key content vocabulary next to definitions (sometimes #2 and #3 will be the same and in that case I expect to see a box AND DEF)

Transcription

1 Unit 6 Text Chemistry I CP 1 Your Key Chemistry Annotation Guide If you are NOT using the following annotation, put in your key to the left of each item. Mr. T s Key Items to be annotated Circle Box Write DEF Underline headings and subheadings. key content vocabulary next to definitions (sometimes # and # will be the same and in that case I expect to see a box AND DEF) important ideas (include captions for visuals). Mr. T s or Write Eq Underline Write Write in the margin next to an important equation. key steps to solving an example problem, or write those steps in the margin. any questions in the margin. a learning objective next to each subheading. Using this handout effectively The studying and annotations and other homework assignments are designed to require a maximum of 45 minutes to an hour of your time. This time limit, however, does not include things like test make-up points or extra credit activities. Students are expected to plan their time. If students wait until the last minute to complete assignments that are assigned multiple days in advance, then the time required to complete the assignment may require more than the 1 hour limit. This document is a work in progress. I have made an attempt to put in as much about the knowledge you need for this unit of study as I can. This document, however, DOES NOT contain everything that you need to know to make 100% on the Unit 4 test. You must also rely on the knowledge that you should have acquired in previous units of study, classroom explanations, your notes, the textbook sections that were assigned for study, and in some cases creative thinking and problem-solving skills. Unit 6: Chemistry I Honors The Standards in This Unit According to the South Carolina Science Standard C-4, students will demonstrate an understanding of the types, the causes, and the effects of reactions. Most of the material in this unit will come from this standard. Under standard South Carolina Science Standards C-4 is a list of indicators and under each of those indicators are supporting documents. All the material in this section stems from these indicators and supporting documents. Since this unit is all about equations let s remind ourselves of a few definitions first. A equation is a symbolic representation of a reaction. The s on the left side of the equation are the reactants and the s on the right side of the equation are the products. A equation can be unbalanced. In the case of an unbalanced equation the formula of the s are shown but the number of atoms on the left side of the equation do not equal the number of atoms on the other side. A balanced equation uses coefficients in front of the formulas in the equation to balance the number of atoms on both sides of the equation (this means to get the number of each kind of atom on the left side of the equation to equal the number of each kind of atom on the right side of the equation). An ionic equation shows the ions in a solution (these are BOTH the polyatomic and monoatomic ions). A complete ionic equation shows all the ions in a solution and a net ionic equation eliminates spectator ions (spectator ions are those that start out as ions and remain as ions when the reaction is done).

2 Unit 6 Text Chemistry I CP Key questions for this next section: 1) How many objects or particles is equal to a mole of those objects or particles? ) For what kind of particles is the mole unit typically used? ) What equality statement can be written from the relationship described in the answer to the last question? 4) What conversion units can be written from the relationship described in the answer to the last question? 5) Why are conversion units powerful? 6) The atomic mass number on the periodic table is NOT the mass of any individual atom, so what is it? 7) What is molar mass? 8) How is molar mass for monatomic elements determined? 9) How is molar mass for compounds determined? 10) In this class, you must be able to express molar mass in an equality statement. Explain how that equality statement must be written. 11) What conversion units can be written from the relationship described in the answer to the last question? 1) Describe the required format for calculating molar mass in this class? Indicator C The mole concept Indicator C-4.4 in the South Carolina Science Standards students should be able to apply the concept of moles to determine the number of particles of a substance in a reaction, the percent composition of a representative compound, the mass proportions, and the mole-mass relationships. In this unit of chemistry students should: Understand that the quantity 6.0 x 10 of any object is defined as a mole of the object. Understand that the atomic mass of a substance, as found on the periodic table, represents the average mass (in atomic mass units) of the naturally occurring isotopes of the element. Understand that the molar mass of a pure substance is the mass (in grams) of one mole of the substance (the molar mass of carbon atoms is the mass (in grams) of one mole of carbon atoms). Understand that the molar mass of an element (measured in grams) is numerically equal to the atomic mass of the element (measured in atomic mass units) Understand that the formula mass is the term used for ionic substances. It is the sum of the atomic masses of all of the elements contained in one formula unit of an ionic compound. Understand that molecular mass is the term used for molecular compounds. It is the sum of the atomic masses of all of the elements in the molecular formula of the substance. Calculate the formula mass or molecular mass of any given compound. Write your learning objective for this section here: Use molar mass, formula mass, or molecular mass to convert between mass in grams and amount in moles of a compound. (see mole chart, pg 5)

