Welcome to Unit 5: Chemical Reactions! Use the gallery below to read about important topics like indicators of chemical change, balancing equations, types of chemical reaction, mole conversions, stoichiometry, empirical and molecular formulas, limiting reactants, and percent yield. Each image contains a description of the concept along with a link to a video about the topic.
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Unit Five - Chemical Reactions and Stoichiometry
Indicators of Chemical Change
Product Test for Hydrogen
Product Test for Oxygen
Product Test for Carbon Dioxide
Chemical Reactions
States of Matter
Chemical Equations
Balancing Chemical Equations
Diatomic Elements
Types of Chemical Reactions
Synthesis Reactions
Decomposition Reactions
Predicting Decomposition Reactions
Decomposition Reaction Examples
Single Replacement Reactions
Activity Series for Predicting Single Replacement Reactions
Single Replacement Reaction Examples
Double Replacement Reactions
Solubility Rules
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Unit Five - Chemical Reactions and Stoichiometry
Unit five covers chemical reactions and stoichiometry. This unit introduces indicators of chemical change, product testing for gases, types of chemical reactions, stoichiometry, empirical and molecular formulas, limiting reactants, and percent yield. -
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Indicators of Chemical Change
The only true way to know that a chemical reaction has taken place is to test the composition of a substance. There are, however, some things to look out for to indicate that a reaction has taken place. Click any image for details and video links. The indicators that a chemical reaction has taken place are color change, temperature change, emitting light, formation of a gas, and formation of a precipitate. A precipitate is a solid that forms during a chemical reaction. -
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Product Test for Hydrogen
There are some simple tests that chemists run to see if certain chemicals are being produced during a chemical reaction. When testing for hydrogen, scientists can put a burning splint into a test tube. If hydrogen is being produced, a popping sound will be observed. The sound is quick and generally only happens one time, but is a distinct popping sound. -
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Product Test for Oxygen
There are some simple tests that chemists run to see if certain chemicals are being produced during a chemical reaction. When testing for oxygen, a scientist can put a glowing splint (meaning that the flame has been put out but the splint is still hot enough to glow) into a test tube. If oxygen is being produced by the reaction in the test tube, the splint will reignite. -
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Product Test for Carbon Dioxide
There are some simple tests that chemists run to see if certain chemicals are being produced during a chemical reaction. When carbon dioxide is bubbled through lime water (calcium hydroxide), the clear solution will turn cloudy. -
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Chemical Reactions
Chemical reactions involve the reorganization of atoms. Bonds are broken and new bonds are formed. The top picture shows the molecular view of what is happening in one particular chemical reaction. Two hydrogen molecules, each consisting of two hydrogen atoms covalently bonded together, are reacting with an oxygen molecule, consisting of two oxygen atoms covalently bonded together. The bonds within these molecules break and rearrange. The result will be two water molecules, each with an oxygen atom surrounded by two hydrogen atoms. New bonds had to be formed in order to produce these new molecules. This information is normally presented with a chemical equation. This chemical equation can be read as ātwo hydrogen molecules react with an oxygen molecule to produce two water molecules.ā Another way to read this would be āHydrogen plus oxygen yields waterā where the term āyieldsā indicates that a chemical reaction has taken place. In a chemical equation, the substances on the left side of the arrow are the reactants. These are the substances present before the chemical reaction takes place. The substances on the right side of the arrow are the products. Products are the substances that have been produced by the rearrangement of the atoms. -
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States of Matter
States of matter are often listed for the substances in a chemical equation. Solid is indicated by the symbol (s), liquid is indicated by (l), and gas is indicated by (g). Water is often used as a solvent in chemistry. Whenever a substance is dissolved in water, (aq) is used, meaning that the substance is aqueous. -
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Chemical Equations
Chemical equations are the simplest way to represent chemical reactions. The reactants and products in a chemical equation are separated by an arrow. The number of each type of atom on the reactants side of the equation must match the number of atoms on the products side. This means that the reaction must be balanced. Subscripts are part of each of the compounds represented in the equation. Subscripts cannot be changed in order to balance an equation. For example, you could not increase the number of oxygens on the products side by adding a 2 to the H2O. This would change the molecule from water to hydrogen peroxide. Coefficients are added to balance equations. Coefficients apply to the entire molecule that they are in front of. For example, the 2 in front of water implies that there are 2 water molecules, each with 2 hydrogens and 1 oxygen. -
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Balancing Chemical Equations
The Law of Conservation of Mass says that mass can be neither created nor destroyed. This means that chemical equations must be balanced. The number of each type of atom on the reactants side of the equation must match the number of that atom on the products side. Coefficients are added to equations to balance them. Subscripts cannot be changed to balance an equation. Coefficients apply to the entire molecule that they are in front of. By adding coefficients to the top equation, each side of the reaction has 2 nitrogens and 6 hydrogens. Take note of the hydrogens. Before the coefficients were added, there were 2 hydrogens on the reactants side of the equation and 3 on the products side. This is a common situation where a common multiple between 2 and 3 should be considered. By adding a 3 has a coefficient for the H2 and a 2 as a coefficient for the NH3, each side of the equation has 6 hydrogens. By adding coefficients to the top equation, each side of the reaction has 3 carbon atoms, 8 hydrogen atoms, and 10 oxygen atoms. Take not of the oxygens. Oxygen appears in 2 compounds on the products side of the reaction. A good strategy for approaching this is to balance oxygen last. Balance the reaction for carbon, then hydrogen, and then oxygen. Whenever an atom appears in multiple places on the same side of the reaction, it is helpful to consider it last. -
