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Welcome to Unit 4 review. This unit will focus heavily on chemical reactions, which are obviously a huge part of chemistry. Specifically we will first introduce reactions, and then look at net ionic equations, representations of reactions, and physical vs. chemical changes. Then we will look at stoichiometry, titration, and then types of chemical reactions, including acid-base reactions, and redox reactions. The first thing we will want to do is to define the basic ways in which
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matter can change forms, so let’s talk about physical and chemical properties as well as physical and chemical changes. First let’s define a physical change. This is when matter changes forms, but the identity of the substance does not change. Recall the phases of matter, solid, liquid, and gas. Now consider water. At cold temperatures, water is a solid, in the form of ice. At room temperature, it’s a liquid, which is what we drink and swim around in.
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At hot temperatures, it’s a gas, which includes water vapor or steam. When matter transforms from one of these phases to another, like when water melts or freezes or boils, this is an example of a physical change, because the identity of the individual water molecules does not change. Each water molecule is exactly the same regardless of the phase they are in, but the way they are arranged in space changes. By contrast, a chemical change is one in which
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the identity of the substance or substances does change. This would require a chemical reaction, whereby chemical bonds break and form, to transform a pure substance into a totally different pure substance. So we begin with some molecule, which contains atoms of specific elements in a particular arrangement, and we end up with a totally different molecule, with atoms in a completely different arrangement. This indicates a chemical change.
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With physical and chemical changes understood, let’s extend this concept. A physical property is a quality that a substance exhibits without having to undergo a chemical change. This would include things like melting point and boiling point, since those transformations are physical changes. They also include things like density and color. These properties can be ascertained without the matter having to undergo any changes at all, whether physical or chemical, which
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means we can observe or measure the properties without performing any chemical reactions that would change the composition of the substance. Physical changes can be used to separate mixtures based on their physical properties. An example of this would be using distillation to separate two liquids based on their differing boiling points, boiling one of them and not the other. Unlike mixtures, however, pure substances can not be separated by physical means, because
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every molecule in a sample of that pure substance has precisely the same set of physical properties, and would react identically to any stimulus. So a pure substance can only be separated by a chemical change. If you are looking at a sample of water, which is a pure substance, it is made of many, many water molecules. Each molecule is made of one oxygen atom and two hydrogen atoms. But we can not
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split water molecules into these elements by a physical process, as each molecule will behave the same way no matter what physical process we subject them to. We can only break apart the chemical bonds that hold the oxygen and hydrogen atoms together by inducing a chemical change. In fact, we can split the water molecules into hydrogen and oxygen molecules by a process called electrolysis. This is a chemical change because chemical bonds in the water molecules are breaking
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and new chemical bonds are forming to generate completely new substances. When this happens, we know chemistry has occurred. Other forms of evidence that chemistry has happened include if heat or light is produced, if a gas forms, if a solid precipitate crashes out of solution, or if there is a color change. Chemical properties are essentially just describing the kinds of chemistry a substance will do. We may be familiar with some of the
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words that describe chemical properties. These include flammability, toxicity, and many others. These words describe the kinds of chemistry a substance will or won’t do. For example, we could examine the likelihood of a particular metal to rust. This is a chemical property, because the process of acquiring rust is due to metal atoms forming new chemical bonds with oxygen atoms from the air. This is an example of a chemical change, where a new compound,
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a metal oxide, is produced. We will see many more examples of chemical properties and corresponding chemical changes as we learn more chemistry. So to summarize, a physical process is one in which the identity of the molecules does not change, and a chemical process is one in which the identity of the molecules does change. Mixtures can be separated into pure substances by physical processes, but pure substances can only be separated into
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smaller components by chemical processes. Now that we understand chemical changes, it’s time to look at actual chemical reactions, which will involve first making sure that we understand the symbols we use to represent them. As we have discussed, a chemical reaction is one in which atoms rearrange their combinations. We start out with one or more reactants, which are placed on the left. Then there is an arrow that indicates a reaction has occurred.