3 Unit 6 Text Chemistry I CP Calculate the number of molecules, formula units, or ions in a given molar amount of a compound. Calculate the percent composition of a given compound. OK, so let s take these one at a time. The relationship between the mole and the number of particles of a substance Understand that the quantity 6.0 x 10 of any object is defined as a mole of the object. This number, 6.0 x 10, is called Avogadro s number, after Amadeo Avogadro. This scientist didn t come up with the number, but he did develop a theory about gases that eventually led to the concept of the mole and allowed later scientists to determine the number. Any object can mean objects such as apples, bricks, or grains of sand. Chemists, however, rarely work with these kinds of objects, but we do work with atoms, ions, molecules, formula units, electrons, and so forth. Rather than call these objects, we typically call these particles. So, more often than not, Avogadro s number and the mole are used to work with these kinds of particles. Since 6.0 x 10 of formula units, atoms, or molecules, or ions is a mole of those objects, this gives us an equality statement: 6.0 x 10 of any object = 1 mole of those same objects. Recall that if we can write an equality statement such as the one above, then we can write different conversion units. These are: 6.0 x 10 of any object 1 mole of any object and 1 mole of those objects 6.0 x 10 of those objects. Recall also that a conversion unit gives us power the power to convert how things are measured without actually changing the amount of those things. Key questions for this next section: 1) How is the number of molecules determined from a given molar amount of a compound? 14) How is the number of formula units determined from a given molar amount of a compound? 15) How is the number of ions determined from a given molar amount of a compound? 16) How is the molar amount of a compound determined from the number of molecules of that compound? Write your learning objective for this section here: 17) How is the molar amount of a compound determined from the number of formula units of that compound? 18) How is the molar amount of a compound determined from the number of ions of that compound?

4 Unit 6 Text Chemistry I CP 4 Using Avogadro s number and dimensional analysis to perform conversion calculations. Calculate the number of molecules, formula units, or ions in a given molar amount of a compound. Calculate the molar amount of a compound from the number of molecules, formula units, or ions. This is based on the fact that there are particles in 1 mole of anything. This statement gives you an equality statement ( molecules of a substance = 1 mole of that substance) and from this relationship you can build the conversion units mentioned earlier. Example for calculating the number of molecules: If you want to find the number of molecules of carbon dioxide (CO ) in 0.7 moles of carbon dioxide, it s done like this molecules CO = 1 mole CO, so You are REQUIRED to show an equality statement in your work. 0.7 mole CO molecules CO molecules CO 1 1 mole CO You are REQUIRED to show all these steps in your work! molecules CO Now, I gave you the formula for carbon dioxide but you should know how to write the formula from the name. If you don t recall how to do that, go back and review unit 4. The process of canceling units of measurement and species as was done above is called dimensional analysis. This form is required in this class. Notice that in the example above, the full answer was rounded in the manner required in this class (with underlining and arrows). Example for calculating the number of formula units: If you want to find the number of formula units in.50 moles of magnesium sulfate (MgSO 4 ), it s done like this formula units of MgSO 4 = 1 mole MgSO 4 so.50 mole MgSO formula units MgSO mole MgSO the answer might look like this on your calculator: 1.505E5 which means formula units MgSO formula units MgSO Example for calculating the number of ions: If you want to find the number of ions formed by moles of magnesium sulfate (MgSO 4 ), it s done like this. Magnesium sulfate is made of 1 magnesium ion (Mg + ) and one sulfate ion (SO 4 ). That s ions altogether. So for every formula unit of magnesium sulfate, ions are formed. 1 formula units of MgSO 4 = ions so mole MgSO 4 moles of ions 1 1 mole MgSO moles of ions.

5 Unit 6 Text Chemistry I CP 5 For every mole of ions, there are ions. So, ions = 1 mole MgSO 4 and 0.00 mole ions ions 1 1 mole ions the answer might look like this on your calculator: 1.04E6 6 6 which means ions ions. Example for calculating the number of moles from molecules: If you want to find the number of moles of water from molecules of water, it s done like this molecules H O = 1 mole H O, so molecules HO 1 mole HO molecules HO 4 Your calculator might show e-4 which is the same as So, the answer should be mole H O mole H O. 4 4 Example for calculating the number of moles from formula units: If you want to find the number of moles of iron(iii) nitrate from formula units of iron(iii) nitrate, it s done like this formula units of Fe(NO ) = 1 mole Fe(NO ), so formula units of Fe NO 1 mole Fe NO formula units of Fe NO Your calculator might show E which is the same as or So, the answer should be mole Fe NO mole Fe NO or mole Fe NO 14 mole Fe NO. Example for calculating the number of formula units from moles: If you want to find the number of formula units in.50 moles of magnesium sulfate (MgSO 4 ), it s done like this formula units of MgSO 4 = 1 mole MgSO 4 so.50 mole MgSO formula units MgSO mole MgSO the answer might look like this on your calculator: 1.505E5 which means formula units MgSO formula units MgSO

6 Unit 6 Text Chemistry I CP 6 Example for calculating the number of ions: If you want to find the number of ions formed by moles of magnesium sulfate (MgSO 4 ), it s done like this. Magnesium sulfate is made of 1 magnesium ion (Mg + ) and one sulfate ion (SO 4 ). That s ions altogether. So for every formula unit of magnesium sulfate, ions are formed. 1 formula units of MgSO 4 = ions so mole MgSO 4 moles of ions 1 1 mole MgSO moles of ions. For every mole of ions, there are ions. So, Key questions for this next section: ions = 1 mole MgSO 4 and 0.00 mole ions ions 1 1 mole ions the answer might look like this on your calculator: 1.04E6 6 6 which means ions ions. 19) What is the correct form with which to start when writing molar mass in this unit? What form will be required in a later unit? 0) How is molar mass for elements determined? 1) What is a formula unit? What kind of compound is this kind of formula typically used? Why is this kind of formula used instead of a formula that shows the actual number of particles bonded together in each unit? ) How is formula mass determined? ) How is molecular mass determined? 4) How is molecular mass different from formula mass and empirical mass? 5) Explain how molar mass, molecular mass, formula mass, and empirical mass are used to make conversions. What is the required format for such conversions? Atomic mass Understand that the atomic mass of a substance, as found on the periodic table, represents the average mass (in atomic mass units) of the naturally occurring isotopes of the element. So, on the periodic table that you have been given, the atomic mass of magnesium is listed as The periodic table does not list the unit of measurement, but you are required to KNOW that the unit that applies to Write your learning objective for this section here: Atomic mass, as it appears on the periodic table, is the average of all the atomic masses of all the isotopes of an element with units of atomic mass units. Atomic mass units have the abbreviation u. An older abbreviation for atomic mass units is amu. that number is the atomic mass unit, and the symbol for the atomic mass unit is amu or u (and u is preferred). If you are asked for the atomic mass of magnesium your answer should be u.

