Unit 3: Stoichiometry Pre-AP Chemistry

Unit 3: Stoichiometry Pre-AP Chemistry
Stoichiometry is the study of the quantitative aspects of chemical formulas and reactions. This chapter we are going to learn about stoichiometry – mastering the topics in this chapter will be essential to all your further quantitative abilities in chemistry. Much of the chemistry course will involve mathematical calculations involving chemicals and the reactions they participate in. Most of this chapter’s notes will be simply working out example problems – it’s hard to learn the math without doing the math. There are a couple examples of each different topic, but if we discuss a topic and you don’t fully understand the examples, or just would like something clarified, please let me know and we will do more examples and discuss it further. This chapter is very important. If you do not understand the concepts here, it will be very difficult for you to understand any of the mathematical concepts in the future of this course.

Diatomic Molecules There are seven elements that occur naturally as diatomic molecules. H2, N2, O2, F2, Cl2, Br2, I2

Characteristics of Chemical Equations
The equation must represent known facts. The equation must contain the correct formulas for the reactants and products. The law of conservation of mass must be satisfied. It must have the same number of atoms of the same kind on both sides. 3

Meaning of Chemical Formulas
Symbol Meaning Composition H2O One molecule of water: Two H atoms and one O atom 2 H2O Two molecules of water: Four H atoms and two O atoms H2O2 One molecule of hydrogen peroxide: Two H atoms and two O atoms 4

Stoichiometric Coefficient
Chemical Equations A chemical equation is a statement in formulas that expresses the identities and quantities of the substances involved in a chemical or physical change. For an equation to depict the amounts reacting and produced accurately, it must be balanced – the same number of each type of atom must appear on both sides of the equation. Reactants are the substances that react during the change and are placed to the left of the yield arrow. Products are the substances that result from the change and are placed to the right of the yield arrow. Equation also gives physical states of the reactants and products. Equation is balanced using stoichiometric coefficients which are numerical multipliers of all the atoms in the formula that follows it. Now we are going to start look at chemical equations which are basically the sentences of chemistry. A chemical equation expresses the amounts of different substances that are involved in a particular chemical or physical change. In order for a chemical equation to be valid, however, it must be balanced, which means that the same number of each atom must appear on either side of the equation. In other words, if four copper atoms were put into the reaction, four copper atoms better come out of the reaction, even if they are combined with something different than what they started with. The reactants in the equation are those substances on the left. These are the substances that react during the change. The products are the substances that are formed from the change and they appear on the right side of the arrow. The arrow itself is read “yields”. An equation is balanced using stoichiometric coefficients which are basically numbers that are put in front of a particular compound and express the number of molecule/formula units of that substance that participate in that reaction. Let’s look at this simple chemical equation for its parts. Let’s check the balancing. To do this, go through each element involved, and calculate the number of atoms on the left and on the right. They should be identical. 2 Mg atoms on left, 2 Mg atoms on right. Two O atoms on left, 2 O atoms on right. The coefficient and subscript both represent the number of atoms of each element. If there is a subscript and a coefficient, the coefficient multiplies with the subscript to give the number of atoms. yields Reactants Product Stoichiometric Coefficient

Physical States in Chemical Equations
Chemical equations give physical states of reactants and products. Physical states appear after the chemical formulas in the equation. Physical State Symbol Solid (s) Liquid (l) Gas (g) Dissolved in water (in aqueous solution) (aq) This equation in words would be read: aqueous hydrochloric acid reacts with solid sodium bicarbonate to yield gaseous carbon dioxide, liquid water, and aqueous sodium chloride.

Additional Symbols Used in Chemical Equations
Used to indicate a reversible reaction Reactants are heated Pressure at which reaction is carried out, in this case 2 atm Pressure at which reaction is carried out exceeds normal atmospheric pressure Temperature at which reaction is carried out, in this case 0 oC Formula of catalyst, in this case manganese (IV) oxide, used to alter the rate of the reaction D 2 atm pressure 0 oC MnO2 7

