Stoichiometry

Table of Contents

• Introduction
• Chemical Equations
• The Law of Definite Proportions, et al.
• The Law of Conservation of Matter and Energy
• Balancing Chemical Equations
• The Mole: A Magic Number
• Limiting Reagent
• Percent Yield
• Percent Composition
• Molarity and Molality
• Titration
• Practice Exercises

Introduction

    In any general chemistry class, a fair amount of time is spent on "stoichiometry."

Stoichiometry is the division of chemistry devoted to quantitative (numerical) analysis of chemicals, their relationships, and their reactions.

    Stoichiometry makes it possible to determine how much of each reactant is needed to produce a given quantity of product in a reaction, to calculate formulas of compounds from decomposition products, to find volumes of gases under given conditions, to label concentrations of solutions, and much more. 

    From here on, it is assumed that you have some basic algebra skills.  If you need a primer on math, we highly recommend http://www.sosmath.com/.  Also, if you have forgotten any dimensional analysis, it is suggested that you read the lesson on that topic.

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Chemical Equations

    You probably know that hydrogen gas is explosive. That statement, though true, does not say very much about hydrogen chemically. For more precise expression, chemists describe chemical reactions (such as the spontaneous oxidation of hydrogen) with chemical equations.

Chemical equations show what reactants are involved in a chemical reaction, what products are produced, and relative amounts of each produced.

    Chemical equations can also show generalized energy transformations (such as absorption or release of heat).  That, however, will not be covered in this lesson.

    A chemical equation is always written with at least one arrow, and is generally written with the symbolic representations of elements and compounds.  If a reaction proceeds in only one direction (that is, none of the products ever revert back to the reactants), the reactants are traditionally written on the left, with an arrow pointing to the right, and the products following.  An example (the explosion of hydrogen gas in an oxygen environment) follows:

    Since it takes time to download a lot of images such as the above, though, we will primarily use the following notation:

2H₂ + O₂ --> 2H₂O

    In reality, there are no reactions that are completely irreversible, that do not have some products going back to reactants.  However, for simplicity, most reactions are considered irreversible, since the amount of product going back to reactant is negligible.  Some reactions, especially dissociation reactions (reactions involving the breaking up of a compound in solution), are very reversible; a lot of product turns back into reactant.  For these reactions a double sided arrow or two half-arrows are used, as the following equations for the dissolution of table salt, sodium chloride (NaCl), show.

    Again, though, in order to reduce page download time, we will primarily use the following notation:

NaCl <--H₂O--> Na⁺ + Cl^-

    Don't worry if you don't understand the meaning of this; right now, just make sure that you can recognize what the different styles of arrows mean.  Basically, a reaction proceeds mostly in whatever direction the arrow points, so a reaction that moves in both directions appropriately has a two-way arrow. For those curious, the H₂O on top of the arrows in the prior equation means that the reaction takes place in water, though the water is neither a reactant or a product. Such a facilitator of a reaction is often called a catalyst, though in this case, chemists would say that the water is the solvent. Again, don't worry if this confuses you, we'll discuss this more later.

    Clearly, there is a lot more than an arrow in a chemical equation.  The symbols should be familiar to you; they are the symbols of the elements off of the periodic table.  At some point, there will be a lesson on chemical bonding that will cover what elements go together to form compounds much more in depth.  For now, though, recognize that symbols not separated by a plus sign are chemically bonded to each other, while plus signs separate chemically individual elements or compounds.

    This leads us into a discussion of...

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The Law of Definite Proportions, et al.

    In science, an idea that has been thoroughly tested and shown to be true over and over again is called a theory (a marked departure from the conventional sense of the word, which more closely means hypothesis). After a long time, and many more tests, the scientific community may promote a theory to the title of law, meaning that it appears to be true under all applicable situations anywhere in the universe.

The Law of Definite Proportions states that any sample of a given compound contains the same elements in the same mass proportions.

