MCAT General Chemistry Review
Part I Review
Chapter 4: Compounds and Stoichiometry
Oh—what is that smell? You and your mother are taking an afternoon walk through the public rose garden, a pleasant activity for a pleasant summer day, and everything was so pleasant until you smelled . . . it, whatever it is. Your mother doesn’t have to wonder long until her own olfactory system is screeching the alarm, too. Oh, it’s just horrible. It smells like rancid almonds. Although your better instincts tell you to walk quickly away, you pause and, holding your breath, crouch down to get a better look. You are a premed, after all, and you did just synthesize cinnamaldehyde in your orgo lab, so your curiosity is understandable.
Scanning the leaves of some plants at the base of the rosebush, you notice a few green bugs whose backs give the impression of a shield. A small stick lies temptingly close to your foot. You can’t resist and so you grab the stick and position it for optimal poking, but just as the pointy end gets within a few inches of the insect, which now senses that its own pleasant afternoon is about to be ruined, it raises its hind quarters in the direction of the stick—that is to say, in your direction—exposing its sophisticated defense weaponry. Stink bugs! you exclaim, as you drop the stick and take what you believe are a sufficient number of steps to remove yourself from the bug’s bombing range.
Our world is filled with aromas and odors, some pleasant and some not so much. What is it that we are smelling when we “smell something”? Well, every odor that we perceive is the result of an interaction between the chemical receptors of our olfactory systems and some chemical compound. A stink bug “stinks” because it produces a highly concentrated solution of volatile compounds that we (and anything else that would dare to disturb it) perceive as malodorous, noxious, and irritating. It’s a pretty ingenious defense system: Make yourself so disgusting, so distasteful, that everyone will leave you alone, or at least think twice before bothering or eating you. Interestingly enough, the primary compounds in the stink bug’s stink bomb are hydrogen cyanide (a highly toxic compound that inhibits cytochrome c oxidase, thereby blocking aerobic respiration) and benzaldehyde.
Wait, benzaldehyde? Yes, benzaldehyde—that relatively simple compound consisting of a benzene ring substituted with an aldehyde functional group. You need to recognize it for the Biological Sciences section of the MCAT. Remember that we classify benzene ring compounds as part of the larger class of molecules called aromatic compounds. Therefore, you can smell benzaldehyde because it is aromatic and, like some other aromatic compounds, it vaporizes at room temperature and reaches your olfactory system as gas particles. Now, you might be surprised to learn that benzaldehyde is the key ingredient in artificial almond extract. At low concentrations, it gives a pleasant aroma of toasted almonds. However, at high concentrations, its odor is that of rotten, stinking almonds, and it is a noxious irritant to skin, eyes, and the respiratory tract. We do not recommend stink bug juice as a suitable cooking substitute for almond extract.
Benzaldehyde is a compound composed of seven carbon atoms, six hydrogen atoms, and one oxygen atom. One mole of benzaldehyde has a mass of 106.12 grams. It can react with other atoms or compounds to form new compounds (pure substances composed of two or more elements in a fixed proportion). They can be broken down by chemical means to produce their constituent elements or other compounds. They are characterized according to the same systems of traits: physical properties and chemical reactivities.
This chapter focuses on compounds and their reactions. It reviews the various ways in which compounds are represented: Empirical and molecular formulas and percent composition will be defined and explained. There is a brief overview of the major classes of chemical reactions, which we will examine more closely in subsequent chapters, and finally, there is a recap of the steps involved in balancing chemical equations with a particular focus on identifying limiting reagents and calculating reaction yields.
Molecules and Moles
A molecule is a combination of two or more atoms held together by covalent bonds. It is the smallest unit of a compound that displays the identifying properties of that compound. Molecules can be composed of two atoms of the same element (e.g., N2 and O2) or may be composed of two or more atoms of different elements, as in CO2, SOCl2, and C6H5CHO (benzaldehyde). Because reactions usually involve a very large number of molecules, far too many to count individually, we usually measure amounts of compounds in terms of moles or grams, using molecular weight to interconvert between these units.
Ionic compounds do not form true molecules because of the way in which the oppositely charged ions arrange themselves in the solid state. As solids, they can be considered as nearly infinite, three-dimensional arrays of the charged particles that comprise the compound. Remember, in Chapter 3 we mentioned that NaCl in the solid state is a 6:6 coordinated lattice in which each of the Na+ ions is surrounded by six Cl- ions and each of the Cl- ions is surrounded by six Na+ ions. As you might imagine, this makes it rather difficult to clearly define a sodium chloride molecular unit. Because no molecule actually exists, molecular weight becomes meaningless, and the term formula weight is used instead. (However, this is a technical distinction over which you ought not to sacrifice too much sleep.)
Ionic compounds form from combinations of elements with large electronegativity differences (that sit far apart on the periodic table), such as sodium and chlorine. Molecular compounds form from the combination of elements of similar electronegativity (that sit close to each other on the periodic table), such as carbon with oxygen.
We’ve mentioned already that the term atomic weight is a misnomer, because it is a measurement of mass, not weight (another distinction not worth any sacrifice of sleep), and the same applies here to our discussion of molecular weight: It’s really a measurement of mass. Molecular weight, then, is simply the sum of the atomic weights of all the atoms in a molecule, and its units are atomic mass units (amu). Similarly, the formula weight of an ionic compound is found by adding up the atomic weights of the constituent ions according to its empirical formula (see the following example), and its units are grams.