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And we end up with one or more products on the right. So in short, with a chemical reaction, chemical bonds break, and new ones form, which will generate entirely new substances. We must keep in mind that any chemical reaction will obey the law of conservation of mass. Atoms do not appear or disappear in a chemical reaction. Every atom present as part of the reactants must be accounted for in the products. This means that substances will react in specific ratios
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called stoichiometric ratios, and any chemical equation must be balanced so as to adequately display these ratios. This involves finding the correct coefficients for each substance involved in the reaction, and we need to find these coefficients in order to accurately describe what is occurring in the reaction. For an example of a balanced chemical equation, let’s take a look at the combustion of methane, or CH4, which is just a fancy way of saying
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that methane reacts with oxygen to produce carbon dioxide and water. Combustion will always involve oxygen as a reactant and these two products. This equation is already balanced, which means that each substance has a coefficient to its left, indicating how many molecules of that substance there are, although if the coefficient is one, it will be omitted. We can see that the coefficients
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are 1, 2, 1, and 2, because these are the number of molecules we will need of each substance in order to ensure that there are the same number of atoms of each element on each side. Counting up the atoms on the reactants side of this reaction, there is one carbon atom, four hydrogen atoms, and four oxygen atoms, since each oxygen molecule has two, and this coefficient tells us there are
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two oxygen molecules. Now looking at the products side, we have one carbon here, we have a total of four hydrogen atoms, because each water molecule has two, and there are two water molecules, and we have four oxygen atoms, two from CO2, and one from each water molecule. Since we have the same number of atoms of each element on both sides of the equation, this equation is balanced. For any unbalanced equation, these coefficients must be determined in order to perform most types of
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quantitative calculations about the reaction. We can now extend this to chemical reactions that occur in aqueous solution. Sometimes these reactions involve ionic compounds that dissociate in solution, and the ions that comprise these compounds may swap to form new compounds. Here is a balanced equation depicting aqueous calcium chloride and silver nitrate reacting with one another to form aqueous calcium nitrate and solid silver chloride. Remember that the aqueous phase
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implies that the compound is dissociated in solution, while this solid is actually water insoluble. We can list the molecular equation as shown here, which shows the ionic substances, but in actuality most of these compounds are not intact, so we also want to be able to write the net ionic equation, which shows the actual chemistry occurring in solution. Let’s split up each aqueous species into its respective ions. Calcium chloride will become a calcium ion and two
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chloride ions, two formula units of silver nitrate will become two silver ions and two nitrate ions, calcium nitrate is actually a calcium ion and two nitrate ions, and silver chloride will remain silver chloride, as it is a solid. Now we have the complete ionic equation. But we may notice that certain ions exist on both sides. These are called spectator ions, because they don’t actively participate in chemistry, they just sit in solution. These can be cancelled out just like
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algebraic variables. Let’s rewrite what is left, which is simply silver ions and chloride ions coming together to form silver chloride. We can divide everything by two to get the least common multiple of each substance, and this will give us the net ionic equation. This shows us that all that is really happening in solution is chloride ions and silver ions are combining to form solid silver chloride, in what is called a precipitation reaction. In this way,
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net ionic equations are the best way to describe the chemistry occurring between ionic solids in aqueous solution, because we can remove all of the extraneous substances that simply sit in solution and don’t do any chemistry. Now that we understand how to interpret chemical equations, we are ready to perform stoichiometric calculations. Calculations involving stoichiometry have to do with the quantitative relationship between the different
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components involved in a chemical reaction. That’s why the coefficients in a balanced chemical equation are called stoichiometric coefficients. They tell us the ratio in which molecules must react to make chemistry happen. If we look at this reaction involving nitrogen and hydrogen producing ammonia, we notice that the stoichiometric coefficients are 1, 3, and 2. These numbers can be applied in a few different contexts. They mean that one nitrogen
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molecule and three hydrogen molecules make two ammonia molecules, but this numerical ratio can be applied at any scale. From dozens of molecules, to thousands, and even moles. Most importantly for chemists, one mole of nitrogen molecules and three moles of hydrogen molecules will make two moles of ammonia molecules. This is useful because those are quantities that are visible in the macroscopic world, and we must learn how to do calculations using these ratios in order to do
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chemistry. A common calculation will involve being given information about the mass of one substance, and being asked about the mass of some other substance. Say we are given the mass of a reactant and asked what mass of product to expect. We would simply convert that mass to moles, using the molar mass of the reactant, then use the stoichiometric ratio to convert to moles of product, meaning the
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coefficients in the balanced equation, and then use the molar mass of the product to convert to mass of product. This would work the same way in the other direction, if asked how many grams of a reactant are required to produce a specific mass of product. In either case, we simply need three conversion factors to go from mass to moles, to moles of the other substance, back to grams. For example, say we are looking at this reaction, involving the combustion of propane,