    When Joseph Proust, a late 18th-century chemist, stated this law, it appeared to satisfy experiments perfectly. However, the discovery of isotopes, atoms of a given element that are the same but with differing numbers of neutrons, posed a problem (albeit minor) for this law.  Since all elements in nature are composed of mixtures of two or more isotopes, and different isotopes have different masses, basing this law on mass makes little sense. However, this definition has stuck, and it works well, provided that any masses used in calculations are averaged to take into account isotopic variation. More on this later.

    To illustrate the Law of Definite Proportions, consider water.  If water were decomposed, that is, broken down into hydrogen and oxygen, and the hydrogen and oxygen were massed (weighed), the ratio of the mass of the oxygen to the mass of the hydrogen would always be the same, regardless of the amount of water that was decomposed.

    Antoine Lavoisier elaborated on the Law of Definite Proportions with his Law of Multiple Proportions.

The Law of Multiple Proportions states that in any compound, the component elements combine together in simple, whole number ratios.

    If we assume that atoms exist, this law follows directly from the Law of Definite Proportions.  Since a given mass of an element will contain a definite number of atoms, and the masses combine together in fixed ratios, it makes sense that the numbers of atoms of each element will be exist in simple, whole number ratios.

    Again, consider water.  The law of definite proportions states that the ratio of the mass of hydrogen to the mass of oxygen in any size sample will be a constant.  Let us therefore assume our sample size to be one molecule of water.  Since, neglecting isotopic variation, each atom of hydrogen will have a fixed mass and each atom of oxygen will have a fixed mass, the ratio of the number of atoms of hydrogen to the number of atoms of oxygen must be a simple ratio.  It happens to be 2:1.  That is why water is given the formula H₂O.

    Here we encounter another piece of information needed to adequately work with chemical equations (remember, that is where this discussion came from).

A subscript number, such as the 2 in H₂O, indicates the number of atoms of the immediately preceding element (in water's case, hydrogen, H) present in a molecule or formula unit of the given compound.

    Don't worry too much right now about the distinction between "molecule" and "formula unit." As we will learn later, the former refers primarily to covalently bonded compounds, while the later refers primarily to ionically bonded compounds. Since this may not make any sense right now, don't worry about it. For the purposes of this lesson, "molecule" and "formula unit" can be used interchangeably.  In a lesson on chemical bonding and molecular structure, this is not the case at all.  Another useful rule when dealing with compounds (in or out of a chemical equation) follows.

A subscript number following a group of atoms in parentheses, such as the 2 in calcium hydroxide, Ca(OH)₂, indicates the number of the preceding group (in the case of calcium hydroxide, OH) present in a molecule or formula unit of the given compound.

    Using these rules, you should be able to determine, for example, how many atoms of each element are in one formula unit of scandium carbonate, Sc₂(CO₃)₃.  The answer is two scandium atoms, three carbon atoms, and nine oxygen atoms (don't forget that each carbonate group has three oxygens, and there are three groups).

    Before we can discuss what the non-subscript numbers in chemical equations are, we must discuss...

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The Law of Conservation of Matter and Energy

The Law of Conservation of Matter and Energy states that the total sum of matter and energy in the universe is fixed.

    Huh? Sum of matter and energy? Well, you've probably heard of Einstein's mass-energy equation, probably the most famous equation in modern history: E = mc². This allows us to treat matter and energy as one (though the mechanics of this equation are outside the bounds of this lesson). Any amount of matter has an energy equivalent, and the converse is also true. Thus, the Law of Conservation of Matter and Energy, a very philosophically large statement (we can define the laws of the entire universe?), can make sense. You may be familiar with the more common and elementary Law of Conservation of Mass:

The Law of Conservation of Mass states that in ordinary chemical and physical reactions, mass is neither created nor destroyed.

    This will work fine for us in this lesson, since we will only talk about "ordinary chemical and physical reactions." The broader law was mentioned so that we do not neglect nuclear reactions, in which mass is always created or destroyed (though some energy is also released or absorbed, in accordance with the prior law).

    We will now expand the Law of Conservation of Mass to establish the following principle:

In ordinary chemical and physical reactions, no element will transmute (change) to another element.