Example: What is the molecular weight of SOCl2?
Solution: To find the molecular weight of SOCl2, add together the atomic weights of each of the atoms.
We defined the term mole in Chapter 1, Atomic Structure, but let’s briefly review it. A mole is a quantity of any thing (molecules, atoms, dollar bills, chairs, etc.) equal to the number of particles that are found in 12 grams of carbon-12. That seems like an awfully strange point of comparison, so all you really need to remember is that “mole” is a quick way of indicating that we have an amount of particles equal to Avogadro’s number, 6.022 × 1023. One mole of a compound has a mass in grams equal to the molecular weight of the compound expressed in amu and contains 6.022 × 1023molecules of that compound. For example, 62 grams of H2CO3 (carbonic acid) represents one mole of the acid compound and contains 6.022 × 1023 molecules of H2CO3. The mass of one mole of a compound, called its molar weight or molar mass, is usually expressed as g/mol. Therefore, the molar mass of H2CO3 is 62 g/mol. You may be accustomed to using the term molecular weight to imply molar mass. Technically this is not correct, but nobody is going to rap your knuckles with a ruler for this minor infraction.
The formula for determining the number of moles of a substance present is
mol = Weight of sample (g)/ Molar weight (g/mol)
Here we mention Avogadro’s number again and can see that the mole is just a unit of convenience, like the dozen is a convenient unit for eggs.
Example: How many moles are in 9.52 g of MgCl2?
Solution: First, find the molar mass of MgCl2.
1(24.31 g/mol) + 2(35.45 g/mol) = 95.21 g/mol
Now, solve for the number of moles.
Equivalent weight and the related concept of equivalents are a source of confusion for many students. Part of the problem may be the context in which equivalents and equivalent weights are discussed: acid-base reactions, redox reactions, and precipitation reactions, all three of which themselves can be sources of much student confusion and anxiety. So let’s start with some very basic discussion, and then, in later chapters, we will see how these concepts and calculations apply to these three types of reactions.
The idea of equivalents is similar to the concept of normality, which we will see when we study Acids and Bases in Chapter 10.
Generally, some compounds or elements have different capacities to act in certain ways, not in terms of their characteristics or behaviors, like electronegativity or ionization energy, but rather the ability of certain elements or compounds to act more potently than others in performing certain reactions. For example, one mole of HCl has the ability to donate one mole of hydrogen ions (H+) in solution, but one mole of H2SO4 has the ability to donate two moles of hydrogen ions, and one mole of H3PO4 has the ability to donate three moles of hydrogen ions. Another example to consider is the difference between Na and Mg: One mole of sodium has the ability to donate one mole of electrons, while one mole of magnesium has the ability to donate two moles of electrons. To find one mole of hydrogen ions for a particular acid-base reaction, we could “source” those protons from one mole of HCl, or we could instead use a half-mole of H2SO4. If we’re using H3PO4, we’d only need one-third of a mole. This is what we mean by “different capacities to act in certain ways.” If we need only one mole of hydrogen ions, it would be a waste to use an entire mole of H3PO4, which could donate three times what we need. We are defining the concept of equivalents: How many moles of the “thing we are interested in” (i.e., protons, hydroxide ions, electrons, or ions) will the number of moles of the compound present produce? One mole of hydrogen ions (one equivalent) will be donated by one mole of HCl, but two moles of hydrogen ions (two equivalents) will be donated by one mole of H2SO4, and three moles of hydrogen ions (three equivalents) will be donated by one mole of H3PO4. Simply put, an equivalent is a mole of charge.
Having discussed the concept of equivalents, we can now introduce a calculation that will be helpful on the MCAT, especially for problems of acid-base chemistry. So far, this discussion has been focused on the mole-to-mole relationship between, say, an acid compound and the hydrogen ions it donates. However, sometimes we need to work in units of mass rather than moles. Just as one mole of HCl will donate one mole (one equivalent) of hydrogen ions, a certain mass amount of HCl will donate one equivalent of hydrogen ions. This amount of compound, measured in grams, that produces one equivalent of the monovalent particle of interest (protons, hydroxide ions, electrons, or ions) is called the gram equivalent weight, and the equation is
Gram equivalent weight = Molar mass/n
where n is the number of protons, hydroxide ions, electrons, or monovalent ions “produced” or “consumed” per molecule of the compound in the reaction. For example, you would need 46 grams of H2SO4 (molar mass = 98 g/mol) to produce one equivalent of hydrogen ions, because each molecule of H2SO4 can donate two hydrogen ions (n = 2). Simply put, an equivalent weight of a compound is the mass that provides one mole of charge.
If the amount of a compound in a reaction is known and you need to determine how many equivalents are present, use the following equation:
Equivalents = Mass of compound (g)/Gram equivalent weight (g)
Finally, we can now introduce the measurement of normality without too much fear of causing the almost-always-fatal head explosion. Normality, as the term suggests, is a measure of concentration. The units for normality are equivalents/liter. A 1 N solution of acid contains a concentration of hydrogen ions equal to 1 mole/liter; a 2 N solution of acid contains a concentration of hydrogen ions equal to 2 moles/liter. The actual concentration of the acidic compound may be the same or different from the normality, because different compounds have different capacities to donate hydrogen ions. In a 1 N acid solution consisting of dissolved HCl, the molarity of HCl is 1 M, because HCl is a monoprotic acid, but if the dissolved acid is H2SO4, then the molarity of H2SO4 in a 1 N acid solution is 0.5 M, because H2SO4 is a diprotic acid. The conversion from normality of acid solution to molarity of acidic compound is
Molarity = Normality/n
where n is the number of protons, hydroxide ions, electrons, or monovalent ions “produced” or “consumed” per molecule of the compound in the reaction.