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or C3H8. Say we begin with 20 grams of propane, and we want to know what mass of water to expect. As we just mentioned, we will need to do a series of three conversions. First we convert the mass of propane to the moles of propane that react, using the molar mass of propane. Then we use the stoichiometric ratio of four water molecules produced for every molecule of propane that reacts, to get the number of moles of water that are produced. Then finally,
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we use the molar mass of water to convert moles into grams, and we have our answer. This is the basic approach that we can use to perform any stoichiometric calculation we need, whether comparing reactants, products, or one of each. We simply consult the stoichiometric ratios inherent in the coefficients of the balanced equation, and then use the relevant molar masses. We should also note that when we perform chemical reactions, it is very unlikely that two substances
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will be present in precisely the right amounts in order to both react completely. Inevitably, one substance will run out first, leaving some of the other unable to react. The reactant that runs out first is called the limiting reactant, or limiting reagent, and the one that has some left over will be called the reagent in excess. Let’s illustrate this concept with a simple analogy. Say we are
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making cheese sandwiches for a family lunch, and we make them according to the following reaction, so to speak. One slice of cheese plus two slices of bread will make one cheese sandwich. So the stoichiometric coefficients of this reaction will be 1 to 2 to 1. Let’s say we look in the fridge and we see that we have 28 slices of bread and 11 slices of cheese. How many sandwiches can we make? 28 slices of bread could potentially make 14 sandwiches, but 11 slices of cheese can only
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make 11 sandwiches. So in this case, cheese will be the limiting reagent. It will get used up completely in order to make the 11 sandwiches, and some bread will be left over, making it our reagent in excess. There will be 6 slices of bread left over. Make sure to note that the limiting reagent will not always be the one present in the smaller amount. If we instead had 15 slices of cheese along with the 28 slices of bread, then all of a sudden bread would
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become the limiting reagent, since that can only produce 14 sandwiches, while 15 slices of cheese can produce 15 sandwiches. So the bread would run out first, leaving one slice of cheese left over. We must consult the stoichiometric ratio of a reaction in order to figure out which substance is the limiting reagent. Let’s now apply this to a real chemical reaction. Silicon nitride is produced according to this equation involving solid silicon and nitrogen
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gas. If we have 2 grams of silicon and 1.5 grams of nitrogen, which will be the limiting reagent? Well the grams tell us nothing, we have to convert to moles. Let’s use the molar masses of silicon and nitrogen to convert these into their respective molar quantities. So two grams of silicon divided by its molar mass gives us this many moles of silicon. Let’s do the same with our mass of nitrogen, and we end up with this many moles of nitrogen. Let’s replace our gram values
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here with the molar quantities we just calculated. Now be sure to remember what we learned with the cheese sandwiches, the limiting reagent is not automatically the substance present in the lesser amount. An extremely common error would be to look at these values and say that nitrogen is the limiting reagent because there is less of it. But this may or may not be the case. We must do some stoichiometric calculations to find out. One strategy we could use would be to take the moles
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of one substance and see how many moles of the other are required for it to react completely. For example, this many moles of silicon times this stoichiometric ratio gives us the number of moles of nitrogen that would be required for all of the silicon to react. Then the question becomes, do we have that many moles of nitrogen available? We do indeed have more than that. This means that silicon would run out first, and is therefore the limiting reagent.
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To be thorough, let’s do the same for nitrogen. Again we multiply by the stoichiometric ratio, being careful to put silicon on top and nitrogen on the bottom, and we get this many moles of silicon that would be required. As expected, we do not have enough, which means the silicon would run out before the nitrogen completely reacts, and again we conclude that it is the limiting reagent, while nitrogen is the reagent in excess. For another strategy, let’s instead see how many
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moles of product each of these reactants could potentially generate, by using the stoichiometric ratios between each substance and the product. This many moles of silicon could potentially generate 0.0237 moles of product. This many moles of nitrogen could potentially generate 0.0268 moles of product. The silicon can not produce as much, it must necssarily run out first, which means that silicon must be the limiting reactant. The moles of product we can expect from such a
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calculation is called the theoretical yield. This is the greatest amount of product that is physically possible for this reaction given the amounts of these reactants. However, in reality, we will never attain this theoretical yield. There will always be unreacted starting material, side products, or material lost due to mechanical imprecision, so we will always get some amount less than this. What we actually get is called the actual yield, and we can use the actual yield to
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calculate the percent yield, which is the the percentage of the theoretical yield that the actual yield represents. To get this, we simply divide the actual yield by the theoretical yield, and multiply by a hundred. This is a useful way to quantify the efficiency of a chemical reaction. With stoichiometry understood, let’s introduce some different types of reactions, so that we are ready to learn about them with more depth in later units.