Key Points for Balancing Chemical Equations
A coefficient operates on all the atoms in the formula that follows it. Chemical formulas cannot be altered. Other reactants or products cannot be added to balance the equation. A balanced equation remains balanced even if all the coefficients are multiplied by the same number. e.g. Here are some key points for balancing equations. First, a coefficient placed in front of a formula operates on all the atoms in the formula that it is attached to. The chemical formulas cannot be altered – meaning you can’t change the subscripts or the elements in the compounds. Additional reactants and products may not be added to help balance the equation. Once you have the reaction, you must keep the same reaction and add only stoichiometric coefficients. A balanced equation will remain balanced even if you multiply all the coefficients by the same number. Let’s examine this equation that is similar to that on the previous slide. Here we multiplied all coefficients by 2. There are still the same number of Mg atoms on either side and the same number of oxygen atoms. Let’s do an example together and then I’ll let you do several to work and then check with me. Balancing equations is sort of like solving a puzzle. This example gives us the reaction in sentence form. We have to translate it first into the chemical formulas. Octane is combusted with oxygen to form water and carbon dioxide. You should have octane and oxygen on the reactant (left) side of the equation and carbon dioxide and water on the right side. Now we have to adjust the number of each atom until all of them balance out.

Balancing Equation Examples
1. 2. 3. 4. 5. 6. 7. 8.

Balancing Equations Examples (cont’d)
NO(g) O2(g)  NO2(g) (NH4)3PO Mg(OH)2  Mg3(PO4)2+ NH4OH AlCl Li2CO  Al2(CO3) LiCl

Writing and Balancing Chemical Equations
Write a balanced equation for the following reactions: Chlorine gas and dissolved sodium bromide react to produce bromine and aqueous sodium chloride. Methane gas reacts with oxygen gas to form carbon dioxide and liquid water. A solution of aluminum sulfate and calcium chloride react to form a precipitate of calcium sulfate. Aqueous aluminum chloride is also produced. 11

Writing and Balancing Chemical Equations (cont’d)
Write balanced equations for each of the following chemical statements: Chunks of sodium react violently with water to form hydrogen gas and sodium hydroxide solution. Aqueous nitric acid reacts with solid calcium carbonate to form carbon dioxide, water, and aqueous calcium nitrate. Phosphorous trifluoride is prepared by the reaction of phosphorous trichloride and hydrogen fluoride. Hydrogen chloride is the other product. Liquid nitroglycerine (C3H5N3O9) (dynamite) explodes to produce a mixture of gases: carbon dioxide, water vapor, nitrogen, and oxygen. Here are some chemical equations for you to balance. Let’s write the basic chemical equation for each of these, and then I will let you take time to balance them.

Types of Chemical Reactions
Synthesis (combination) reaction A + B  AB Decomposition reaction AB  A + B Single Replacement Reaction A + BC  AC + B Double Replacement Reaction AB + CD  AD + CB Acid-Base (Neutralization) Reaction HX + BOH  BX + HOH Combustion (of hydrocarbon) Reaction CH + O2  CO2 + H2O element compound compound element use activity series to predict driving force…water, gas, or precipitate compound compound compound compound Objective: To classify a chemical reaction as one of the following types: combination (synthesis), decomposition, single replacement, double replacement, or neutralization. acid base salt water 13

Synthesis Reactions 2H2 + O2  2H2O O O O O 2 Na + Cl2  2 NaCl
General form: A B  AB element or element or compound compound compound 2H O  H2O O O O O H H H H H H H H 2 Na Cl  NaCl Na Cl Na+ Cl - Cl Cl - Na+ Na 14

Decomposition Reactions
General form: AB A + B compound two or more elements or compounds 2H2O  H O2 O O H H O O H H H H H H 2 NaCl  Cl Na Na Na+ Cl - Cl Cl - Na+ Cl Na 15

Single and Double Replacement Reactions
Single-replacement reaction Mg CuSO4  MgSO Cu General form: A BC  AC B Double-replacement reaction CaCO HCl  CaCl H2CO3 General form: AB CD  AD CB 16

Activity Series Element Reactivity More Reactive Halogen Reactivity Li
Rb K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H2 Cu Hg Ag Pt Au F2 Cl2 Br2 I2 More Reactive 17

Single-Replacement Reactions
General Form Activity Series A BC AC B Li Rb K Ba Ca Na Mg Al Mn Zn Cr Fe Ni Sn Pb H2 Cu Hg Ag Pt Au F2 Cl2 Br2 I2 Fe CuCl2 Can Fe replace Cu? Zn HNO3 Can Zn replace H? MgCl Br2 Can Br replace Cl? 18

Predicting Single Replacement Reactions
2 Mg + AlCl3  Al + MgCl2  The question we must ask is can the single element replace its counterpart? Metal replaces metal or nonmetal replaces nonmetal. 3 3MgCl2 + 2Al Can magnesium replace aluminum? No reaction Can aluminum replace magnesium? 19