    This is almost as powerful a statement as the prior two laws. This abolishes the ancient idea of finding a chemical means for converting lead to gold. It also makes more clear what was evident from the prior law, that all atoms on one side of a chemical equation must exist on the other side of the chemical equation. This simple idea leads into...

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Balancing Chemical Equations

    The Law of Conservation of Mass dictates that anything on one side of a chemical equation (for example, any atoms present in the reactants in a reaction) must also be on the other side (the products). This does not say anything about what atoms are bonded to what - compounds can break apart, form, or rearrange, but no element can disappear or change to a different element. It therefore becomes apparent that not all chemical equations are valid. For example, consider the oxidation (explosion) of hydrogen gas once more, in this incorrect equation:

 H₂ + O₂ --> H₂O

    Do you see the problem? Count the number of atoms of each element on each side of the equation. There are two hydrogen atoms on both sides, but there is only one atom of oxygen on the right, and two on the left. To solve this, we cannot just chop the subscript off of the left oxygen, for oxygen is diatomic, it only exists in nature bonded to itself, in groups of two atoms. To solve this problem, and balance the chemical equation, we add coefficients to the equation.

Coefficients, numbers before a compound or element in a chemical equation, indicate how many discrete units of the subsequent element or compound are used or produced in the reaction.

    What does discrete mean? Basically, whereas a subscript number indicates how many of the preceding element or group of elements are chemically bonded to each other in a molecule or formula unit of a compound, a coefficient indicates how many individual formula units or molecules of the subsequent compound or element are needed for the reaction to go to completion. The coefficients are always whole numbers, indicating that...

In any chemical reactions, the reactants combine together in simple, whole number ratios to form products in simple, whole number ratios.

    Let us now consider the correct equation for the explosion of hydrogen gas:

2H₂ + O₂ --> 2H₂O

    There are now the same number of atoms of each element on either side. Furthermore, this equation tells us the for every two molecules of diatomic hydrogen gas, one molecule of diatomic oxygen is needed, producing two molecules of water. We can make this equation even more informative as follows:

2H₂ (g) + O₂ (g)--> 2H₂O (l)

    The letters in parentheses tell us the state of the preceding reactant or product. A "g" means gas, "l" means liquid, "s" means solid, and "aq" means aqueous (dissolved in water). We now know that gaseous hydrogen and oxygen will react to form liquid water. Note that it is not mandatory that these letters be written.

    Now, for the main part of our discussion in this section. How to balance equations. There are several basic rules that help when balancing, though practice (such as working through the problems at the bottom of the page) will help most.

1. Is the equation already balanced?

2. Identify any elements that are present only in one compound on either side. Balance these compounds first, for the coefficients will have a fixed ratio dependent on the number of atoms of the element in each.

3. If everything would work if one of the compounds had a fractional coefficient (i.e. 1.5), multiply all the coefficients by a given factor to eliminate the fraction. For example, if a desired coefficient were 1.5, multiply everything by two, so that coefficient goes to 3.

4. Keep in mind that in most cases ionic compounds will break up only into their ions.  Rarely will polyatomic ions split up, so they can be balanced as a group.

5. Charge must balance in a compound, unless one is balancing polyatomic ions!

6. In general, balance hydrogen second-last and oxygen last.

    The last two rules involve charge balancing, a topic we have not yet covered. If you are not familiar with ionic chemistry, consider reviewing a general chemistry text. At some point we will have a lesson on ions on ION, but right now all we have is a table of ions and other useful ideas.