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Diatomic Elements
There are 7 elements on the periodic table that are not found alone in nature. These atoms are more stable bonded to another atom of themselves, then they are alone. They are referred to as diatomic elements. Whenever a diatomic element is used in a chemical reaction, it should receive a subscript of a ā2.ā The diatomic elements are hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine, and iodine. One way to remember the diatomic elements is that they make the shape of an apostrophe 7 on the periodic table, with hydrogen as the apostrophe. The ā7ā begins with element number 7 (nitrogen) and there are 7 of the elements in all. Another way to remember the diatomic elements is with an acronym: I (iodine) Brought (bromine) Clothes (chlorine) From (fluorine) Old (oxygen) Navy (nitrogen) Home (hydrogen). -
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Types of Chemical Reactions
There are many types of chemical reactions, but we will look at some of the more common types. The five types covered in this unit are synthesis reactions, decomposition reactions, single replacement reactions, double replacement reactions, and combustion reactions. Note that many textbooks refer to double replacement reactions as metathesis reactions. -
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Synthesis Reactions
Synthesis reactions occur when two or more reactants combine to form one product. The general formula is A + B yields AB. It is common the be given the reactants in a reaction and be asked to predict the products. Whenever new products are formed, the cation should be written first, followed by the anion. The charges of the ions should be crossed down to get the subscripts of the new compound. For example, when magnesium and oxygen combine to form magnesium oxide, consider that magnesium typically has a charge of +2 and oxygen typically has a charge of -2. The positively charged ion should be written first. The charges should be crossed down to give Mg2O2, and then reduced to the lowest form to give MgO. Notice that oxygen had a 2 for a subscript on the left side of the equation (because it is a diatomic element). This 2 did not transfer to the products side. Never bring subscripts from one side of the equation to the other. The subscripts come from crossing down charges. The reaction can then be balanced using coefficients. Another example is aluminum and sulfur combining to form aluminum sulfide. Aluminum has a charge of +3 and sulfur is -2. These charges are crossed down to form Al2S3. The reaction is then balanced by adding coefficients in front of the aluminum and sulfur on the reactants side of the equation. -
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Decomposition Reactions
Decomposition reactions take place when one product breaks down to form 2 or more products. The general formula for decomposition reactions is AB yields A + B. The next picture shows some patterns that occur in decomposition reactions. -
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Predicting Decomposition Reactions
Decomposition reactions occur when one reactant breaks down to form multiple products. There are some patterns that help predict how this will occur. This image shows some of the ways that decomposition reactions break down. In these guidelines, āMā stands for metal and āNMā stands for nonmetal. In the case of metal carbonates, the reactant breaks down to form a metal oxide and carbon dioxide. The charges on the metal oxide would need to be crossed down to find subscripts. If the charges are crossed down correctly, this reaction should be balanced any coefficients. For metal hydrogen carbonates (often called metal bicarbonates) the products are a metal oxide, water, and carbon dioxide. The charges on the metal oxide would need to be crossed down to find subscripts. Metal hydroxides decompose to produce a metal oxide plus water. The charges on the metal oxide would need to be crossed down to find the subscripts. Metal chlorates decompose to form a metal chloride and oxygen. The charges on the metal chloride would need to be crossed down to find the subscrits. The oxygen has a 2 for a subscript because it is a diatomic element. Acids that contain oxygen (often called oxyacids) decompose to form a nonmetal oxide and water. In these guidelines, āMā stands for metal and āNMā stands for nonmetal. -
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Decomposition Reaction Examples
Decomposition reactions occur when one reactant breaks down to form multiple products. Use the patterns in the previous picture to help predict the products of decomposition reactions. The first example is the decomposition of potassium carbonate. Potassium is a metal. Metal carbonates decompose to form metal oxides plus carbon dioxide. The charges should be crossed down to determine the subscripts for the metal oxide. Potassium has a charge of 1+ while oxygen has a charge of 2-, making potassium oxide K2O. Always check to make sure the reaction is balanced. If you have crossed charges correctly, the decomposition of carbonates should always be balanced without any additional coefficients. The second example is potassium hydroxide. Potassium is a metal. Metal hydroxides decompose to form metal oxides plus water. The charges should be crossed down to determine the subscripts for the metal oxide. Potassium has a charge of 1+ while oxygen has a charge of 2-, making potassium oxide K2O. Always check to make sure the reaction is balanced. In this case, a 2 was needed in front of potassium hydroxide to balance the equation. The third example is the decomposition of sodium chlorate. Sodium is a meta. Metal chlorates decompose to form metal chlorides plus oxygen. The charges should be crossed down to determine the subscripts for the metal chloride. Sodium has a charge of 1+ while chlorine has a charge of 1-, making sodium chloride NaCl. Oxygen is a diatomic element causing it to have a subscript of a 2 when it is by itself. Always check to make sure the reaction is balanced. In this case, a 2 is needed in front of sodium chlorate, a 2 in front of sodium chloride, and a 3 in front of oxygen to balance the equation. -