7 Unit 6 Text Chemistry I CP 7 The relationship between the mole and the sum of atomic masses in Understand that the molar mass of a pure substance is the mass (in of one mole of the substance (the molar mass of carbon atoms is the in grams of one mole of carbon atoms). This is basis upon which we calculate the molar mass of a substance. The molar mass of carbon is g/mole. Look up carbon on your periodic table. The numeric value comes from the atomic mass for carbon on the periodic table. The molar mass of monoatomic oxygen is g/mole. Because there are oxygen atoms is atmospheric oxygen (the oxygen in the air), the molar mass of atmospheric oxygen (O ) is g/mole. In other words, we multiply g/mole by and we get g/mole. The molar mass of carbon dioxide is the sum of the atomic masses of all the atoms in the formula (CO ): g/mole. In this class you are required to calculate molar mass and write an equality statement to justify the molar mass conversion units used in stoichiometric calculations. An equality statement is simply a mathematical equation where you have one number with a measurement unit on one side of the equals sign along with a species and another number formula grams) mass with a different measurement unit on the other side of the equals sign along with that same species. So, if the molar mass of atmospheric oxygen (O ) is g/mole, then we can write: and that is an equality statement g O = 1 mole O From the equality statement above you can write different conversion units: g O 1 mole O and 1 mole O g O 8 O Oxygen C Carbon You should know that the following elements are ALWAYS treated as diatomic gases when not ly combined with other elements (i.e. pure): hydrogen [H (g)], nitrogen [N (g)], oxygen [O (g)], fluorine [F (g)], and chlorine [Cl (g)]. Additionally, bromine is typically a diatomic liquid [Br (l)] and iodine is can be diatomic as both a gas or a solid [I (g) or I (s)]. You MUST know these formulas and you must know the physical states of all but iodine. Lastly, you MUST know that we treat carbon dioxide as a gas [CO (g)] unless you are told otherwise. Molar mass is the sum of all the atomic masses with units of grams of all the atoms in the formula of any species. This sum is equal to one mole of that same species. Molar mass is also often written as a fraction with the sum in grams on the top and moles on the bottom. Notice that we simply put one side of the equality statement on the top and the other side of the equality statement on the bottom.

8 Unit 6 Text Chemistry I CP 8 Calculate the formula mass or molecular mass of any given compound. The required format for calculating molar mass in this class (meaning, if you don t do it I count off points of tests, homework, and class work) for carbon dioxide is The required format for calculating molar mass in this class. C: = O: = g CO = 1 mol CO In the example above, rounding was NOT required but when it IS required it MUST BE shown (this includes arrows and underlining). If you don t remember the rules or correct format for rounding in multiplication and division, review the rules for rounding in Unit 1. From the equality statement above, conversion units can be written. They are: Practice Problems: 1. Calculate the molar mass for water g CO 1 mole CO and 1 mole CO g CO. Write conversion units for the molar mass of water.. Calculate the molar mass for calcium chlorate [Ca(ClO ) ]. 4. Write conversion units for the molar mass of calcium chlorate. The relationship between the mole and the sum of the atomic mass of an element Understand that the molar mass of an element (measured in grams) is numerically equal to the atomic mass of the element (measured in atomic mass units) Elements on the periodic table typically have their atomic numbers, symbol, name, and atomic mass listed. In the example on the right, iron s atomic mass is atomic mass units (symbol u). The molar mass of iron, therefore, is grams per each mole. It is often stated in this manner, The molar mass of iron is , or The molar mass of iron is grams per unit mole. In this class we will treat molar mass as an equality statement in which the atomic mass in grams is equal to one mole of the element. Example: g Fe = 1 mole Fe Further, the sum of the molar masses of the molecular formula, empirical formula, or formula unit of a compound is also equal to 1 mole of that compound. Example: As you can see from the segment of the periodic table on the right, the sum of all the atomic masses of the atoms in carbon dioxide (CO ) (the sum of the atomic masses of 1 carbon atom and oxygen atoms) is u. atomic atomic atomic atomic mass of mass of + mass of + mass of = carbon carbon oxygen oxygen dioxide u u u = u 6 C Carbon N Nitrogen Fe Iron O Oxygen