Combustion Reactions Hydrocarbon + O2  CO2 + H2O
Combustion of propane: C3H8(g) O2(g)  CO2(g) H2O(g) Combustion of octane: C8H18(g) + O2(g)  CO2 (g)+ H2O(g) Combustion of glucose: C6H12O O2  CO2(g) + H2O(g) Combustion of iron: Fe + O2  Fe2O3 20

Sequential Chemical Reactions
Many times, a product for one reaction becomes a reactant for the next reaction in a sequence of reactions. If one product is completely used up in a following reaction, it does not show up in the net (overall) reaction. To determine the overall (net) equation: Write the sequence of balanced equations. Adjust the equations arithmetically to cancel out the common substance. Add the adjusted equations together to obtain the overall balanced equation. A lot of times in nature, the product for one reaction becomes a reactant in a sequence of reactions. We see this every day as a chain reaction – like you were late to school so you got tardy roundup so you missed a quiz in your first hour – so you had to stay after school to take it which made you miss your bus and that is why you ended up walking home!! Sequences of reactions are often written as a balanced overall (net) reaction for a whole series of reactions summed up into one. It’s like if I said, you were late to school so you had to walk home. It doesn’t show you all the intermediate steps but it gives you the starting point and the end result. Pay attention to the note here that says if a product formed in one reaction is completely used up in the following reaction, it won’t appear in the overall reaction. Think of it this way. If you go to the casino with $20 bucks, win big in a game and get up to $100 and then lose terrible and walk out with $10, your net change will be -$10. There will be no evidence of the big win and the $100 – the only people who will know will be the ones who experienced it – or heard you telling it. Here are some steps to writing a balanced overall equation. First, you must write out the sequence of balanced equations. It is best to write them one under the other. Next, adjust the equations arithmetically (by multiplying an equation through by a common multiplier) in order cancel out any substances that appear on both the product and reaction sides. Then, add the equations together by bringing down all the reactants that are left on the left side and all the products remaining on the right side. Let’s look at an example of this using a reaction for a previous example and the next step in the sequence.

Sequential Reactions Example
In the first step of a sequential chemical reaction solid copper (I) sulfide is heated with oxygen gas to form powdered copper (I) oxide and gaseous sulfur dioxide. In the next step, copper (I) oxide reacts with powdered carbon to yield copper metal and carbon monoxide gas. Write a balanced overall equation for the two-step sequence.

The Mole Concept Avogadro’s Number = 6.022 x 1023 Objectives:
To state the value of Avogadro’s number: x1023. To state the mass of Avogadro’s number of atoms for any element by referring to the periodic table. To relate the moles of a substance to the number of particles. Avogadro’s Number = x 1023 23

How Big is a Mole? One mole of marbles would cover the entire Earth
(oceans included) for a depth of three miles. One mole of $100 bills stacked one on top of another would reach from the Sun to Pluto and back 7.5 million times. 6.02 x 1023 softballs = volume of Earth 6.02 x 1023 Olympic shot puts = mass of Earth 6.02 x 1023 atoms H laid side by side would encircle Earth ~1,000,000 times It would take light 9500 years to travel from the bottom to the top of a stack of 1 mole of $1 bills. 24