    Let us now try balancing the following equation:

NaCl + H₂SO₄ --> Na₂SO₄ + HCl

    This reads as "sodium chloride reacts with sulfuric acid to yield sodium sulfate and hydrochloric acid." Is this already balanced? No, once can immediately see that the number of sodium atoms on either side is not equal. We now go to rule number two above. Sodium and chlorine (well, actually, the chloride ion) are each present in only one compound on either side. Since NaCl and HCl both contain only one chloride, their coefficients must be the same. There are two sodium atoms in Na₂SO₄, so its coefficient must be half of that of NaCl. However, this would leave us with a coefficient of .5 for Na₂SO₄, not a whole number. We are not sure that everything would work, though, so we don't apply rule three just yet. Instead we add a two in front of NaCl, giving us the following:

2NaCl + H₂SO₄ --> Na₂SO₄ + HCl

    But this is not balanced either. Now there are two chlorides on the left, and only one on the right. So we hang a two in front of the HCl:

2NaCl + H₂SO₄ --> Na₂SO₄ + 2HCl

    Voila! The equation is balanced! There is one more basic rule of which you must be aware:

In chemical equations, as in algebraic equations, coefficients are reduced to the lowest possible whole numbers.

    It is for this reason that the aforementioned equation will always be in the form written, never, for example, as:

4NaCl + 2H₂SO₄ --> 2Na₂SO₄ + 4HCl

    As you can see, you can easily divide all the coefficients by two and still have simple, whole number ratios.

    Once you have mastered the art of balancing equations, you can save time by using the online equation balancer, but remember, you won't always have this utility available!

    But what good is balancing equations? We don't easily work with individual atoms or molecules in the lab, we work with liters and grams. How do we get from molecules to grams? The answer is something called...

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The Mole: A Magic Number

    No, we're not talking about the animal. A mole is a lot like a dozen. You can have a dozen eggs, a dozen fish, a dozen atoms. Likewise, you can have a mole of eggs, a mole of fish, and all the time a mole of atoms. A mole is a lot bigger than a dozen, though. Specifically, a mole specifies a Avogadro's Number of something.

A mole of any element or compound contains Avogadro's Number of atoms, molecules, or formula units, 6.022 x 10²³.

    How does this work? Basically, the mole derives from our system of atomic masses (specifically, the mass of the isotope carbon-12), the desired result being the above fact. We won't talk about the derivation of the mole.

    This is all well and good, but how does this help us? There is another magic property of the mole that is of far greater importance to us:

The mass in grams of one mole of any element or compound is numerically equivalent to the mass in AMU (or Daltons) of one atom, molecule, or formula unit of the given element or compound.

    This is truly powerful. We can now say that one mole of atoms of carbon-12 weighs exactly 12 grams. We have a relationship between mass in the "real world" and atomic mass. At this point, we should note that

The mass of a compound is equal to the sum of the masses of its atoms.

    So, the mass of one molecule of water in AMU (and thus the weight in grams of one mole, or 6.022 x 10²³ atoms, of water) is 2 x 1.00794 (the mass of an average hydrogen atom) plus 15.9994, the mass of an average oxygen atom, or 18.0153. There is a molar mass calculator in the utilities section to help you calculate this. It also does percent composition, something we will talk about in due course.

    Since a mole of any element or compound contains the same number of atoms, molecules, or formula units, moles can be directly related to coefficients in chemical equations. Simply consider any coefficients to be the number of moles of the reactant needed to produce a given number of moles of product, and you can easily figure out the mass of reactants needed to yield a given mass of product.  Here is an example to clarify this discussion:

    Practice solving these stoichiometric problems with some of the questions at the bottom of this lesson.

    It should be noted that there is one more important magic property of the mole: one mole of an ideal gas at standard temperature and pressure occupies 22.4 L of space. This fact is more related to the gas laws, however, and will be discussed in that lesson.

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Limiting Reagent

    Rarely does one perform a chemical reaction in which all the reactants are in exactly the right stoichiometric ratios.  For example, if a reaction calls for three moles of reactant X and two moles of reactant Y to form two moles of product Z, and one has slightly less than two moles of X and slightly more than three moles of Y, it is obvious that the amount of product Z produced is limited by the amount of X, since there is not enough to make the ratio work correctly.

The limiting reagent of a chemical reaction is the reactant that, if the reaction proceeded, would run out first and thus cause the reaction to stop.