There is a real benefit to working with equivalents and normality because it allows a direct comparison of quantities of the “thing” you are most interested in. Face it, in an acid-base reaction, you really only care about the hydrogen ion and/or the hydroxide ion. Where they come from is not really your primary concern. So it is very convenient to be able to say that one equivalent of acid (hydrogen ion) will neutralize one equivalent of base (hydroxide ion). The same could not necessarily be said to be true if we were dealing with moles of acidic compound and moles of basic compound. For example, one mole of HCl will not completely neutralize one mole of Ca(OH)2, because one mole of HCl will donate one equivalent of acid but Ca(OH)2 will donate two equivalents of base.
Conceptually, that was the most challenging discussion in this chapter. Let’s move on to review ways in which compounds are represented.
Representation of Compounds
There are different ways of representing compounds and their constituent atoms. We’ve already reviewed a couple of these systems (Lewis dot structures and VSEPR theory) in Chapter 3. In organic chemistry, it is common to encounter “skeletal” representations of compounds, called structural formulas, to show the various bonds between the constituent atoms of a compound. Inorganic chemistry typically represents compounds by showing the constituent atoms without representing the actual bond connectivity or atomic arrangement. For example, C6H12O6 (which you ought to recognize is the molecular formula for glucose) tells us that this particular compound consists of 6 atoms of carbon, 12 atoms of hydrogen, and 6 atoms of oxygen, but there is no indication in the molecular formula of how the different atoms are arranged or how many bonds exist between each of the atoms.
LAW OF CONSTANT COMPOSITION
The law of constant composition states that any pure sample of a given compound will contain the same elements in the identical mass ratio. For example, every sample of H2O will contain two hydrogen atoms for every one oxygen atom, or in terms of mass, for every one gram of hydrogen, there will be eight grams of oxygen. We’re sorry to say that it doesn’t matter whether the water you drink is sourced from deep springs in the Fiji Islands or flows out of a tap connected to a shallow well. The law of constant composition says that water is water.
EMPIRICAL AND MOLECULAR FORMULAS
There are two ways to express a formula—the elemental composition—for a compound. The empirical formula gives the simplest whole number ratio of the elements in the compound. The molecular formula gives the exact number of atoms of each element in the compound and is usually a multiple of the empirical formula. For example, the empirical formula for benzene is CH, while the molecular formula is C6H6. For some compounds, the empirical and molecular formulas are the same, as in the case of H2O. For the reasons previously discussed, ionic compounds, such as NaCl or CaCO3, will only have empirical formulas.
The percent composition by mass of an element is the weight percent of a given element in a specific compound. To determine the percent composition of an element X in a compound, the following formula is used:
% Composition = (Mass of X in formula/Formula weight of compound) × 100%
You can calculate the percent composition of an element by using either the empirical or the molecular formula; just be sure to use to the appropriate mass measurement:
Percent composition is a common way for stoichiometry to be tested on the MCAT. Practice these problems to build up speed and efficiency for Test Day.
formula weight for empirical formula or molar mass for molecular formula. Formula weight is simply the mass of the atoms in the empirical formula of a compound.
You can determine the molecular formula if both the percent compositions and molar mass of the compound are known. The following examples demonstrate such calculations.
When there are two methods for approaching a problem, be well versed in both. Knowing multiple ways to solve a problem will help you tackle questions efficiently. help you tackle questions efficiently.
Example: What is the percent composition of chromium in K2Cr2O7?
Solution: The formula weight of K2Cr2O7 is
Example: What are the empirical and molecular formulas of a compound that contains 40.9 percent carbon, 4.58 percent hydrogen, and 54.52 percent oxygen and has a molar mass of 264 g/mol?
Method One: First, determine the number of moles of each element in the compound by assuming a 100-gram sample; this converts the percentage of each element present directly into grams of that element. Then convert grams to moles:
Next, find the simplest whole number ratio of the elements by dividing the number of moles by the smallest number obtained in the previous step.
Finally, the empirical formula is obtained by converting the numbers obtained into whole numbers (multiplying them by an integer value).
C1H1.33O1 × 3 = C3H4O3
C3H4O3 is the empirical formula. To determine the molecular formula, divide the molar mass by the formula weight. The resultant value is the number of empirical formula units in the molecular formula.
The empirical formula weight of C3H4O3 is
Method Two: When the molar mass is given, it is generally easier to find the molecular formula first. This is accomplished by multiplying the molar mass by the given percentages to find the grams of each element present in one mole of compound, then dividing by the respective atomic weights to find the mole ratio of the elements:
Thus, the molecular formula, C9H12O9, is the direct result.
The empirical formula can now be found by reducing the subscript ratio to the simplest integral values.
The molecular formula is either the same as the empirical formula or a multiple of it. To calculate the molecular formula, you need to know the mole ratio (this will give you the empirical formula) and the molecular weight (molecular weight ÷ empirical formula weight will give you the multiplier for the empirical formula to molecular formula conversion).