9 Unit 6 Text Chemistry I CP 9 Therefore, since the sum of the atomic masses of all the atoms in the formula for carbon dioxide is equal to u, and the molar mass is numerically equal to that, then the molar mass of CO is grams per unit mole: Formula mass Molar mass in fraction form and should be thought of as g CO 1 mol CO Molar mass in equality statement form g CO = 1 mole CO. Understand that the formula mass is the term used for ionic substances. Formula mass is the sum of the atomic masses of all of the elements contained in one formula unit of an ionic compound. So, formula mass is determined in the same manner as any molar mass but a formula mass is the mass of the atomic units in the formula unit. The formula unit is the empirical formula of an ionic or covalent network solid compound. It is the lowest whole number ratio of ions or atomic particles represented in an ionic compound. We use the term formula mass because ionic compounds have huge and varying numbers of atoms in any particle, but the ratio in the formula does not change. So, for ionic compounds, the molar mass is also the formula mass. Here is what is meant by varying numbers of atoms in any particle. Many molecular compounds have pretty simple ratios of atom particles. For example, a water molecule has hydrogen atoms and one oxygen atom bonded together to form a single molecular particle with the formula H O. An ionic compound, however, typically forms a crystal in which the actual ratio of ionic particles bonded together is not a simple ratio. In a small crystal of table there may be a million (1,000,000) sodium ions (Na + ) and a million chloride ions (Cl ) bonded together. The crystal next to that one in the salt shaker might have 500,000 sodium ions (Na + ) and 500,000chloride ions (Cl ) bonded together. So, a simple formula for an ionic crystal is simply not possible. Instead, we use a formula unit for ionic compounds. This formula unit communicates the simplest, whole number ratio of the ionic particles bonded together. For table salt, that s NaCl with one sodium ion and one chloride ion. The molecular mass Understand that the molecular mass is the term used for molecular compounds. It is the sum of the atomic masses of all of the elements in the molecular formula of the substance. We use the term molecular mass because in covalent compounds that form molecules the coefficients indicate the exact numbers of atoms in each particle (molecule). So, for molecular formulas, the molar mass is also the molecular mass. Be careful! Remember that a molecular formula communicates with subscript numbers the actual number of atoms in each molecule. There are also empirical formulas which communicates with subscript numbers the simplest, whole number ratio of the atoms in each molecule. You might use the term empirical mass for the molar mass of an empirical formula. In a later unit you will be expected to use this molar mass fraction form. For this unit, it is required that you start with the equality statement form and build he appropriate fraction from that.

10 Unit 6 Text Chemistry I CP 10 Using molar mass, molecular mass, or formula mass to make conversions. Use molar mass, formula mass, or molecular mass to convert between mass in grams and amount in moles of a compound. This is part of the process of performing stoichiometric calculations such as determining the limiting and excess reactant, theoretical yield, and amount of excess reactant. Example: If you want to find the number of moles of carbon dioxide in 10.0 g of carbon dioxide, it s done like this 10.0 g CO 1 mole CO g CO mole CO 0.7 mole CO The format for the calculation above is called dimensional analysis. This is the format that is REQUIRED for ANY CONVERSIONS in this class. See the mole map later in this unit. Be careful! Notice that in the example above, the full answer was rounded in the manner required in this class (with underlining and arrows). This is required for all work done in this class. Indicator C-4.5 Indicator C-4.4 in the South Carolina Science Standards students should be able to predict the percent yield, the mass of excess, and the limiting reagent in reactions. In this unit of chemistry students should: Perform stoichiometric calculations Mass-mass Limiting reactant Percent yield Note: In this class students are expected to determine the kinds of calculations and the determination of excess reactant remaining by interpreting a single word problem. Below is a step-by-step system for solving these problems. Key questions for this next section: 6) What is the definition of reaction stoichiometry? 7) What is the key step in reaction stoichiometry? 8) What is necessary in order to write a molar ratio? 9) What do molar ratios allow you and I to covert? OK, so let s take these one at a time. Using molar mass to calculate percent composition of a compound Perform stoichiometric calculations Write your learning objective for this section here: In this class stoichiometry means reaction stoichiometry as opposed to composition stoichiometry. Stoichiometry is the calculation of quantitative (measurable) relationships of the reactants and products in a reaction based upon a balanced equation.

11 Unit 6 Text Chemistry I CP 11 The key step in the stoichiometric process is the molar ratio conversion. Stoichiometry is math that we do in chemistry that requires that you balance a equation. We need the balanced equation to write a molar ratio (or mole ratio). Recall that the coefficients in a balanced equation can be thought of as 1) the actual number of formula units or ) the number of moles of each formula unit. The molar ratio is based on the nd of those concepts. A molar ratio is a fraction in which the coefficient for one substance in the equation is listed in the numerator of a fraction with the unit mole and the formula from the balanced equation and the coefficient for another substance in the equation is listed in the denominator with the unit mole and the formula from that same balanced equation. Here s what it looks like: If we put silver nitrate in water and dissolve it and we put magnesium chloride in water and dissolve that, and then we put to solutions together we can write a equation that will describe the reaction that might possibly occur. Before going on, see if you can write a formula equation based only on the information above. A formula equation is an unbalanced equation that does not include the physical state symbols OK, if you were successful you should have an equation that looks like this: AgNO + MgCl AgCl + Mg(NO ) If we use the solubility rules we can determine the identity of a solid that might form and thereby determine whether the reaction that we will get a precipitate (a solid that forms in the liquids). Before going on see if you can add the physical states of the substances in the equation based on the information in the solubility rules and predict if there is a driving force. OK, if you were successful you should have an equation that looks like this: AgNO (aq) + MgCl (aq) AgCl(s) + Mg(NO ) (aq) As you can see, silver chloride forms a solid and that gives us a driving force for the reaction. OK, so we now need a balanced equation in order the write the molar ratios. We balance equations using the atom inventory process AgNO (aq) + MgCl (aq) AgCl(s) + Mg(NO ) (aq) Ag: 1 1 N: 1 O: 6 6 Mg: 1 1 Cl: 1 If you are given 6.00 moles of silver nitrate and you want to find how many moles of magnesium chloride are needed to completely react with that, you will need a molar ratio to do that. Two molar ratios can be written from the coefficients in front of silver nitrate and magnesium chloride. They are:

12 Unit 6 Text Chemistry I CP 1 moles AgNO 1 mole MgCl and 1 mole MgCl moles AgNO So, to convert 6.00 moles of silver nitrate to moles of magnesium chloride, you want to put that number, unit of measurement, and species (the species is the, ion, atom, etc) over 1 and multiply that fraction by the molar ratio that will allow you to cancel the units and species that you want to cancel, leaving you with the units and species that you want. Here s how that works: 6.00 mole AgNO 1 mole MgCl 1 mole AgNO.00 moles MgCl This is why a balanced equation is necessary. Without it, you cannot construct a proper molar ratio. Molar ratios are the step in stoichiometric calculations that allows you and I to convert from one species to another. Notice in the problem above that the unit of measurement in the answer is the same unit of measurement in the starting amount: mole. The species in the answer is different from the species in the starting amount. Practice Problems: 1. Based on the following equation, how many moles of hydrochloric acid are needed to react with 0.64 moles of potassium permanganate? KMnO HCl Cl + MnO + 4 H O + KCl. If sufficient hydrochloric acid is used to react completely with 1.5 moles of magnesium, how many moles of hydrogen will be produced? Key questions for this next section: 0) What are the steps in a mass-to-mass calculation? 1) In a mass-to-mass calculation, what do you do with the measured amount of the starting species? ) In a mass-to-mass calculation, what do you do to convert the measured amount of the starting species to moles? ) In a mass-to-mass calculation, what do you do to convert the moles of the starting species to moles of the ending species? 4) In a mass-to-mass calculation, what do you do to convert the moles of the ending species to the mass of the ending species? 5) What is a mole map? How is a mole map used? Write your learning objective for this section here: 6) If you ve set up your mass-to-mass stoichiometric calculation in dimensional analysis form, what will happen that will help you in solving your problem?

13 Unit 6 Text Chemistry I CP 1 Stoichiometry: performing mass-to-mass calculations Mass-mass With a mass-to-mass stoichiometric calculation, we add a molar mass conversion before and after the molar ratio. The general pattern of mass-to-mass stoichiometry is: put that starting amount over 1, calculate the molar mass of starting species construct molar mass conversion for starting species with mass on bottom and moles on top construct molar ratio (from the balanced equation) so that moles of the starting species in on the bottom and moles of the ending species in on the top calculate the molar mass of ending species construct molar mass conversion for ending species with moles on bottom and mass on top solve the problem (don t forget to round properly) For example, if you have 1.5 g of silver nitrate (AgNO ) and you want to find out what mass of silver chloride will be produced from that, you will: put that starting amount (1.5 g AgNO ) over 1, calculate the molar mass of silver nitrate to convert the mass of silver nitrate to moles of silver nitrate, convert the moles of silver nitrate to moles of silver chloride using the molar ratio above, and lastly you will calculate the molar mass of silver chloride to convert the moles of silver chloride to mass of silver chloride. See the mole map at the end of this unit and follow the pattern from top to bottom of the vertical center. On the mole map, Mass of Substance A would be your 1.5 g of silver nitrate, and the Mass of Substance B is the mass of silver chloride that you are trying to find. Now to make all this happen, we must have a balanced equation which we determined performed in the last section. AgNO (aq) + MgCl (aq) AgCl(s) + Mg(NO ) (aq) To perform the calculation described above ( find out what mass of silver chloride will be produced ) we will be making an assumption that might not be true in all cases. The assumption is that there s plenty of magnesium chloride (MgCl ) to completely react with all of the silver nitrate (AgNO ). We ll perform calculations later in which there may not be enough of a certain reactant. For now, we ll just assume that there is plenty of magnesium chloride. OK, so let s put that starting amount over 1: 1.5 g AgNO 1

14 Unit 6 Text Chemistry I CP 14 Now, calculate the molar mass of silver nitrate: Ag: = N: = O: = g AgNO = 1 mol AgNO Now we use the equality statement g AgNO = 1 mol AgNO to build a conversion unit and convert the mass of silver nitrate to moles of silver nitrate: 1.5 g AgNO 1 mol AgNO Now, cancel units and calculate an answer: g AgNO 1.5 g AgNO 1 mol AgNO g AgNO mol AgNO mol AgNO. Now, refer back to the original problem. The problem did NOT ask for you to find the moles of silver nitrate. This is just the 1 st of calculations needed to find the answer to this problem. The next step requires that we 1 st have a balanced equation. Look back at the balanced equation. We can see that we have moles of silver nitrate and moles of silver chloride. With this information we are ready to find the moles of the product that we are looking for: silver chloride. We ll put the answer to the last problem over one. This will be our new starting amount. We ll convert the moles of silver nitrate to moles of silver chloride using the molar ratio from the balanced equation AND we ll keep cancelling units and species: mol AgNO mol AgCl 1 mol AgNO mol AgCl. Now, refer back to the original problem again. The problem did NOT ask for you to find the moles of silver chloride, either. This is the nd calculation needed to find the answer to this problem. The next step requires that we convert the moles of silver chloride to the mass of silver chloride. Do make this conversion we need the molar mass of silver chloride. This is done in the manner previously shown: Ag: = Cl: = g AgCl = 1 mol AgCl Now we will convert moles of silver chloride to the mass of silver chloride by using the equality statement 14.1g AgCl = 1 mol AgCl to build another conversion unit: mol AgCl 14.1 g AgCl 1 1 mol AgCl g AgCl 1.16 g AgCl