    When dealing with chemical reactions, it is often convenient to consider one or more of the reactants to be in excess.  This means that no matter how much of the other reactants are present, there will always be enough of the reactant in excess present for the reaction to proceed.  Thus, the reactant in excess can never be the limiting reagent.

    Calculating which reactant in a reaction is the limiting reagent is simple enough.

To calculate which reactant, if any, is the limiting reagent, arbitrarily choose one of the reactants not in excess.  Convert all units to moles, and then use a ratio to multiply (upward or downward) the number of moles of each reactant until the chosen reactant's mole quantity is equal to its coefficient in the equation.  If all the reactants' mole quantities are equal to their respective coefficients, there is no limiting reagent and all reactants will be consumed completely.  If any one of the reactants' mole quantities are lower than their respective coefficients, there is a limiting reagent and it is the reactant whose mole quantity is farthest below its respective coefficient.

    It is difficult to visualize this in words, so we use an example/problem:

    Knowing the limiting reagent allows one to perform more accurate stoichiometric calculations.  It is important to know that

When calculating stoichiometric quantities in reactions with limiting reagents, all calculations must be dependent on the limiting reagent, not reagents in excess.

    We can apply this to another example/problem, an extension of the previous problem:

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Percent Yield

In reality, no reaction proceeds in exact accordance with stoichiometric calculations.

    Why is this? There are a number of reasons. Reactions are dependent on the seemingly random motion of particles; if any particle fails to come near the correct particle(s) the reaction is imperfect. Impurities are inevitably present that skew mass readings and might even affect a reaction. Reaction rates can vary with environmental variables. Regardless, in general, stoichiometry lets chemists predict relatively accurately various properties of a reaction, not the least of which is the amount of product expected.

    Again, let us consider the combustion of isopropyl alcohol.  Above, we calculated that 1.32 g CO₂ would be produced by the combustion of 6.00 g of isopropyl alcohol.  In reality, if this experiment were performed, even in the best conditions in the most high-tech laboratory, with the purest reactants, one would be lucky to produce 1.30 g.  As a rule, 

Reactions generally produce less product than theoretically expected.

An easy way to describe how well a reaction proceeds (that is, how close to theoretical calculations an experiment is) is percent yield.

Percent Yield is a percentage equivalent to 100 x (actual yield / calculated yield).

The yields are in terms of grams or moles (since one is just a decimal multiple of the other).  Thus, in our combustion reaction, 100 x (1.30 g / 1.32 g) = 98.4%.  This would be extremely good if this was an actual percent yield.

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Percent Composition

    Another useful percentage calculation is what percentage of a compound a given element comprises.  This is called percent composition.

The Percent Composition of compound is found, for each element, by multiplying the average atomic mass of a given element by the number of atoms of that element present in each molecule or formula unit of a compound, and dividing the product by the mass of the molecule or formula unit.  The answer is then multiplied by 100.

    This is somewhat harder to digest than percent yield.  Let us use an example/question:

    The percent composition calculation is especially useful for determining empirical formulas (formulas that indicate relative amounts of elements; CH₄ and C₂H₈ are both valid and unique compounds that have the same empirical formula [in this case, the former actually is the empirical formula]).  Percent compositions can be converted to molar quantities to experimentally determine empirical formulas.  Such a problem can be found at the bottom of this lesson.

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Molarity and Molality

Molarity and Molality are both expressions of the concentration of a solution.  Molarity is defined as the number of moles of solute per liter of solution while molality is defined as the number of moles of solute per kilogram of solvent.

    These two methods of measuring concentration appear frequently.  Generally, molarity is most popular.  

    At the time of writing, a popular chemical supply company listed in its online catalog, "Hydrochloric Acid (12M)."  This means that the solution sold is a 12 molar solution, meaning that there are 12 moles of HCl per liter of H₂O.

    Normality is another concentration expression frequently used by supply companies.  Normality is normally used in conjunction with acids and bases, and relates to the degree to which a compound ionizes in solution. That is, how many H⁺ or OH^- ions result when a compound is dissolved. This gets somewhat complicated, so we will not discuss it further. Normality is usually expressed with an upper case N.