Types of Chemical Reactions
This section reviews the major classes of chemical reactions. As it has probably already become apparent to you in your inorganic and organic chemistry classes, it would be quite impossible to memorize every single individual reaction that could occur. Fortunately, there is no need to memorize any reaction, as long as you take the time now and throughout your preparation for the MCAT to learn and understand the recognizable patterns of reactivities between compounds. Some classes of compounds react in very “stereotyped” ways, or at least appear to react in stereotyped ways, because of the MCAT’s focus on a particular subset of a chemical’s reactivities. The MCAT has a tendency to typecast certain compounds, in spite of repeated protests made by the compounds’ talent agents.
A combination reaction has two or more reactants forming one product. The formation of sulfur dioxide by burning sulfur in air is an example of a combination reaction.
S (s) + O2 (g) SO2 (g)
Combination reactions generally have more reactants than products.
A + B C
A decomposition reaction is the opposite of a combination reaction: A single compound reactant breaks down into two or more products, usually as a result of heating or electrolysis. An example of decomposition is the breakdown of mercury (II) oxide. (The [delta] sign over a reaction arrow represents the addition of heat.)
Sometimes one atom in a molecule gets tired of the other (“I love you, but I’m not in love with you…”), and divorce becomes inevitable when something better comes along. This is a single-displacement reaction: an atom (or ion) of one compound is replaced by an atom of another element. For example, zinc metal will displace copper ions in a copper sulfate solution to form zinc sulfate.
Zn (s) + CuSO4 (aq) Cu (s) + ZnSO4
Single-displacement reactions are often further classified as redox reactions, which will be discussed in great detail in Chapter 11, Redox Reactions and Electrochemistry. Not to carry the analogy to the point of ridiculousness, but just to make the point: Cu in CuSO4 has an oxidation state of +2, but when it leaves the compound, it gains two electrons (the Cu+2 is reduced to Cu)—you may think of this as Cu+2 getting an alimony settlement or gaining back its dignity for leaving that no-good cheater, SO42-. On the other side, Zn loses its dignity (in the form of two electrons) when it desperately throws itself into the arms of SO42-.
Well, if you think atoms are acting scandalously in single-displacement reactions, just wait until you get a load of the atomic depravity in double-displacement reactions. In double-displacement reactions, also called metathesis, elements from two different compounds swap places with each other (hence, the name double-displacement) to form two new compounds. This type of reaction occurs when one of the products is removed from the solution as a precipitate or gas or when two of the original species combine to form a weak electrolyte that remains undissociated in solution. For example, when solutions of calcium chloride and silver nitrate are combined, insoluble silver chloride forms in a solution of calcium nitrate.
CaCl2 (aq) + 2AgNO3 (aq) Ca(NO3)2 (aq) + 2AgCl (s)
Neutralization reactions are a specific type of double displacement in which an acid reacts with a base to produce a salt. For example, hydrochloric acid and sodium, hydroxide will react to form sodium chloride and water.
HCl (aq) + NaOH NaCl (aq) + H2O
Decomposition reactions generally have more products than reactants.
C A + B
Acids and bases (which we will see in Chapter 10) combine in neutralization reactions to produce salts and water.
Net Ionic Equations
Just as in our discussion of equivalents and normality, in which we admitted that we really care about the acid and the base functionalities themselves rather than the compounds that are donating the hydrogen protons and hydroxide ions, here we have another opportunity to confess our little secrets. Come on, say it with us—you’ll feel better getting this off your chest: When it comes to many reactions, there are certain species that we find boring and want to ignore. For example, in many reactions, such as displacements, the ionic constituents of the compounds are in solution, so we can write the chemical reaction in ionic form. In the previous example involving the reaction between zinc and copper (II) sulfate, the ionic equation would be
Zn (s) + Cu2+ (aq) + SO42- (aq) Cu (s) + Zn2+ (aq) + SO42- (aq)
You’ll notice that the SO42- (aq) is just hanging, not really doing anything. It’s not taking part in the overall reaction but simply remains in the solution unchanged. We call such species spectator ions. They’re like boring people who go to parties and just stand around, taking up space. Because the SO42- ion isn’t doing anything of interest, we can ignore it and write a net ionic reaction showing only the species that actually participate in the reaction:
Zn (s) + Cu2+ (aq) Cu (s) + Zn2+ (aq)
Net ionic equations list only the cool people at the party who are actually doing fun stuff. They are important for demonstrating the actual reaction that occurs during a displacement reaction.
It is unlikely that you will come across a question that explicitly asks you to balance an equation. However, you will need to recognize unbalanced reactions and quickly add the necessary coefficients. Look at the
1) charge on each side; and
2) number of atoms of each element.
Balancing a checkbook is a dying art. In this age of automated and computer banking, it’s easy enough to monitor the account balance without taking the time to record by hand each and every transaction. Nevertheless, the art of balancing chemical equations is one in which you must be skilled, and you ought to expect the MCAT to test your understanding of the steps involved. Because chemical equations express how much and what type of reactants must be used to obtain a given quantity of product, it is of utmost importance that the reaction be balanced so as to reflect the laws of conservation of mass and charge. The mass of the reactants consumed must equal the mass of products generated. More specifically, you must ensure that the number of atoms on the reactant side equals the number of atoms on the product side. Stoichiometric coefficients, which are placed in front of the compound, are used to indicate the number of moles of a given species involved in the reaction. For example, the balanced equation expressing the formation of water is
2 H2 (g) + O2 (g) 2 H2O (l)
The coefficients indicate that two moles of H2 gas must be reacted with one mole of O2 gas to produce two moles of water. In general, stoichiometric coefficients are given as whole numbers.