15 Unit 6 Text Chemistry I CP 15 This, now, is the answer to the original prompt. If you ve set up your mass-to-mass stoichiometric calculation in dimensional analysis form (as seen above), then all the units of measurement and species will be cancelled out except the species and units of measurement that you want. If it doesn t work out that way, you need to back and check your set up. So, you can see how important it is to include the correct units of measurement and the correct species in every step of the problem. Recall: The dimensional analysis process of canceling units of measurement and species as was done above is required in this class. Now, it s possible to combine these problems into just one problem. 1.5 g AgNO 1 1 mol AgNO g AgNO mol AgCl mol AgNO 14.1 g AgCl 1 mol AgCl g AgCl Let s make sure that we can see how that is done: Original set of problems: Problem #1 1.5 g AgNO 1 mol AgNO g AgNO mol AgNO mol AgNO. Problem # mol AgNO mol AgCl 1 mol AgNO mol AgCl. Problem # mol AgCl 14.1 g AgCl 1 1 mol AgCl 1.16 g AgCl g AgCl Original set of problems combined into 1: 1.5 g AgNO 1 Problem #1 1 mol AgNO g AgNO Problem # mol AgCl mol AgNO Problem # 14.1 g AgCl 1 mol AgCl g AgCl You are allowed to use either stoichiometric conversion systems: the one that breaks the problem into individual problems or the one that combines all the conversions into a single conversion problem.

16 Unit 6 Text Chemistry I CP 16 Now the calculations described above are based on the assumption that the amount of product silver chloride is based on the amount of silver nitrate. If silver nitrate is the limiting reactant in the problem, then the calculations described above would be practical. If you don t know whether silver nitrate is the limiting reactant then we need to be able to perform the kind of calculations described in the next section. Practice Problems: 1. What mass of SrF can be prepared from the reaction of 10.0 g Sr(OH) with excess HF?. How many grams of oxygen will be produced if. g of xenon difluoride reacts with excess water according to the following unbalanced equation? XeF (g) + H O(l) Xe(g) + HF(aq) + O (g) Key questions for this next section: 7) What is a limiting reactant? 8) In a reaction in which there is a limiting reactant, what will be left when the reaction runs its course? 9) What is the process we use to give us the information needed to determine which reactant is the limiting reactant in a reaction/equation? 40) Once you complete the mass-to-mass calculation how is the answer used to determine which reactant is the limiting reactant in a reaction/equation? 41) Why do we need to know which reactant is the limiting reactant? 4) Once the limiting reactant is determined what do we do with it? 4) What amount of the limiting reactant (the amount calculated or the amount given in the problem) is used to calculate the theoretical yield of the product? 44) What is a theoretical yield and how is it different from an actual yield? Determining which reactant is the limiting reactant Limiting reactant In the example above we assumed that there was more than enough of the reactant magnesium chloride to completely use up all of the silver nitrate. This means that the magnesium chloride was in excess. In the real world, we often don t know which of the reactants is in excess or which one might be a limiting reactant. A limiting reactant is the species that that runs out first in a reaction, leaving excess reactant remaining with the products that were produced. Once the limiting reactant runs out, not more products can be produced. Here s a potential limiting reactant problem: If a solution containing g silver nitrate (AgNO ) is placed into a container with a solution containing g sodium sulfate (Na SO 4 ), what is the maximum amount of the solid product that can be produced? To answer the question above, we must 1 st identify the solid product and to do that we must set up a equation. Let s start with the reactants: AgNO (aq) + Na SO 4 (aq) Write your learning objective for this section here:

17 Unit 6 Text Chemistry I CP 17 To predict the product you need to recognize that the reactants above are soluble salts. Using your knowledge of reaction classifications you can infer that a double replacement reaction is possible. If a double replacement reaction does take place the products will be the result of silver ion (Ag + ) combining with the sulfate ion (SO 4 ) and sodium ion (Na + ) combining with the nitrate ion (NO ). A double replacement reaction follows the pattern that was described in unit 5: Reactants Products Illustration reaction #: AB + CD AD + CB Illustration reaction #: AB + CD CB + AD -part -part -part -part The result is: Reactants Products AgNO (aq) + Na SO 4 (aq) AgSO 4 (s) + NaNO (aq) -part -part -part -part You should be able to use the solubility rules to determine that silver sulfate is the solid product that we are looking for (unit 5). You should also be able to determine the oxidation numbers on the silver ion, the sulfate ion, the sodium ion, and the nitrate ion. You should be able to use those oxidation numbers to balance the formulas for the products. Now the product formulas need to be balanced. Silver has a 1+ charge and sulfate has a charge, so we need silver ions for every sulfate ion to create a neutrally charged compound: Ag SO 4. Sodium has a 1+ charge and nitrate has a 1 charge, so we need 1 sodium ion for every nitrate ion to create a neutrally charged compound: NaNO. The result is: Reactants Products AgNO (aq) + Na SO 4 (aq) Ag SO 4 (s) + NaNO (aq) -part -part -part -part Now, we re ready to balance the equation. Recall that we always use an atom inventory to balance equations: Reactants Products AgNO (aq) + Na SO 4 (aq) Ag SO 4 (s) + NaNO (aq) -part -part -part -part Ag 1 N 1 1 O Na 1 S 1 1 Now we re ready to determine which of the reactants is the limiting reactant. There are several ways to do this. The traditional way to do it is to determine the molar ratio of