    Molality is used when solutions expand or contract significantly due to temperature changes (in general, the higher the temperature, the more volume a solution will occupy).  It becomes evident that a system of measurement based on volume is flawed in this situation, so molality, based entirely on unchanging masses and mole quantities, it used.  Molality is usually expressed with an lower-case italicized m.

    Let's try a practice problem:

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Titration

    A chemistry teacher is preparing a lab (that he found at www.nitrogenorder.org) for his class that involves 1M, 2M, and 4M hydrochloric acid solutions.  He forgets, however, to label the beakers since he has to deal with an emergency in the room next door (his comrade accidentally turned on the emergency shower and can't shut it off).  When he returns he can't remember which beaker contains which solution.  All of the precise electronic balances were stolen the day before by a rogue substitute teacher, so he can't simply mass equal quantities of each solution to determine which contains the most solute and which contains the least solute.  What can he do, aside from discarding all the solutions and starting over?  He can perform a titration to determine quantitatively the solution concentrations.

    A titration involves carefully measuring how much of a solution is required to completely react with another solution.  In the hypothetical situation above, the teacher could pour an exact volume of a base (of known concentration) into a flask containing a negligible amount of universal indicator solution (a solution that changes color depending on acidity [pH]), and then drip, using a buret or titration column, one of the HCl solutions into the flask.  When the indicator shows that the solution is neutral, the reaction has gone exactly to completion, and he can use the volume of known concentration base (found by looking at the buret or titration column) to determine the concentration of the acid solution.  This is the subject of our practice exercise:

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Practice Exercises

Chemical Equations

1. What things must be present in a true chemical equation?  Answer

The Law of Definite Proportions, et al.

1. What are isotopes and why does their existence complicate the Law of Definite Proportions?  Answer
2. Using the argument that all compounds are made of atoms, justify the Law of Multiple Proportions.  Answer
3. What is the difference between H₂O and HOH? Between H₂O and OH₂?  Answer

The Law of Conservation of Matter and Energy

1. Why does the equation Ca(OH)₂ --> Ca + H₂O violate the Law of Conservation of Mass?  Answer
2. Does the equation Si + O₂ --> SO₂ conform to the basic laws of matter and energy? Why or why not?  Answer

Balancing Chemical Equations

1. Without using the equation balancer, balance the equation N₂O₅ --> NO₂ + O₂.  Answer
2. Without using the equation balancer, balance the equation for the "combustion" of rocket fuel, hydrogen: H₂ + O₂ --> H₂O.  Answer
3. Without using the equation balancer, balance the equation for the combustion of the fuels and oxidizer used in the Apollo Lunar Module rockets: hydrazine, dimethylhydrazine, and nitrogen tetroxide: NO₄ + N₂H₂ + N₂(CH₃)₂  --> CO₂ + N₂ + H₂O. This information came from the Smithsonian Institute. Answer

The Mole: A Magic Number

1. How much does one mole of the radioactive isotope strontium-90 weigh?  Answer
2. How many molecules of the exotic noble gas compound xenon hexafluoride are in one mole of the compound?  Answer
3. How much does 5.3 x 10¹⁹ molecules of the silicon and glass etching agent hydrofluoric acid weigh?  Answer
4. Ammonia reacts with nitric acid to form the fertilizer and high explosive ammonium nitrate according to the equation NH₃ + HNO₃ --> NH₄NO₃. If 20 g of ammonia reacted with an excess of nitric acid, how many grams of ammonium nitrate were formed?  Answer
5. Nitric oxide, NO, is an important signaling molecule in cells.  It is also a major component of smog, produced during the high temperature combustion reactions in engines, according to the equation N₂ + O₂ --> 2NO.  If an engine completely combusts all the nitrogen in a 100 g sample of air that is 78% nitrogen by mass, with an excess of oxygen provided by an oxidizing agent added to the gasoline, how many grams of nitric oxide are produced?  Answer