The steps you take to balance a chemical reaction are necessary to ensure you have the correct recipe. You wouldn’t want to bake a cake using a recipe that didn’t properly balance the amounts of flour, eggs, sugar, and butter, and you wouldn’t want to conduct an experiment or chemical process without the balanced equation.
When balancing equations, focus on the least represented elements first and work your way to the most represented element of the reaction (usually oxygen or hydrogen).
Let’s review the steps involved in balancing a chemical equation, using an example.
Example: Balance the following reaction.
C4H10() + O2(g) CO2(g) + H2O()
Solution: First, balance the carbons in the reactants and products.
C4H10 + O2 4 CO2 + H2O
Second, balance the hydrogens in the reactants and products.
C4H10 + O2 4 CO2 + 5 H2O
Third, balance the oxygens in the reactants and products.
2 C4H10 + 13 O2 8 CO2 + 10 H2O
Finally, check that all of the elements, and the total charges, are balanced correctly. If there is a difference in total charge between the reactants and products, then the charge will also have to be balanced. (Instructions for balancing charge are found in Chapter 11.)
Applications of Stoichiometry
Once you’ve balanced the chemical equation, you then have a very valuable tool for solving many chemical reaction problems on the MCAT. Perhaps the most useful bit of information to glean from a balanced reaction is the mole ratios of reactants consumed to products generated. Furthermore, you can also generate the mole ratio of one reactant to another or one product to another. All these ratios can be generated by a comparison of the stoichiometric coefficients. In the example involving the formation of water, we now understand that for every one mole of hydrogen gas consumed, one mole of water can be produced, but that for every one mole of oxygen gas consumed, two moles of water can be produced. Furthermore, we see that, mole-to-mole, hydrogen gas is being consumed at a rate twice that of oxygen gas.
Stoichiometry problems usually involve at least a few unit conversions, so you must be careful when working through these types of problems to ensure that units cancel out appropriately, leaving you with the desired unit(s) of the answer choices. Pay close attention to the following problem, which demonstrates a clear and easy-to-follow method for keeping track of the numbers, the calculations, and the unit conversions.
Because the MCAT is a critical-thinking test, we might see application and analysis questions related to limiting reactants or percent yields.
Example: How many grams of calcium chloride are needed to prepare 72 g of silver chloride according to the following equation?
CaCl2 (aq) + 2 AgNO3 (aq) Ca(NO3)2 (aq) + 2 AgCl (s)
Solution: Noting first that the equation is balanced, 1 mole of CaCl2 yields 2 moles of AgCl when it is reacted with 2 moles of AgNO3. The molar mass of CaCl2 is 110 g, and the molar mass of AgCl is 144 g.
Thus, 27.5 g of CaCl2 are needed to produce 72 g of AgCl.
When the quantities of two reactants are given on the MCAT, we should expect to have to figure out which is the limiting reactant.
If you recall some of the experiments that you ran in your general chemistry lab, you know that rarely did you ever use stoichiometric quantities of compounds. Remember that stoichiometric coefficients are usually whole numbers and refer to the number of moles of the particular reactants and products. Some common compounds, like CaCO3, have molar masses in excess of 100 g/mol, and using such quantities per student would make for a very, very expensive lab. Rarely, then, are reactants added in the exact stoichiometric proportions as shown in the balanced equation. As a result, in most reactions, one reactant will be used up (consumed) first. This reactant is known as the limiting reactant because it limits the amount of product that can be formed in the reaction. The reactant that remains after all the limiting reactant is used up is called the excess reactant.
For problems involving the determination of the limiting reactant, you must keep in mind two principles:
1. All comparisons of reactants must be done in units of moles. Gram-to-gram comparisons will be useless and maybe even misleading.
2. It is not the absolute mole quantities of the reactants that determine which reactant is the limiting reactant. Rather, the rate at which the reactants are consumed (the stoichiometric ratios of the reactants) combined with the absolute mole quantities determines which reactant is the limiting reactant.
Example: If 28 g of Fe react with 24 g of S to produce FeS, what would be the limiting reagent? How many grams of excess reagent would be present in the vessel at the end of the reaction?
The balanced equation is Fe + S FeS.
Solution: First, determine the number of moles for each reactant.
Since 1 mole of Fe is needed to react with 1 mole of S, and there are 0.5 moles Fe for every 0.75 moles S, the limiting reagent is Fe. Thus, 0.5 moles of Fe will react with 0.5 moles of S, leaving an excess of 0.25 moles of S in the vessel. The mass of the excess reagent will be
Sometimes chemistry lab professors like to torture students by grading them on the purity of their products and the yields of their experiments. The enjoyment for the professor comes from watching crazed premed students desperately scraping their glassware in the forlorn hope that they might be able to capture just enough errant product to increase yield by a few measly percentage points. The “yield” of a reaction is either the amount of product predicted (theoretical yield) or obtained (raw or actual yield) when the reaction is carried out. Theoretical yield is the maximum amount of product that can be generated, predicted from the balanced equation, assuming that all of the limiting reactant is consumed, no side reactions have occurred, and the entire product has been collected. Theoretical yield, as your experience has most certainly taught you by now, is rarely ever attained through the actual chemical reaction. Actual yield is the amount of product that you are actually able to obtain. The ratio of the actual yield to the theoretical yield, multiplied by 100 percent, gives you the percent yield, and this number is the fragile thread by which so many premeds fear their future careers hang.
Percent yield = (Actual yield/Theoretical yield) × 100%
An experimental-type passage that involves a chemical reaction may include a pseudo-discrete question that involves finding the percent yield.