The Mole The mole (mol) is the SI unit for the amount of substance.
Defined as the amount of a substance that contains the same number of entities as there are atoms in exactly 12 g of carbon-12. This number is called Avogadro’s number and is equal to 6.022x1023 entities. 1 mol of a substance represents a fixed number of chemical entities and has a fixed mass. The atomic mass of an element expressed in amu is numerically the same as the mass of 1 mole of atoms of the element expressed in grams. e.g. 1 O atom has a mass of amu, and 1 mole of O atoms has a mass of grams The molecular mass (or formula mass) of a compound expressed in amu is numerically the same as the mass of 1 mole of the compound expressed in grams. e.g. 1 molecule of H2O has a mass of amu, and 1 mole of H2O has a mass of g The mole maintains the same mass relationship between macroscopic samples as exists between individual chemical entities. The mole relates the number of chemical entities to the mass of a sample of those entities. First up is an all-important concept – we are learning a new unit, one that will be used extensively. The mole (abbreviated mol) is the unit we use to express an amount. Now, as you can see, there is a very specific definition to what a mole is. It was developed as the number of atoms in 12 g of carbon This number is 6.022x10^23 and is known as Avagadro’s number. Thus, a mol is really just a number of items, like a “dozen” means there are 12 of some item. A mole means there is 6.022x10^23 of some item. The mol was based on a measurement with carbon 12 but it will be used with all the other elements and compounds. A mole of a substance represents a specific number of atoms or molecules (avagadro’s number) but does a mol of carbon have the same mass as a mol of hydrogen? No, even though the number of atoms is the same, a hydrogen atom weighs considerably less than a carbon atom (because of the number of protons/neutrons/electrons). Now we are going to start looking at where we will use this information the most – the periodic table. On the periodic table, the atomic mass of each element is listed underneath the element symbol. This atomic mass is given in amu. This value is equivalent to the mass of 1 mol of atoms of the element in grams. This is more easily understood when we look at it in practice. Oxygen has an atomic mass of amu. Thus, 1 mole of O atoms has a mass of g. This is called the molar mass and we would express it as g/mol. We use the same concept with molecules or formula units (just means ionic compound units). If we were to determine the atomic mass of water, we would have to add up the atomic mass of 2 hydrogen atoms and 1 oxygen atom. This value would equal amu. Similarly, we express the molar mass of water as g / mol. We use the mole in order to more easily work with large samples of things instead of individual atoms. In the real world, we work in grams. It is hard to measure the gram in 10^-25 or so with our instruments. The mole, however, maintains the same relationship (as we saw in molar mass vs atomic mass) and allows us to understand them in values we can measure easily. The last statement basically reiterates what we have discussed, but it is important to your understanding that the mole relates number of atoms or chemical entities (which can be atoms, molecules, ions, formula units, particles, etc.) to the mass of a sample of those same entities.

Molar Mass The molar mass (M) of a substance is the mass per mole of its entities (atoms, molecules, or formula units). Molar mass has units of g/mol Determining molar mass of: Elements  use the atomic mass of the element from the periodic table and note whether the element occurs naturally as individual atoms or as molecules Compounds  add the molar masses of the atoms of the elements in the formula The subscripts in a formula refer to the individual atoms (or ions) as well as to moles or atoms (or ions). Find the molar mass of the following elements/compounds: Sulfur Hydrogen Ammonia (NH3) Molar mass. This is something we discussed on the previous slide. It is simply mass per mole of entities. In other words, how much do 6.022x10^23 silicon atoms weigh? The units are g/mol. We discussed this on the previous slide as well, but we are going to work out a few more examples of finding molar mass. For elements, we simply have to observe the periodic table atomic mass measurements. Now there are a few elements for which the molar mass requires a little more than just the atomic mass. These elements are those that occur naturally as diatomic molecules, meaning they occur as two atoms bonded together. There are 8 of these that we see commonly. They are the five halogen gases (F, Cl, Br, I, At), nitrogen, hydrogen, and oxygen. To help you remember, aside from hydrogen, they make a 7 on the periodic table. Thus, the molar mass for O2 (oxygen) would be (2*16.00) or g/mol. For compounds, you simply add up the molar masses of all the atoms of the elements that make it up. We are going to look at some formulas and find the molar mass of each now. Sulfur occurs naturally as a single atom, so we simply need to read the value from the periodic table  g/mol Hydrogen occurs naturally as a diatomic molecule, so we need to multiply the atomic mass by 2  g / mol Ammonia is a compound. We need to add the molar mass of one nitrogen to that of three hydrogens  g/mol

Molecular Weight and Molar Mass
Molecular weight is the sum of atomic weights of all atoms in the molecule NaCl has a molecular weight of 58.5 a.m.u. Composed of a single molecule of NaCl Molar mass is the molecular weight in grams NaCl has a molar mass of 58.5 grams Composed of 6.02x1023 formula units (molecules) of NaCl 27

The Molar Mass & Number of Particles in 1-Mole Quantities
Substance Molar Mass Number of Particles in One Mole Carbon (C) g x 1023 C atoms Sodium (Na) g x 1023 Na atoms Iron (Fe) g x 1023 Fe atoms NaF (preventative g x 1023 NaF formula units for dental cavities) CaCO3 (antacid) g x 1023 CaCO3 formula units C6H12O6 (glucose) g x 1023 glucose molecules C8H10N4O2 (caffeine) g x 1023 caffeine molecules 28

Summary of Mass Terminology
Definition Unit Isotopic Mass Mass of an isotope of an element Amu Atomic Mass (Atomic Weight) Average of the masses of the naturally occurring isotopes of an element weighted according to their abundance Molecular (or formula) mass Sum of the atomic masses of the atoms (or ions) in a molecule (or formula unit) Molar mass Mass of 1 mole of chemical entities (atoms, ions, molecules, formula units) g/mol Here is a simple table to help keep the different masses we’ve discussed clear. The first three are from chapter 2. Isotopic mass is the mass of an atom of an isotope of an element. The atomic mass that we see on the periodic table is the weighted average of the naturally occurring isotopes. The molecular mass is simply the sum of all the atomic masses of the atoms in a molecule or formula unit. The units on all three of these were amu. Molar mass is the mass of 1 mole of chemical entities of a particular substance and the units are in g/mol.