Limiting Reagent

1. Why is it important to consider limiting reagents when calculating stoichiometric values?  Answer
2. Ideal combustion, resulting in only carbon dioxide and water, rarely happens. In general, at the very least, some carbon monoxide is also produced.  The real combustion of methane is more closely represented, then, by the equation 3CH₄ + 5O₂ --> 6H₂O + CO₂ + 2CO.  0.1 mol CH₄ is allowed to react with 3.2 g O₂. What is the limiting reagent, if any? Answer
3. Using the data from problem two, how many grams of carbon monoxide will be produced?  Answer
4. Elemental zinc, Zn, reacts violently with elemental sulfur, S₈, according to the equation 8Zn + S₈ --> 8ZnS.  If 12.0 g of zinc react with 12.1 x 10²⁴ atoms of sulfur, what is the limiting reagent, if any? Answer
5. Using the data from problem four, how many moles of zinc sulfide, ZnS, are produced?  Answer

Percent Yield

1. What are two possible reasons for a reaction producing less product(s) than theoretically calculated?  Answer
2. If an industrial process is capable of producing ammonia, NH₃, at a percent yield of 83%, and the chemical engineers' calculations suggest that the process should produce 3.2 metric tons per hour, how many metric tons of ammonia can be reasonably expected per hour?  Answer
3. If a 72% efficient carbon dioxide, CO₂, production process produces 1200 kg of carbon dioxide in reality, how much was theoretically expected?  Answer
4. Iron (III) oxide, Fe₂O₃, reacts with elemental aluminum in a brilliant explosion, forming elemental iron and aluminum oxide according to the equation Fe₂O₃ + 2Al --> Al₂O₃ + 2Fe.  If 15.0 g of Fe₂O₃ react with an excess of aluminum, and careful analysis shows that 7.65 g of Al₂O₃ is produced, what is percent yield of the reaction? Answer

Percent Composition

1. Without using the molar mass calculator's composition feature, what is the percent composition, by mass, of the complex plant pigment chlorophyll-a, C₅₅H₇₂MgN₄O₅? Answer
2. A similar pigment found in plants, chlorophyll-b, has the formula C₅₅H₇₀MgN₄O₆. A large pure sample of one of the pigments is found to contain 2.802 g nitrogen, 3.9998 g oxygen, 3.636 g hydrogen, and 33.029 g carbon. Is the sample chlorophyll-a or chlorophyll-b? Answer
3. 100.0 g of a hydrocarbon are combusted completely and ideally (the only combustion products are carbon dioxide and water). A laboratory analysis shows that about 338 g of CO₂ and about 69.20 g of H₂O are produced. What is empirical formula of the hydrocarbon? Answer
4. If the hydrocarbon in problem three has a molecular mass of about 52.08 g/mol, what is the molecular formula of the hydrocarbon?  Answer

Molarity and Molality

1. Why is molarity generally used more than molality?  Answer
2. When is molality particularly useful?  Answer
3. Which has the higher ratio of water molecules to dissolved ions, a 1M or 5M solution of Mg(OH)₂? Answer
4. What is the molarity of a solution consisting of 1 L water and 35 g CaCl₂?  Answer
5. How many moles of stearic acid, CH₃(CH₂)₁₆COOH, are present in 500 mL of a 2M solution (where carbon tetrachloride [tetrachloromethane], CCl₄, is the solvent)?  Answer
6. What is the molality of a solution formed by adding 27 g of stearic acid to 450 mL of carbon tetrachloride at room temperature? It may be helpful to use information from problem five and the assumption that carbon tetrachloride has a density of about 1.5 g/mL at room temperature.  Answer

Titration

1. Potassium iodide, KI, reacts with lead (II) nitrate, Pb(NO₃)₂, according to the equation 2KI + Pb(NO₃)₂ --> 2KNO₃ + PbI₂. A chemist adds a solution of unknown concentration KI to 100 mL of 1M Pb(NO₃)₂. He finds that it takes 42.3 mL of KI for the reaction to go to completion. What was the concentration of the KI solution? Answer

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