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Single Replacement Reactions
Single replacement reactions take place when a single element in a compound gets replaced. There are two general formulas for single replacement reactions. Choose which formula to us based on if the element being replaced is a metal or a nonmetal. When the element being replaced in the reaction is a metal, the general formula is A + BC yields AC + D. When the element being replaced in the reaction is a nonmetal, the general formula is BC + D yields BD + C. Keep in mind that for compounds, metals are always written first, with nonmetals second. In these general formulas A and B are metals, while C and D are nonmetals. Single replacement reactions can be recognized by the fact that they begin with an element by itself plus a compound. -
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Activity Series for Predicting Single Replacement Reactions
Single replacement reactions can take place when an atom replaces an ion in a compound. A reaction does not always take place just because two elements come in contact with one another. We have a tool to help determine when a single replacement reaction will take place. The tool is the activity series. There is an activity series for metals and a separate activity series for halogens. To use the series, find the element that is trying to replace the ion in the compound. In your single replacement reaction, this would be the atom that starts off by itself. If that element is higher in the series than the element it is trying to replace, the reaction will take place. Elements higher in the series are more reactive and can replace elements lower in the series. The next image has examples where the activity series has been used to determine if single replacement reactions will take place. This activity series comes from the North Caroline Chemistry Reference Tables. -
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Single Replacement Reaction Examples
The activity series (shown in the previous image) can be used to determine if a single replacement reaction will take place. When using the activity series, look at the reactants side of the single replacement reaction. Find the element that is by itself in the activity series. Letās call this the replacing element. If the replacing element is a metal, then it is trying to replace the metal in the compound. Like replaces like. If the replacing element is higher in the activity series then the element that it is trying to replace, the reaction will take place. If the replacing element is lower, then the reaction will not take place. In the first example, lithium (the element by itself on the reactants side) is a metal. Lithium will try to replace the metal in the compound ā sodium in this case. Lithium is higher in the activity series than sodium is. This reaction will take place. On the products side of the reaction, lithium is now paired with chlorine in the compound and sodium is by itself. When predicting products, charges should always be considered to help determine the subscripts of any compounds that are formed. In this case, lithium has a charge of +1, chlorine is -1, no subscripts are needed for lithium chloride. It is also important to ensure that equations are balanced. In the second example, fluorine (the element by itself on the reactants side) is a nonmetal. Fluorine will try to replace the nonmetal in the compound ā iodine in this case. Fluorine is higher in the activity series than iodine is. This reaction will take place. On the products side of the reaction, fluorine is now paired with magnesium in the compound and iodine is by itself. When predicting products, charges should always be considered to help determine the subscripts of any compounds that are formed. In this case, magnesium has a charge of +2, fluorine is -1, these charges should be crossed down to give MgF2. It is also important to ensure that equations are balanced. In the third example, lead (the element by itself on the reactants side) is a metal. Lead is trying to replace the metal in the compound ā calcium in this case. Lead is lower in the activity series than calcium. This reaction will not take place. Lead cannot replace calcium in a compound. In the fourth example, iodine (the element by itself on the reactants side) is a nonmetal. Iodine is trying to replace the nonmetal in the compound ā chlorine in this case. Iodine is lower in the activity series than chlorine. This reaction will not take place. Iodine cannot replace chlorine because it is lower in the activity series. -
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Double Replacement Reactions
Double replacement reactions are often called metathesis reactions. Double replacement reactions take place when two elements are replaced. The ions in each compound are swapped. The general formula for a double replacement reaction is AB + CD yields AD plus CB. When predicting the products of double replacement reactions, keep in mind that the cation is always first in a compound, followed by the anion. Charges should always be considered when determining subscripts on a compound. Reactions should always be balanced. In the first example magnesium oxide (AB) reacts with potassium chloride (CD). Think of it as the positive ions swapping places. The reaction yields magnesium chloride (AD) plus potassium oxide (CB). In the forming of the products, charges were crossed down to get subscripts. In the second example sodium bromide (AB) reacts with calcium sulfide (CD). The reaction yields sodium sulfide (AD) plus calcium bromide (CB). In the forming of the products, charges were crossed down to get subscripts. In the third example potassium hydroxide (AB) reacts with sodium chloride (CD). The reaction yields potassium chloride (AD) plus sodium hydroxide(CB). Notice that the polyatomic ion āhydroxideā was present. Hydroxide stays together on both sides of the reaction. It is the āBā in the general formula. In the forming of the products, charges were crossed down to get subscripts. -
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Solubility Rules