Converting Moles of Elements
To convert between amount (mol) and mass (g), use the molar mass (g/mol). To convert between amount (mol) and number of entities, use Avogadro’s number (entities/mol). For atoms that occur as molecules, use the molecular formula to find atoms/mol. To convert between number of entities and mass, first convert to number of moles. There are several different conversions that you can use with the molar mass value to achieve different desired quantities. I am going to talk about these three conversion processes briefly, but I don’t expect you to understand it well until we see it in action on some example problems. We are going to employ our “railroad track” method for these calculations. I hope that you will see how lining up the units will make memorizing these instructions worthless. First, if you want to convert between amount (in mol) and mass (in g) of a substance, simply use the molar mass of the substance. If you want to convert between amount (in mol) and number of atoms or particles, use Avagadro’s number (look up on the wall above the boards). If you want to go from atoms/particles to mass, you have to go through the moles first. You will see examples of each of these in the upcoming examples.

Molar Volume at STP MOLES LITERS OF GAS AT STP (22.4 L/mol) MASS IN
GRAMS MOLES NUMBER OF PARTICLES Molar Mass (g/mol) 6.02  1023 particles/mol Molarity (mol/L) LITERS OF SOLUTION 31

Molar Mass and Converting Example Questions
How many grams of Ag are in mol of Ag? How many Fe atoms are in 95.8 g of Fe? What is the mass in grams of 3.22x1020 Mn atoms? This problem asks to convert an amount in mol of silver to grams of silver. We must use the molar mass. This problem asks to determine number of chemical entities in a specific mass of iron. First we must go through moles! This problem asks for mass in a certain number of atoms. Again, we must go through moles. This problem wants to know how many moles are in a specific mass of carbon. We need the molar mass. This problem uses a compound of ammonium carbonate. It asks for formula units – which are equivalent to atoms – in a specific mass. Just like above, we have to go through moles, and we must calculate the molar mass of the compound. This final problem asks for the mass in a specific number of molecules of a compound. We must go through moles again. It also asks how many phosphorous atoms there were in the sample. Now if we just look at the formula for the compound, we see that there are 4 phosphorous atoms per molecule. That same ratio is true in moles as well. There are 4 moles of P per mol of compound.

Molar Mass and Converting Example Questions
How many moles of carbon are in 315 mg of carbon? How many formula units are in 41.6 g of ammonium carbonate? What is the mass (in g) of 4.65 x 1022 molecules of tetraphosphorus decaoxide? How many P atoms are in this sample? This problem asks to convert an amount in mol of silver to grams of silver. We must use the molar mass. This problem asks to determine number of chemical entities in a specific mass of iron. First we must go through moles! This problem asks for mass in a certain number of atoms. Again, we must go through moles. This problem wants to know how many moles are in a specific mass of carbon. We need the molar mass. This problem uses a compound of ammonium carbonate. It asks for formula units – which are equivalent to atoms – in a specific mass. Just like above, we have to go through moles, and we must calculate the molar mass of the compound. This final problem asks for the mass in a specific number of molecules of a compound. We must go through moles again. It also asks how many phosphorous atoms there were in the sample. Now if we just look at the formula for the compound, we see that there are 4 phosphorous atoms per molecule. That same ratio is true in moles as well. There are 4 moles of P per mol of compound.