Example: What is the percent yield for a reaction in which 27 g of Cu is produced by reacting 32.5 g of Zn in excess CuSO4 solution?
Solution: The balanced equation is as follows:
Zn (s) + CuSO4 (aq) Cu (s) + ZnSO4 (aq)
Calculate the theoretical yield for Cu.
Finally, determine the percent yield.
When we are given an excess of one reagent on the MCAT, we know that the other reactant is the limiting reagent. Be sure to take advantage of these easy cues when they appear on Test Day!
We began our consideration of compounds with a particularly odoriferous one: benzaldehyde. As a compound, it is made from constituent atoms of different elements in a set ratio defined by its empirical or molecular formula. Each molecule of a compound has a defined mass that is measured as its molecular weight. The mass of one mole of any compound is determined from its molar mass in the units of grams/mole. We reviewed the basic classifications of reactions commonly tested on the MCAT: combination, decomposition, single-displacement, and double-displacement reactions. Furthermore, we are now confident in our understanding of the steps necessary to balance any chemical reaction; we are ready to tackle more stoichiometric problems in preparation for Test Day.
Before moving to the next chapters discussing chemical kinetics and thermodynamics, let us offer our heartiest congratulations to you. If you have been reading these chapters in order, you have now completed one-third of this general chemistry review! Take note of this, in part, because it is an important milestone in your progress toward success on Test Day and you should be proud of your accomplishments, but mostly because these first four chapters have introduced you to the fundamental concepts of chemistry—everything from the structure of the atom and trends of the elements to bonding and the formation of compounds. The understanding you have gained so far will be the foundation for your comprehension of even the most difficult general chemistry concept tested on the MCAT. Keep moving forward with your review of general chemistry; don’t get stuck in the details. Those will come to you best through the application of the basic principles to MCAT practice passages and questions.
And remember, now that you’ve read this chapter, the next time somebody says, “Oh, what stinks?” your response can be more than just, “Oh, sorry.”
CONCEPTS TO REMEMBER
A compound is a pure substance composed of two or more elements in a fixed proportion. Compounds can react with other elements or compounds to form new compounds and be broken down by chemical means to produce their constituent elements or other compounds, which can themselves go on to become involved in other reactions.
Molecular weight is the mass in amu of the constituent atoms in a compound, given by the molecular formula, which gives the exact number of atoms of each element in a compound. Empirical formula weight is the mass of the constituent atoms in a compound’s empirical formula, which is the smallest whole number ratio of the elements in a compound. Molar mass is the mass in grams of one mole (6.022 × 1023 molecules) of a compound.
An equivalent is a measure of capacity to react in a certain way. One equivalent is an amount of a chemical compound equal to one mole of hydrogen ions or hydroxide ions in acid-base reactions or one mole of electrons in redox chemistry. Gram equivalent weight is the mass in grams of a compound that will yield one equivalent of hydrogen ions, hydroxide ions, or electrons. Normality is the ratio of equivalents per liter.
Combination reactions occur when two or more reactants combine to form one product.
Decomposition reactions occur when one reactant is chemically broken down into two or more products, usually by heat or electrolysis.
Single-displacement reactions occur when an atom or ion of one compound is replaced by an atom or ion of another element.
Double-displacement reactions occur when elements from two different compounds trade places with each other to form two new compounds. Neutralization reactions are a specific type of double displacement in which an acid reacts with a base to produce a solution of salt and water.
Net ionic equations ignore spectator ions to focus only on the species that actually participate in the reaction.
The steps for balancing chemical equations are as follows:
—Balance the nonhydrogen and nonoxygen atoms.
—Balance the oxygens.
—Balance the hydrogens.
—Balance charge if necessary.
Balanced reactions are essential for calculating limiting reactant (the reactant that will be consumed first in a chemical reaction) and yields. Theoretical yield is the maximum amount of product that can be formed, assuming all limiting reactant is consumed. Actual yield is the amount of product collected from a chemical reaction. Percent yield is the ratio of actual yield divided by theoretical yield, multiplied by 100 percent.
EQUATIONS TO REMEMBER
1. Which of the following best describes ionic compounds?
A. Ionic compounds are formed from molecules containing two or more atoms.
B. Ionic compounds are formed of charged particles and are measured by molecular weight.
C. Ionic compounds are formed of charged particles that share electrons equally.
D. Ionic compounds are three-dimensional arrays of charged particles.
2. Which of the following has a formula weight between 74 and 75 grams per mole?
3. Which of the following is the gram equivalent weight of H2SO4?
A. 98.08 g/mol
B. 49.04 g/mol
C. 196.2 g/mol
D. 147.1 g/mol
4. Which of the following molecules CANNOT be expressed by the empirical formula CH?
5. In which of the following compounds is the percent composition of carbon closest to 63 percent?
6. What is the most accurate characterization of the reaction shown?
Ca(OH)2 (aq) + H2SO4 (aq) CaSO4 (aq) + H2O ()
7. In the reaction shown, if 39.03 g of Na2S are reacted with 113.3 g of AgNO3, how much, if any, of either reagent will be left over once the reaction has gone to completion?
Na2S + 2 AgNO3 Ag2S + 2 NaNO3
A. 41.37 g AgNO3
B. 13.00 g Na2S
C. 26.00 g Na2S
D. 74.27 g AgNO3
8. Using a given mass of KClO3, how would one calculate the mass of oxygen produced in the following reaction, assuming it goes to completion?