A precipitate is a solid that forms during a chemical reaction. Very often, in double replacement reactions, one of the products forms a precipitate. We have a tool to determine when a precipitate will form. These solubility rules are derived from the North Carolina Chemistry Reference Tables. The solubility rules are based off of the anion in the compound you are studying. Find the anion in the list to determine if it is in the soluble or the insoluble column. Then look at the cation to see if it is one of the exceptions listed for your anion. Consider sodium chloride for example. The chlorides are listed under the soluble column. The exceptions to the rule about the chlorides are silver, lead, and mercury(I). Sodium was not listed as one of the exceptions. Sodium chloride is soluble. This means it will dissolve in water and should have (aq) for āaqueousā written beside it in a chemical reaction. Consider barium sulfate. The sulfates are listed under the soluble column. The exceptions to the rule about the sulfates are calcium, strontium, barium, mercury, lead(II), and silver. Barium is one of the exceptions causing barium sulfate to be insoluble. This means that it will not dissolve in water. A precipitate will form. Barium sulfate would have (s) for āsolidā written beside it in a chemical reaction. -
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Precipitation Reactions
Precipitation reactions occur when a solid (or precipitate) is produced during a chemical reaction. Solubility rules are provided on the previous slide to help determine if a substance will be soluble or insoluble in water. Soluble substances will dissolve in water and are given āaqā as a state of matter. This stands for āaqueous.ā Insoluble substances will not dissolve in water and are given āsā as a state of matter. This stands for āsolid.ā In the first example, lead(II) nitrate and potassium iodide are combined. Both of these are aqueous substances. Solid lead(II) iodide and aqueous potassium nitrate are formed during the reaction. The solubility rules would be used to determine which substances are aqueous versus solid. The reaction follows the pattern for a double replacement (or metathesis reaction). The general formula for this type of reaction is AB + CD ļ AD + CB. In a lab environment, you would be combining two liquid solutions. You would see a cloudy substance start to form and settle out of the solution. Over time it will become more obvious that this cloudy substance is a solid. In the second example, barium hydroxide and potassium sulfate are combined. Both of these are aqueous substances. Solid barium sulfate and aqueous potassium hydroxide are formed during the reaction. These examples are both considered precipitation reactions because they produce a solid product. -
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Types of Equations (Molecular, Complete Ionic, Net Ionic)
The type of equation that most students are used to seeing is a molecular equation. It shows all of the compounds present while a chemical reaction is taking place. It also usually shows the states of matter for the compounds. In this example, the molecular equation is aqueous barium hydroxide plus aqueous potassium sulfate yields solid barium sulfate plus aqueous potassium hydroxide. What may be new information for some of us is that some of the elements listed in a molecular equation do not really participate in the chemical reaction. They are present, but remain unchanged during the reaction. This is easier to understand if you consider what really happens when you drop those aqueous solutions into your beaker. The fact that they are aqueous means that they dissolve. For an ionic compound, dissolving means breaking apart into the ions that form them. The aqueous barium hydroxide breaks apart into barium ions and hydroxide ions. These ions separate from each other and can move freely in the solution. A complete ionic equation shows all of the ions that are present. The aqueous ionic substances in the molecular equation are shown as separate particles with a charge. The charged particles are all considered aqueous. Notice that the solid remains together in the complete ionic equation. Some of the ions appear exactly the same on both the reactants and the products side of the complete ionic equation. This means that they were unchanged during the chemical reaction. We call these substances spectator ions. They do not participate in the chemistry taking place, but rather watch it take place like a spectator. The final type of equation is a net ionic equation. This only shows the chemistry taking place. It only shows the particles that were affected by the chemical reaction. Crossing out the spectator ions in the complete ionic equation (or crossing out the ions that look exactly the same on both sides of the reaction) gives the net ionic equation. In this case, the net ionic equation shows that the barium ion reacts with the sulfate ion to produce solid barium sulfate. -
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Combustion Reactions
Combustion reactions take place when a hydrocarbon burns in the presence of oxygen to produce water plus carbon dioxide. A hydrocarbon is a molecule made up of hydrogen and carbon. The general formula for a hydrocarbon is CnH2n+2. This means that if you know the number of carbons in a hydrocarbon, you can predict the number of hydrogens. In the first example, the hydrocarbon had one carbon atom. Plugging 1 in for ānā in the hydrocarbon general formula tells us that there will be 4 carbon atoms. In combustion reactions, the hydrocarbon is always reacted with oxygen (which is a diatomic). Water and carbon dioxide are always the products. This reaction is balanced by adding a 2 for the coefficient for both oxygen and water. When balancing combustion reactions, it is best to balance the carbon first, then the hydrogen, and finally the oxygen. The next example will demonstrate a problem that is often encountered balancing these reactions and how to solve it. In the last example, there are 6 carbon atoms. Plugging the 6 in for ānā in ā2n+2ā tells us that there are 14 hydrogens in the hydrocarbon. The rest of the reaction looks just like every other combustion reaction. The hydrocarbon reacts with oxygen and produces water plus carbon dioxide. Balancing this reaction proves to be a little tricky. Starting with the carbon, you would put a 6 in front of carbon dioxide. Then you would balance the hydrogens by placing a 7 in front of water. The problem happens with the oxygen. At this point there would be 2 oxygens on the reactants side and 19 on the products side. An odd number on one side and an even number on the other side indicates an issue. This can be fixed by multiplying all of the coefficients by 2. The beauty of these reactions is that the products are always the same. The reaction just needs to be balanced. -
