Unit Conversion
Example 1.7
Based on the following reaction, how many grams of water would form from
0.33moles C4H10O reacting with an unlimited amount of oxygen gas?
C4HioO(g) + 6 02(g) * " 4C02(g) + 5H20(g)
A. 18.00 grams
B. 24.00 grams
C. 30.00 grams
D. 36.00 grams
Solution
With an excess of oxygen, the limiting reagent in this reaction is C4H10O. The
amount of water formed is determined by the 0.33 moles of C4H10O reactant.
Using the balanced equation, the ratio of H20 to moles C4H10O is 5 :1, so 1.667
moles of water are formed. At 18 grams per mole, this means that fewer than 36
grams but more than 27 grams are formed. This makes choice C the best answer.
0.33molesC4H10Ox 5molesH2Q x 18gH2Q =ix5x 18gH20 =30gH20
1 mole C4H10O 1 mole H20 3
Example 1.8
How many liters of C02(g) result from the complete decomposition of 10.0grams
of CaC03(s) to carbon dioxide and calcium oxide at STP?
CaC03(s) *~ CaO(s) + C02(g)
A. 1.12 liters
B. 2.24 liters
C. 3.36 liters
D. 4.48 liters
Solution
You are asked to determine the amount of product from a known quantity of
reactant. The first step in problems of this type is to make sure the reaction is
balanced. In this case, it is already balanced. The mole ratio of the two
compounds is 1 : 1. The required conversion involves changing from mass
reactant, to moles reactant, to moles product, and finally to volume product.
This is one variation of unit conversion via mole ratio calculation. In addition,
there is the "g - m - m - g" conversion and the "v - m - m - g" conversion. You
need three steps to go from grams reactant to the target (liters product). Units
are important here. The units for the mass of reactant is grams. You need to
multiply mass by molesand divide by grams. This is the same as dividing by the
MW. The second step is to read the mole ratio from the balanced equation. In
this reaction, the mole ratio is 1:1 (the units of both numerator and denominator
are moles). The third and final step is to convert from moles product into liters
product (i.e., multiply by liters and divide by moles.) This is done by
multiplying by the molar volume of the product gas, which at STP (standard
temperature and pressure) is 22.4 liters.
10gramsCaCO3X l™leCaC03 x 1mole CO; x22.4 liters CQ2
100gramsCaC03 lmoleCaC03 lmoleCOfc
= 10x22.4 = 2.24 liters C02
100
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