2 KClO3 2 KCl + 3 O2
9. Aluminum metal can be used to remove tarnish from silver when the two solid metals are placed in water, according to the following reaction.
3 AgO + 2 Al 3 Ag + Al2O3
This reaction is a
I. double-displacement reaction.
II. single-displacement reaction.
III. redox reaction.
IV. combination reaction.
A. II only
B. I and III
C. II and III
D. IV only
10. Which of the following type(s) of reaction(s) generally has/have the same number of reactants and products?
I. Single-displacement reaction
II. Double-displacement reaction
III. Combination reaction
A. I only
B. II only
C. II and III
D. I and II
11. Which of the following is the correct net ionic reaction for the reaction of copper with silver nitrate?
A. Cu + AgNO3 Cu(NO3)2 + Ag
B. Cu + 2 Ag+ + NO3- Cu2+ + 2 NO3- + 2 Ag+
C. 2 Ag+ + 2 NO3- 2 NO3- + 2 Ag+
D. Cu + 2 Ag+ Cu2+ + 2 Ag
12. In the process of photosynthesis, carbon dioxide and water combine with energy to form glucose and oxygen, according to the following equation.
CO2 + H2O + Energy C6H12O6 + O2
What is the theoretical yield, in grams, of glucose if 30.00 grams of water are reacted with excess carbon dioxide and energy, in the balanced equation?
A. 50.02 grams glucose
B. 300.1 grams glucose
C. 30.03 grams glucose
D. 1,801 grams glucose
13. One way to test for the presence of iron in solution is by adding potassium thiocyanate to the solution. The product when this reagent reacts with iron is FeSCN2+, which creates a dark red color in solution via the following net ionic equation.
Fe3+ (aq) + SCN- FeSCN2+
How many grams of iron sulfate would be needed to produce 2 moles of FeSCN2+?
A. 400 grams
B. 800 grams
C. 200 grams
D. 500 grams
Small Group Questions
1. Can the Law of Constant Composition be applied to solutions?
2. What is the purpose of calculating a reactant’s gram equivalent weight?
Explanations to Practice Questions
Ionic compounds are composed of atoms held together by ionic bonds. Ionic bonds associate charged particles with disparate electronegativities; for example, sodium (Na+) with chloride (Cl-). In ionic bonds, electrons are not really “shared” but are rather donated from the less electronegative atom to the more electronegative atom. As a result, ionic compounds are not formed from true molecules, as are covalent compounds. (A) and (B) both describe covalent compounds; their smallest unit is a molecule, which is typically described in terms of molecular weight and moles. In contrast, ionic compounds are measured using “formula weights” and are made of three-dimensional arrays of their charged particles, as indicated in (D). (C) is incorrect because ionic compounds do not share electrons equally; equal sharing occurs in covalent bonds.
Of the compounds listed, only (A) and (C) are ionic compounds, which are measured in “formula weights.” The other options, (B) and (D), are covalent compounds and thus are measured in “molecular weight.” This clues us in to the fact that we don’t really even need to examine choices (B) and (D). (A) consists of potassium (39.0983 g) plus chloride (35.453 g), which has a total weight of 74.551 grams. (C) is made up of 2 lithiums (2 × 6.941 g) and 2 chlorides (2 × 35.4527 g), whose sum exceeds 75 grams.
First, it is helpful to know the molecular weight of one mole of H2SO4, which is found by adding the molecular weight of the atoms that constitute the molecule: 2 × (molecular weight of hydrogen) + 1 × (molecular weight of sulfur) + 4 × (molecular weight of oxygen) = 2 × 1.00794 g + 32.065 g + 4 × 15.9994 g = 98.078 g. Next, you must understand what the term gram equivalent weight means. Gram equivalent weight is the weight (in grams) that would release one acid equivalent. Because H2SO4 has two hydrogens per molecule, the gram equivalent weight is 98.078 g/mole divided by 2, or 49.039 g/mole. All of the other answer choices are multiples of this number.
The definition of an empirical formula is a formula that represents a molecule with the simplest ratio, in whole numbers, of the atoms/elements comprising the compounds. In this case, given the empirical formula CH, any molecule with carbon and hydrogen atoms in a 1:1 ratio would be accurately represented by this empirical formula. Benzene, C6H6 (A), is thus correct, as is ethyne, C2H2 (B). (D) has eight carbon atoms and eight hydrogen atoms, making it correct as well. (C) is incorrect, and thus the right answer, because it has three carbon atoms while having four hydrogens. Both its molecular and empirical formulas would be C3H4, because this formula represents the smallest whole number ratio of its constituent elements.
The percent composition of any given element within a molecule is equal to the molecular mass of that element in the molecule, divided by the formula or molecular weight of the compound, times 100%. In this case, acetone, C3H6O, has a total molecular weight of (12.0107 g × 3 + 1.00794 g × 6 + 15.994 g × 1) = 58.074 g/mol, of which 12.0107 g × 3 = 36.0321 g/mol is from carbon. Thus, the percent composition of carbon is 63.132%. With this calculation serving as an example, you can calculate the percent composition for ethanol (C2H6O; MW = 41.023 g/mol) to be 58.556%; for C3H8 (MW = 44.096 g/mol) to be 81.713%; and for methanol (CH4O; MW = 32.036 g/mol) to be 37.491%. These calculations make it clear that although both acetone (A) and ethanol (B) have percent compositions of carbon close to 63%, acetone is closer (within 1%).