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First let’s look at acid-base reactions. Under the Bronsted-Lowry model, these involve proton transfer. Here, we are referring to a positive hydrogen ion as a proton, because if protium loses its lone electron, all that is left is a singular proton. An acid will donate a proton, and a base will accept a proton, and this proton transfer from acid to base is an acid-base reaction. A strong acid is one that will ionize completely, such that every molecule
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of acid donates its acidic proton to some base. A weak acid is one that will ionize only partially, such that some molecules ionize and others do not. We can describe acid strength by using the acid ionization constant, or Ka. This will make more sense later after we discuss equilibria, but every acid has a different Ka value, and this gives us the ratio of products to reactants for an acid-base reaction. A large value means mostly products and a stronger
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acid. A smaller value means mostly reactant and a weaker acid. We can do the same thing for a base, reporting its strength with a Kb value. Some substances can act as either an acid or base, and these are called amphoteric. Water is such a substance. It can donate a proton to become its conjugate base, the hydroxide ion, or it can gain a proton to become its conjugate acid, the hydronium ion. We should be able to write out acid-base conjugate pairs by adding or
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removing a proton, depending on whether something is acting as an acid or base. We will talk a lot more about acids and bases later, but for now we can use this understanding to introduce the concept of titration in the context of stoichiometry. With stoichiometry we are trying to determine the concentration of a solution, which we call the analyte. This could be an acid or a base. We do this by reacting with a titrant,
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which will be the other of these two substances. We add just enough titrant to reach something called the equivalence point, where a substance called an indicator will change colors, indicating that the analyte has been precisely neutralized. A stoichiometric amount of titrant has been added so as to react with precisely the amount of analyte present. We can then use the volume of titrant added to find the moles of titrant added, which from the stoichiometric ratios gives us
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the moles of analyte present, and we can use volume of analyte to then get the concentration. Another type of reaction is the precipitation reaction. When two soluble ionic salts react in solution to produce an insoluble solid, it will crash out of solution and sit at the bottom of the flask. Finally we must discuss oxidation-reduction reactions, or redox reactions. These are reactions in which electrons are transferred from
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one substance to another. The substance that loses electrons has been oxidized, and the one that gains electrons has been reduced. It can be confusing to associate a gain of electrons with reduction, so think of it as a reduction of charge. We can also remember the mnemonic OIL RIG, if that is helpful in remembering the definitions of oxidation and reduction. The way we keep track of things with redox reactions is by assigning oxidation numbers. Every
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element in a compound will receive an oxidation number. Pure elements have an oxidation number of zero, unless they are monoatomic ions, in which case their oxidation number equals their formal charge. The oxidation numbers for a compound must add up to zero if it is neutral, or they must add up to the formal charge if it is a polyatomic ion. The most reliable way to assign these is to know that hydrogen is always +1 and oxygen is always -2, except for in certain rare
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circumstances we will not discuss. From these two numbers, we can usually get the other numbers we need. Then when looking at a reaction, the element which has had its oxidation number increase was oxidized, and the one which has had its oxidation number decrease was reduced. Here is an example, I2O5 + 5CO yields I2 + 5CO2. We can immediately place a zero above I2, since that is a pure element. For the reactants, we can put -2 above each oxygen atom. To find
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the other oxidation numbers, we must take into account how many oxygen atoms there are. With CO, oxygen is -2, so carbon must be +2 in order for the oxidation numbers to add up to zero. With I2O5, five times -2 is -10, so the iodines must add up to +10, and since there are two of them, they are each +5. Then for CO2, there are two oxygens, each of which are -2, for a total of -4, so carbon must be +4 in order for the oxidation numbers to add up to zero. We can see that carbon
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has gone from +2 to +4, so it has been oxidized. Iodine went from +5 to 0, so it has been reduced. An important subclass of redox reactions we should be able to recognize is the combustion reaction. This is when a species reacts with oxygen gas to generate water and carbon dioxide. The thing that reacts with oxygen is always getting oxidized, and spotting oxygen gas as a reactant is an easy way to recognize this type of reaction. Many of these concepts will return in
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upcoming units, but for now that concludes our Unit 4 review. I’ll see you in Unit 5.
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