18 Unit 6 Text Chemistry I CP 18 reactants from the masses of each and compare that ratio to the molar ratio in the balanced equation. After years of teaching, I ve come to the conclusion that the high school students that I teach more readily understand another method and that method is what will be explained here. Step 1 is to perform a mass-to-mass calculation from one reactant to another. Previously, we calculated from the mass of a reactant to the mass of a product, but here we want to calculate from the mass of one reactant to the other. It doesn t matter which reactant we start with, so I will start with the 1 st one. Let s write as much of the equation as we can based on the information that we already have. We already know the molar mass ratio for silver nitrate (from the previous section) and we have a balanced equation from which to determine a molar ratio. Starting amount over 1 Conversion unit built from the molar mass equality statement Molar ratio built from the balanced equation g AgNO 1 1 mol AgNO g AgNO 1 mol Na SO 4 mol AgNO Before we can put in the last conversion unit we need to calculate the molar mass of sodium sulfate: Na: = S: =.065 O: = g Na SO 4 = 1 mol Na SO 4 Now that we have an equality statement (14.04 g Na SO 4 = 1 mol Na SO 4 ) we can enter a conversion unit to the calculation: Starting amount over 1 Conversion unit built from the molar mass equality statement Molar ratio built from the balanced equation Conversion unit built from the molar mass equality statement g AgNO 1 1 mol AgNO g AgNO 1 mol Na SO 4 mol AgNO g Na SO 4 1 mol Na SO g Na SO 4 OK, this is point where we determine the identity of the limiting reactant and here s how: if the amount of reactant that you calculated is greater than the amount given to you in the problem, then that reactant is the limiting reactant. In our example, the amount of Na SO 4 that was calculated is 4.6 g. The amount of Na SO 4 that was given to you in the problem was g. So, the amount calculated is less than the amount given to you in the problem so Na SO 4 IS NOT the limiting reactant; it s the excess reactant. Since Na SO 4 is the excess reactant, then the other reactant AgNO is the limiting reactant.

19 Unit 6 Text Chemistry I CP 19 If we had started our calculation above with sodium nitrate, we would have found that the amount calculated is more than the amount given in the problem. Once the limiting reactant is determined, then you can find the answer that the problem asks you to find which is to calculate to the amount of solid product that can be produced. This is another mass-to-mass calculation problem and to do this we start our calculations (yes, we re starting a whole new mass-to-mass calculation) with the limiting reactant as the starting amount. Rule for determining the limiting reactant: Complete a mass-to-mass calculation from one reactant to the other, If the amount of reactant that you calculated is greater than the amount given to you in the problem the ending species is the limiting reactant. If the amount calculated is less than the amount given to you in the problem the ending species is the excess reactant g AgNO 1 1 mol AgNO g AgNO 1 mol Ag SO 4 1 mol AgNO Before we can put in the last conversion unit we need to calculate the molar mass of silver sulfate: Ag: = S: =.065 O: = g Ag SO 4 = 1 mol Ag SO g AgNO 1 mol AgNO 1 mol Ag SO g AgSO 4 1 mol AgSO g AgNO mol AgNO g Ag SO 4 So the amount of solid product that can be produced is 10.1 g Ag SO 4. This is the theoretical yield. The theoretical yield is the maximum amount of product that can be produced from the limiting reactant amount. It s called a theoretical yield because it s not possible to recover such a perfect result. As will be discussed in the next section, the actual yield is the amount that is actually recovered and it will be some amount less than the theoretical yield. Key questions for this next section: 45) What is the percent yield? 46) How is the percent yield calculated? 47) What do we mean by the phrase amount of excess reactant remaining? 48) How is the amount of excess reactant remaining determined? 49) How do we determine the amount of excess reactant used? Write your learning objective for this section here:

20 Unit 6 Text Chemistry I CP 0 Determining the percent yield of a product Percent yield Once a theoretical yield has been calculated, it is sometimes important to know how efficient the recovery process is. For this we use a percentage yield calculation. Percent yield is the actual yield divided by the theoretical yield multiplied by 100. Actual Yield Theoretical Yield 100 = Percentage Yield In the problem described above, if the reaction was performed in the lab and the experimenters were able to recover 8.5 g of silver sulfate then the percentage yield would be: 8.5 g Ag SO g Ag SO = % 81.4% yield of silver sulfate. There is one more calculation that can be made; the amount of excess reactant remaining when the reaction has finished. This also involves a mass-to-mass calculation. For this calculation we also start with the limiting reactant, but we calculate to the other reactant the excess reactant. This allows us to determine just how much of the excess reactant is actually used up in the reaction. We have already determined how much of the excess reactant has been used up; it s the 4.6 g of sodium sulfate that we calculated above. To find the amount of excess reactant remaining, we simply take this amount and subtract it from the amount given: g Na SO 4 given in the problem 4.6 g Na SO 4 used up in the reaction 5.8 g Na SO 4 excess reactant remaining after the reaction is finished The term typically used for the excess reactant used up in the reaction is the excess reactant used. The term typically used for the excess reactant remaining after the reaction is finished is the excess reactant remaining or the excess reactant that remains. Had we not been lucky enough to have started with the limiting reactant in our 1 st calculation we would have had to perform that mass-to-mass calculation to determine the amount of excess reactant actually used up the in the reaction. And then we would have subtracted the amount of excess reactant actually used up the in the reaction from the amount of excess given in the problem. OK, let s review all the steps in the problem solution process explained above: Step 1: Make sure the equation is balanced. If you have an unbalanced equation presented to you, then you need to balance the equation first. If you are only given the reactants you will have to predict the products and balance the product formulas (you cannot properly balance a equation unless the formulas are balanced).

21 Unit 6 Text Chemistry I CP 1 o You may only be given the word formulas for the reactants, in which case you must be able to write the symbolic formula from the word formula and balance those formulas as well. o In order to predict products you must know how to classify equations. Step : Find the limiting reactant (also called the limiting reagent or limiting factor). In this unit, this determination will always be a mass-to-mass calculation. This mass-to-mass calculation will start with one reactant and end with the other reactant. o The rule is that if the calculated amount (called a theoretical amount) of the reactant at the end of the mass-to-mass calculation is greater than the amount given in the problem, then the species at the end of the calculation is the limiting reactant. o If the calculated amount (called a theoretical amount) of the reactant at the end of the mass-to-mass calculation is smaller than the amount given in the problem, then the species at the end of the calculation is the excess reactant. o One reactant is virtually always a limiting reactant and the other reactant is the excess reactant. To find a limiting reactant you must know how to calculate the molar mass (or molar weight) of the reactants and products AND you must know how to determine a molar ratio. o Determine molar mass by writing out the formula, such as H O for water, and look up the weight of each atom on the periodic table. For example, multiply the hydrogen atom weight by two and add it to the weight of oxygen. Use only the molar mass calculation format required and approved for this class. o The molar ratio comes from the coefficients and formulas in the balanced equation. Step : Calculate the theoretical yield or how much will be synthesized under perfect conditions. In this unit, this determination will always be a mass-to-mass calculation. Step 4: Know the actual yield, or the amount of product truly synthesized in the original experiment. This involves no additional calculations. It is simply the amount measured at the end of the experiment or given to you in a word problem. Step 5: Calculate percentage yield (mass of actual yield divided by mass of theoretical yield multiplied by 100 percent). Step 6: Calculate the amount of excess reactant remaining after the reaction is complete. In this unit, this determination will always start with a mass-to-mass calculation. You have a 50:50 chance that this mass-to-mass calculation was performed in step above. You must start with the limiting reactant and calculate the theoretical amount of excess reactant. If you did this in step, then go straight to the next sentence. If no, you must perform the mass-tomass calculation from the limiting reactant to the excess reactant species to determine the amount of excess reactant used. Finally, you subtract the theoretical amount of excess reactant from the given amount of excess reactant to determine the excess reactant that remains.

22 Unit 6 Text Chemistry I CP Key questions for this next section: 50) What is percent composition? 51) How is the number of formula units determined from a given molar amount of a compound? 5) How is the number of ions determined from a given molar amount of a compound? 5) How is the molar amount of a compound determined from the number of molecules of that compound? 54) How is the molar amount of a compound determined from the number of formula units of that compound? 55) How is the molar amount of a compound determined from the number of ions of that compound? Using molar mass to calculate percent composition Calculate the percent composition of a given compound. Percent composition simply means the percent mass of each element in a compound. Since we were using carbon dioxide as an example earlier, let s continue with that for now. As was said earlier the atomic mass of carbon is u, the atomic mass of oxygen is u, and the sum of the atomic masses of both oxygen atoms in carbon dioxide is u, and the total atomic mass of carbon dioxide (the sum of the atomic masses) of 1 carbon and oxygen atoms is u. To find the percent composition you simply take the mass of carbon and divide by the entire mass of a carbon dioxide molecule then multiply by 100, and then you take the total mass of oxygen and divide by the entire mass of a carbon dioxide molecule then multiply by 100. The percent composition for carbon dioxide comes in parts: 1) the percent mass of carbon and ) the percent mass of oxygen. The percent composition for any compound comes in as many parts as there are elements in the formula. Example: Percent composition of carbon dioxide Write your learning objective for this section here: Example: The required format for calculating percent composition begins with the same format required for calculating molar mass in this class (meaning, if you don t do it I count off points of tests, homework, and class work). So for carbon dioxide you begin with C: = Mass of all carbon atoms The required format for calculating O: = Mass of all oxygen atoms molar mass and the beginning of g CO percent composition in this class. = 1 mol CO Mass of the carbon dioxide molecule From the calculations above, you have the mass of all the carbon atoms and the mass of all the oxygen atoms and the mass of the entire compound.