This reaction is a classic example of a neutralization reaction, in which an acid and a base react to form water and a new aqueous compound. Although this reaction may also appear to fit the criteria for a double-displacement reaction, in which two molecules essentially “exchange” with each other, neutralization is a more specific description of the process. A single-displacement reaction is typically a redox (reduction/oxidation) reaction in which one element is replaced in the molecules; thus (A) and (D) are incorrect.
In this question, you are first given the masses of both reactants used to start the reaction. To figure out what will be left over, we must first determine which species is the limiting reagent. First, determine the molecular weight of each of the reactants: Na2S = 78.05 g/mol; AgNO3 = 169.9 g/mol. We find that we are given 0.5 mol Na2S for the reaction and 0.6669 mol AgNO3. Because we need two molar equivalents of AgNO3 for every mole of Na2S, AgNO3 is the limiting reagent, and the correct answer choice will be in grams of Na2S. Next, determine how much of the Na 2S will be left over by figuring out how much will be used if it reacts with all of the AgNO3.
Use half as much Na2S as AgNO3:
[(1.000 mol Na2S)/(2.000 mol AgNO3)] × 0.6669 mol AgNO3 = 0.3334 mol Na2S
Then subtract this amount of reagent used from the total available:
0.5000 mol Na2S-0.3334 mol Na2S = 0.1666 mol excess Na2S
Finally, determine the mass that this represents:
0.1666 mol excess Na2S × 78.05 g/mol Na2S = 13.00 g Na2S
(A) and (D) are incorrect because there will be no remaining AgNO3. (C) is the mass in 0.33 mol of Na2S, and you would have gotten this if we mixed it up with 0.166 mol.
You’re told that you must begin with some given mass of KClO3, x grams. To use it to determine a mass of another product, we must convert it to moles. Thus far, you would have
which is equal to the number of moles of KClO3.
This first step eliminates (C).
Next, convert the number of moles of KClO3 to the number of moles of oxygen, according to the balanced equation presented in the question stem:
Putting both steps together, the equation thus far is
This second step eliminates (B) because it does not use the correct molar ratio between KClO3 and O2. The final step is to convert the number of moles of oxygen to a mass, in grams of O2. At this point in your equation, the number of moles is in the numerator, and you want the number of grams of oxygen in the numerator, so multiply your ratio by the molar mass of oxygen, as shown in (A).
In the reaction, there is a single displacement (II), with the silver in silver oxide being replaced by the aluminum to form aluminum oxide. This single-displacement reaction also necessitates a transfer of electrons in a reduction/oxidation reaction or “redox” reaction (III). Therefore, the correct answer is both II and III, (C). A double-displacement reaction (I) typically takes two compounds and causes two displacements, whereas only one occurs in this question. Combination reactions (IV) generally take two atoms or molecules and combine them to form one product, usually with more reactants than products, which also is not the case in this question.
Typically, single-displacement (or oxidation/reduction) reactions and double-displacement reactions both have the same number of reactants and products. For example, single-displacement reactions are often of the following form (M = metal 1; M’ = metal 2; A = anion):
M + M’A M’ + MA
In this type of reaction, M takes the place of M’ in combining with an anion. Typically, a process of oxidation and reduction of the involved metals enables this. Double-displacement reactions also tend to have the same number of reactants and products, represented by this type of reaction (C = cation 1; C’ = cation 2; A = anion 1; A’ = anion 2):
CA + C’A’ C’A + CA’
In this reaction, the two compounds essentially “swap” anions/cations, beginning and ending with two compounds.
Combination reactions typically have more reactants than products, represented by the following reaction:
A + B C
A net ionic equation represents each of the ions comprising the compounds in the reactants and products as individual ions, instead of combining them as molecules. Thus, (A) is not a net ionic reaction. The term net means that the correct answer does not include any spectator ions (ions that do not participate in the reaction). In this reaction, nitrate (NO3-) remains unchanged. You can eliminate any answer choice that includes it, which leaves only (D).
This is a popular equation that you’ve probably seen before. What’s missing from it are the coefficients! This is an unbalanced equation. To get anywhere with this problem, you have to balance it first.
6CO2 + 6H2O + energy C6H12O6 + 6O2
The theoretical yield is the amount of product synthesized if the limiting reagent is completely used up. This question therefore asks how much glucose is produced if the limiting reagent is 30 grams of water. First, calculate the number of moles of water represented by 30 grams by dividing by its molecular weight (18.01 g/mol). You’ll have a lot of conversion factors, so wait until the end to multiply it all out.
Next, add conversion factors to find the equivalent number of moles of glucose.
Finally, convert moles of glucose into grams of glucose by multiplying by its molecular weight (180.2 grams / mol glucose).
After crossing out equal terms from the numerator and denominator, you are left with this:
Dividing 18 and 6 into our numerators will yield 50 grams of glucose.
What you are shown is a net ionic equation. To answer this question, work backward from the amount of product to the reactant with its spectator ions (in this case, sulfate).
First, for every two moles of FeSCN we create, we must react two moles of Fe3+. Therefore, you’re looking for the mass of iron sulfate that can provide two moles of iron.
Next, determine the charge on sulfate, which is -2 and iron, which in this case is +3. Therefore, iron sulfate must be present as follows:
The molecular formula tells you that each mole of iron sulfate releases two moles of atomic iron. Therefore, you only need one mole of iron sulfate for this reaction, which means the molar mass of iron sulfate is the answer.