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The Mole and Avogadro's Number
Counting atoms can be tricky because there are SO many of them ā even in small samples. The concept of the mole makes this easier. It provides a bridge between the microscopic world and the laboratory. A mole is simple a number ā it just happens to be a really large number: 6.022x1023. This is also called Avogadroās number. This really large number becomes handy when we are trying to count something really tiny ā like atoms. We count in units of moles rather than in units of atoms. Consider an aluminum can with a mass of 14 grams. There are roughly 310,000,000,000,000,000,000,000 atoms in this aluminum can. Dealing with this many zeros leaves too much room for error. A chemist would divide that large number of atoms by Avogadroās number and report it as 0.52 moles of atoms. Since the mole is just a number, it can be used to count anything that is tiny. It can be used to count atoms, molecules, particles, ions, or formula units. -
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Converting from Atoms to Moles
Here is an example where we will convert from atoms to moles. The question is: How many moles are in 1.2x1021 atoms? Always start with the number and unit you are given in the question. In this case, start with 1.2x1021 atoms. Set up a conversion factor that will convert from atoms to moles. Conversion factors can be flipped upside down when making conversions. Youāll notice in the next slide that the conversion factor used here is flipped. To determine how to list the conversion factor, always put the unit you start with on the bottom of the next fraction. This causes that unit to cancel out. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. Be cautious when dividing by a number in scientific notation. It is important to either put the number in parenthesis, or to use the āEEā function on the calculator. To enter this into the calculator using the EE function, I would press: 1.2 EE 21 divided by 6.022 EE 23. Many calculators have the āEEā button on the same button as the comma ā meaning you need to press ā2ndā then āEE.ā -
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Converting from Moles to Atoms
Here is an example where we will convert from moles to atoms. The question is: How many atoms are in 3.9 moles? Always start with the number and unit you are given in the question. In this case, start with 3.9 moles. Set up a conversion factor that will convert from moles to atoms. Conversion factors can be flipped upside down when making conversions. Youāll notice in the previous slide that the conversion factor used here is flipped. To determine how to list the conversion factor, always put the unit you start with on the bottom of the next fraction. This causes that unit to cancel out. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. -
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Converting Between Moles and Grams
Moles serve as a bridge between the microscopic world of atoms and the laboratory world. In order for this to be useful, we need to be able to convert to something we can measure on a scale ā like grams. The average atomic mass on the periodic table is a conversion factor between grams and moles. Think of this number as having units of āgrams per mole.ā This number is often called the molar mass. Nitrogen has an average atomic mass of 14.01, based on the periodic table. This means that there are 14.01 grams per every one mole of nitrogen. This is a conversion factor that can be flipped. It can be written as 14.01 grams per 1 mole, or as 1 mole per every 14.01 grams. -
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Converting From moles to Grams
Here is an example where the molar mass from the periodic table is being used as a conversion factor from moles to grams. The question is: How many grams of lead are in 1.42 moles of lead? Always start with the number and unit you are given in the question. In this case, start with 1.42 moles of lead. Set up a conversion factor that will convert from moles to grams. This conversion factor is the molar mass from the periodic table. This number has units of grams per mole. Conversion factors can be flipped upside down when making conversions. Youāll notice in the next slide that the conversion factor used here is flipped. To determine how to list the conversion factor, always put the unit you start with on the bottom of the next fraction. This causes that unit to cancel out. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. -
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Converting From Grams to Moles
Here is another example where the molar mass from the periodic table is being used as a conversion factor, but this time we are converting from grams to moles. The question is: How many moles of zinc are in 5.78 grams of zinc? Always start with the number and unit you are given in the question. In this case, start with 5.78 grams of zinc. Set up a conversion factor that will convert from grams to moles. This conversion factor is the molar mass from the periodic table. This number has units of grams per mole. Conversion factors can be flipped upside down when making conversions. Youāll notice in the previous slide that the conversion factor used here is flipped. To determine how to list the conversion factor, always put the unit you start with on the bottom of the next fraction. This causes that unit to cancel out. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. -
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Converting From Grams to Atoms
This is an example of how to convert from grams to atoms. This is a two step process. We have one conversion factor that allows us to convert from grams to moles (the molar mass off of the periodic table), and then another conversion factor that allows us to convert from moles to atoms (Avogadroās number). Here is the question: How many atoms are in an aluminum can with a mass of 14.1 grams? Always start with the number and unit you are given in the question. In this case, start with 14.1 grams of aluminum. Set up a conversion factor that will convert from grams to moles. This conversion factor is the molar mass from the periodic table. This number has units of grams per mole. To determine how to list the conversion factor, always put the unit you start with on the bottom of the next fraction. This causes that unit to cancel out. To set up the second conversion factor, pay attention to the units on the first conversion factor. Whatever unit is on the top of your last conversion factor should be the bottom unit of the next conversion factor. In this case, that would be moles of aluminum. Use Avogadroās number to convert from moles of aluminum to atoms of aluminum. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. -
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Converting From Atoms to Grams
This is an example of how to convert from atoms to grams. This is a two step process. We have one conversion factor that allows us to convert from moles to atoms (Avogadroās number), and then another conversion factor that allows us to convert from grams to moles (the molar mass off of the periodic table). Here is the question: How many grams are in a sample of copper containing 1.25x1024 atoms? Always start with the number and unit you are given in the question. In this case, start with 1.25x1024 atoms of copper. Set up a conversion factor that will convert from moles to grams. This will be Avogadroās number. To determine how to list the conversion factor, always put the unit you start with on the bottom of the next fraction. This causes that unit to cancel out. To set up the second conversion factor, pay attention to the units on the first conversion factor. Whatever unit is on the top of your last conversion factor should be the bottom unit of the next conversion factor. In this case, that would be moles of copper. Use the molar mass of copper from the periodic table to convert from moles to grams. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. -
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Molar Mass
Molar mass measures the number of grams found in one mole of an element. The units are grams/mole. For a compound, the molar mass is calculated by adding up the average atomic mass of all the elements in the compound. In the case of nitrogen dioxide, there is one nitrogen with a mass of 14.01, and two oxygens each with a mass of 16.00. Adding these gives a molar mass of 46.01 grams per mole of nitrogen dioxide. This is a conversion factor that can be used to convert between grams and moles. As a conversion factor it can be flipped depending on the desired units of the question. In this example the molar mass can be used as 46.01 grams per 1 mole, or as 1 mole per 46.01 grams. The next pictures will give more examples of calculating molar mass and demonstrate how it can be used as a conversion factor. -
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Molar Mass Practice
Molar mass is the number of grams in one mole of an element. It is calculated by adding up the average atomic masses off of the periodic table for all of the atoms in a compound. In the example of Cl2, there are 2 chlorines each with a mass of 35.45 grams/mole, giving a molar mass of 70.90 grams per mole. In the example of calcium nitride, there are 3 calcium atoms each with a mass of 40.08 grams/mole. Calcium contributes 120.2 gams to a mole of the compound. Calcium nitride has 2 nitrogens, each with a mass of 14.01 grams/mole. Nitrogen contributes 28.02 grams to one mole of the compound. Combining the elements gives a total molar mass of 122.2 grams per mole for calcium nitride. In the example of calcium phosphate, the two outside of the parenthesis is distributed to the atoms inside the parenthesis. This means that calcium phosphate has 3 calcium atoms, 2 phosphorous atoms, and 8 oxygen atoms. Summing up the average atomic masses for these atoms gives a total molar mass of 310.1 grams per mole for calcium phosphate. -
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Molar Mass as a Conversion Factor
Molar mass is the number of grams in one mole of an element. It is calculated by adding up the average atomic masses off of the periodic table for all of the atoms in a compound. Molar mass is a conversion factor that allows you to convert between grams and moles of a substance. This example asks you to determine how many grams of calcium chloride are in 0.325 moles of calcium chloride. Begin with the number and the unit given in the question (0.325 moles). Then a conversion factor is needed to switch from moles to grams. This conversion factor is the molar mass. The molar mass of calcium chloride is calculated by adding the average atomic mass values for 1 calcium atom and 2 chlorine atoms. These numbers are obtained from the periodic table. The molar mass of calcium chloride is 110.98 grams per mole. Molar mass can be written as grams per mole, or as moles per gram depending on the question. In this case, the question begins with units of moles. In order to make moles cancel, moles should be on the bottom of the conversion factor. The molar mass is entered as 110.98 grams divided by 1 mole. This equation is solved by multiplying by any numbers on the top and dividing by any numbers on the bottom. In this case, 0.325*110.98/1 gives a final answer of 36.1 grams. Keep an eye on the units. Notice that moles has cancelled, leaving the answer in the desired units of grams. -
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Empirical Versus Molecular Formulas
There are two types of formulas used to represent chemical compounds. Empirical formulas are in their lowest possible form. The subscript numbers cannot be reduced any further. These are commonly used for ionic compounds where the subscripts represent the ratio of atoms present rather than the actual number of atoms present. Molecular formulas give the actual number of atoms present in a molecule. They do not necessarily have to be in the lowest possible form. N2O4 is a molecular formula. The subscripts (2 and 4) are both divisible by 2. Reducing these to the lowest possible form gives NO2, which is an empirical formula. C6H14 is a molecular formula. The subscripts (6 and 14) are both divisible by 2. Reducing these to the lowest possible form gives C3H7, which is an empirical formula. C6H12O6 is a molecular formula. The subscripts (6, 12, and 6) are all divisible by 6. Reducing these to the lowest possible form gives CH2O, which is an empirical formula. -
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Calculating Empirical Formulas
Here is an example where we will calculate an empirical formula using three steps. The question is: A compound contains 38.67% carbon, 16.22% hydrogen, and 45.11% nitrogen. Find the empirical formula. The first step for calculating an empirical formula is to write the percentages in units of grams. You can do this because you are assuming a 100 gram sample. It is not necessary to move the decimal. Simply write whatever the percentage is in units of grams. In this example, there are 38.67 grams of carbon, 16.22 grams of hydrogen, and 45.11 grams of nitrogen. The second step is to convert those gram values to moles. This means dividing by the molar mass off of the periodic table for each of the elements involved. In this example, there were 3.220 moles of carbon, 16.09 moles of hydrogen, and 3.220 moles of nitrogen. The third step is to divide all of the mole values by the smallest mole value. In this example, 3.220 was the smallest mole value. Each of the mole vales were divided by 3.220. These numbers become the subscripts in your equation. If they are whole numbers (like they are in this example), the work is done. Use them as the subscripts. Our empirical formula is CH5N. If they are not whole numbers, they will need to be ratioed up to make them whole numbers. For example if one of them were 0.5 (same as Ā½), I could multiply all of them by 2 (the denominator in that fraction) to make them whole numbers. Be sure to multiply all elements by the same amount. -
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Calculating Molecular Formulas
A molecular formula is some multiple of an empirical formula (which is always in the lowest possible form). Two pieces of information are needed to calculate a molecular formula: the empirical formula and the molar mass of the molecular formula. Here is the question: A compound has a molar mass of 220.1 g/mole and an empirical formula of CO2. What is the molecular formula? We will be taking the molar mass of the molecular formula and dividing it by the molar mass of the empirical formula. This will give us a whole number that tells us what factor the empirical formula needs to be ratioed up by. The mass given in the question is for the molecular formula. The question also gives us the empirical formula. We can use the molar masses off of the periodic table to determine the molar mass of the empirical formula. In this case dividing the molar mass of the molecular formula by the molar mass of the empirical formula gave us a value of 5. The empirical formula was CO2. This needs to be multiplied by a factor of 5 giving C5O10 as the molecular formula. This was a case where the empirical formula was provided. In other examples (like on the next slide), it is necessary to calculate the empirical formula and then calculate the molecular formula. -
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Calculating Empirical and Molecular Formulas
In this example, we have been asked to calculate the molecular formula but we are not given the empirical formula. We need to follow the steps for calculating an empirical formula, and then use it to calculate a molecular formula. Here is the question: Nicotine contains 74.0% carbon, 8.70% hydrogen, and 17.3% nitrogen. It has a molar mass of 162.2 g/mole. What is the molecular formula of nicotine? Two pieces of information are needed to calculate a molecular formula: the empirical formula and the molar mass of the molecular formula. The mass given in the question is for the molecular formula. In order to find the empirical formula, follow these three steps: Write the percentages as grams (assuming a 100 gram sample). Convert to moles. Divide all of the mole values by the smallest mole value. The first step for calculating an empirical formula is to write the percentages in units of grams. You can do this because you are assuming a 100 gram sample. It is not necessary to move the decimal. Simply write whatever the percentage is in units of grams. In this example, there are 74.0 grams of carbon, 8.70 grams of hydrogen, and 17.3 grams of nitrogen. The second step is to convert those gram values to moles. This means dividing by the molar mass off of the periodic table for each of the elements involved. In this example, there were 6.16 moles of carbon, 8.63 moles of hydrogen, and 1.23 moles of nitrogen. The third step is to divide all of the mole values by the smallest mole value. In this example, 1.23 was the smallest mole value. Each of the mole vales were divided by 1.23. These numbers become the subscripts in your equation. The empirical formula is C5H7N. Find the molar mass of the empirical formula by adding up the masses off of the periodic table for 5 carbons, 7 hydrogens, and 1 nitrogen. We are now ready to find the molecular formula. In this case dividing the molar mass of the molecular formula by the molar mass of the empirical formula gave us a value of 2. The empirical formula was C5H7N. This needs to be multiplied by a factor of 2 giving C10H14N2 as the molecular formula. -
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Mole Ratios
Mole ratios are used to convert between substances within a balanced chemical equation. The beauty of this is that it allows us to switch substances. We can start with information about one of the substances in the reaction, and calculate information about another. Mole ratios are needed for stoichiometry, which is converting between substances within a balanced chemical equation. The mole ratio comes from the coefficients. In the example shown, the mole ratio for oxygen to water is 5:4. This is a conversion factor that can be flipped to meet our needs. It can be written as 5 moles of oxygen for every 4 moles of water, or as 4 moles of water for every 5 moles of oxygen. -
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Stoichiometry - Converting From Moles to Moles with a Mole Ratio
This is an example where mole ratios are used to convert between substances within a balanced chemical equation (also known as stoichiometry). The question is: Using the chemical equation shown, how many moles of water can be produced when 2.05 moles of oxygen are consumed? Always start with the number and the unit you are given in the question ā in this case 2.05 moles of oxygen. Focus on the units. Whatever unit you start with should go on the bottom of the next fraction. We are going to use a mole ratio to switch from moles of oxygen to moles of water. Oxygen needs to be on the bottom so that it will cancel out. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. Notice that moles of oxygen will cancel out leaving us with 1.64 moles of water. -
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Stoichiometry - Converting From Grams to Grams with a Mole Ratio
This is an example where mole ratios are used to convert between substances within a balanced chemical equation (also known as stoichiometry). The question is: Using the chemical equation shown, how many grams of CO2 can be produced when 8.45 grams of C3H8 are consumed? We are asked to convert from grams of one substance to grams of another substance. In order to switch substances, we must be in units of moles. This is a 3 step calculation. We will first convert from grams of C3H8 to moles of C3H8 using the molar mass off of the periodic table. Then we will convert from moles of C3H8 to moles of CO2 using a mole ratio. At that point, we have the correct substance but not the correct unit. A final step is needed to convert from moles of CO2 to grams of CO2 using the molar mass off of the periodic table. Follow the units through the question. Notice that we start with the number and unit given in the question. Whenever setting up a conversion factor, the unit on the bottom matches the unit on the top of the previous conversion factor. This causes it to cancel out. Also notice, that every time we convert from grams to moles, the substance stays the same. It is only when we convert from moles to moles that we can switch substances. A mole ratio is required to switch substances. The numbers in the mole ratio are the coefficients from the balanced equation. Once the conversion factors are set up, multiply by any numbers on the top and divide by any numbers on the bottom. -
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Stoichiometry - Converting From Grams to Grams - Color Coded
This is the same example from the previous slide, but has been color coded by substance. Notice that the only conversion factor that has two different substances present is the mole ratio. Gram to mole conversion factors cannot be used to switch substances. You must be in moles to convert from one substance to another. The mole ratio uses the coefficients from the balanced chemical equation. -
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Limiting and Excess Reactants
In a chemical reaction, typically one of the reactants runs out before the other reactant does. This is called the limiting reactant. It limits the amount of products that can be produced. If you begin this reaction with 4 hydrogen molecules and 4 oxygen molecules, hydrogen runs out and limits the number of water molecules that can be produced. Hydrogen is the limiting reactant. Notice that there are 2 unreacted oxygen molecules remaining after the reaction. Oxygen is the excess reactant. -
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Solving a Limiting Reactant Problem
Limiting reactant problems can be recognized by the fact that information is given for multiple reactants. Consider this example: How many grams of ammonia can be produced when 10.0 grams of nitrogen and 10.0 grams of hydrogen react? We have been given information about both nitrogen and hydrogen. This is our clue that this is a limiting reactant question. We are going to go through some extra steps to determine which reactant is limiting, then we use the information about the limiting reactant to answer the question. The first step in determining the limiting reactant is to convert to moles. In this case, we were given 10.0 grams of both nitrogen and hydrogen. Both of these have been converted to moles by dividing by the molar mass from the periodic table. We have 0.357 moles of nitrogen and 4.96 moles of hydrogen. The second step is to divide the mole value by the coefficient in the balanced chemical equation. There were 0.357 moles of nitrogen, divided by the coefficient of 1. There were 4.96 moles of hydrogen, divided by a coefficient of 3. This gave the vales of 0.357 for nitrogen and 1.65 for hydrogen. These numbers are somewhat meaningless in terms of units, however they are helpful in the limiting reactant process. Whichever number is the smallest is the limiting reactant. In this case, 0.357 is smaller than 1.655, making nitrogen the limiting reactant. This makes hydrogen the excess reactant. These numbers are not used for anything else, other than determining the limiting and excess reactants. Once we have determined the limiting reactant, we use the information about the limiting reactant to answer the original question. In the third step, we start with the original grams of the limiting reactant, and converted it to the grams of desired product. In this case, we started with 10.0 grams of nitrogen, converted it to moles of nitrogen using the molar mass from the periodic table, converted to moles of ammonia using coefficients from the balanced chemical equation, and then converted to grams of ammonia using the molar mass from the periodic table. The final answer is 12.2 grams of ammonia. -
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Percent Yield
Percent yield is a calculation that would likely be used after carrying out a lab. It compares what you expected to produce to what you actually produced. The formula for percent yield is actual yield divided by theoretical yield, multiplied by 100. The actual yield is what was actually measured in the lab. The theoretical yield is what stoichiometry predicted that you would make in the lab. The theoretical yield is sometimes called the expected yield. In this example, the theoretical yield is 2.74 grams. The actual yield is 2.57 grams. This produced a percent yield of 93.8%. Notice that the units are the same for the actual yield and the theoretical yield. These units cancel out. -
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Percent Yield Example
This is a percent yield example where we will compare what we expected to produce in a lab, to what we actually produced in the lab. Here is the question: 4.39 grams of methane were burned in the presence of excess oxygen producing 9.02 grams of water. What is the percent yield of the reaction? The actual yield is 9.02 grams of water. The question does not give the theoretical yield. We need to calculate it using stoichiometry. Keep in mind that the actual yield and the theoretical yield need to have the same unit. Since the actual yield is in units of grams of water, the theoretical yield should be in grams of water. We need to take the information given in the question (4.39 grams of methane) and convert it to grams of water. This will be the theoretical yield. This involves converting from grams of methane to moles of methane using the molar mass off of the periodic table, then converting from moles of methane to moles of water using the coefficients from the balanced chemical equation, then converting from moles of water to grams of water using the molar mass off of the periodic table. The theoretical yield comes from the stoichiometry. In this case it is 9.86 grams of water. The percent yield is calculated by dividing the actual yield by the theoretical yield and multiplying by 100. In this case, the lab produced 91.4% of the amount of product expected.