Mass Percent from the Chemical Formula
Use the molecular mass and chemical formula to find the mass percent of any element in a compound. Example: What is the mass percent of each element in glucose (C6H12O6)? How many grams of carbon are in g of glucose? Next, we are going to look at mass percent. This concept determines how much of a compound’s mass is made up of each element that is part of it. To perform the calculations for mass percent, we must find the molecular mass and have the chemical formula. We are going to look at glucose for this problem. Now, at the bottom of this slide is a formula for finding mass %. However, if you will understand the units, it will make a lot of the memorization go away. Ok so this problem asks for the mass percent of each element in glucose. Ok, first we need to figure out the total mass of one mol of this compound (or the mass of one of its molecules). So we add up all the atoms that are in it. Six carbons, 12 hydrogen, six oxygen. This value should come out as g/mol. Ok, well we want to know what percent of that was contributed by carbon. How much of that mass was made up of the carbon atoms. Well we used 6 carbons and that mass totaled g/mol. So if we want to find a percentage, you just have to put the amount that the item contributed over the total substance. To get a percentage, you will want to multiply this value by This value should come out as 40%. Go ahead and do oxygen and hydrogen as well. When you are done, what is a simple calculation you can do to make sure your answer makes sense? Add up the percentages – they should equal 100% or you did something wrong!. Ok, the next question is still about the glucose we just worked with. It asks how many grams of carbon are in g of glucose. So the mass they just gave us was the mass of the entire substance. The mass percent does not change based on the amount so we have two pieces to our puzzle. Let’s do a little math to figure this out. It should be 40% of the g.

Mass Percent from the Chemical Formula
What is the mass percent of each element in water? How many grams of oxygen are in g of water? Next, we are going to look at mass percent. This concept determines how much of a compound’s mass is made up of each element that is part of it. To perform the calculations for mass percent, we must find the molecular mass and have the chemical formula. We are going to look at glucose for this problem. Now, at the bottom of this slide is a formula for finding mass %. However, if you will understand the units, it will make a lot of the memorization go away. Ok so this problem asks for the mass percent of each element in glucose. Ok, first we need to figure out the total mass of one mol of this compound (or the mass of one of its molecules). So we add up all the atoms that are in it. Six carbons, 12 hydrogen, six oxygen. This value should come out as g/mol. Ok, well we want to know what percent of that was contributed by carbon. How much of that mass was made up of the carbon atoms. Well we used 6 carbons and that mass totaled g/mol. So if we want to find a percentage, you just have to put the amount that the item contributed over the total substance. To get a percentage, you will want to multiply this value by This value should come out as 40%. Go ahead and do oxygen and hydrogen as well. When you are done, what is a simple calculation you can do to make sure your answer makes sense? Add up the percentages – they should equal 100% or you did something wrong!. Ok, the next question is still about the glucose we just worked with. It asks how many grams of carbon are in g of glucose. So the mass they just gave us was the mass of the entire substance. The mass percent does not change based on the amount so we have two pieces to our puzzle. Let’s do a little math to figure this out. It should be 40% of the g.

Determining the Formula of an Unknown Compound
For a compound whose formula is not known, the masses of the elements in the compound may be used to determine the empirical formula for the compound. Compound is decomposed into its elements. The masses of these elements are determined and converted to moles of the elements. Mole ratios between the elements are converted to whole number subscripts. Example: Elemental analysis of a sample of an ionic compound gave the following results: 2.82 g Na, 4.35 g Cl, and 7.83 g O. What is the empirical formula and name of the compound? A lot of times, a chemist will come be working in the lab trying to identify some compound. If I just gave you a white powder, you would have no idea what it was. It could be a myriad of different things, some harmless and some very harmful. So a chemist will use his analytical skills to determine what elements are in the compound and in what ratio they combine. In the experimental stage, the compound is broken down into its individual elements, and the masses of these elements are measured. Now, we have learned that we can convert the masses of elements into moles using what? Molar mass! Now the important thing to understand when you are doing this is there is no mass ratio that is expressed by a chemical formula. The subscripts for the elements only show the ratio of atoms or moles (since a mole is a number of atoms), so we must be eventually working in moles. Let’s do an example to help clarify this. Ok, so we’re in lab and we break this compound down into sodium, chlorine, and oxygen. The masses of each of these are given here. We have to figure out the empirical formula for the compound and then name it. Well the naming is easy after we find the formula. Notice that it says to find the empirical formula. Remember back that that is the simplest ratio of the elements, not necessarily the actual number. Ok so we could write a basic NaClO formula right now, but we need to know the subscript relationships to each other. The first thing we need to do is get on common ground between the elements by converting their masses into moles using the molar masses. After doing this, we see that there are the same number of moles of sodium and chlorine so those have to be in even amounts. There are more moles of oxygen however. We have to find how many more moles of oxygen in comparison to the number of moles of sodium and chlorine. If we compare the two values, we see that the moles of oxygen is 4 times greater than those of sodium and chlorine. So if we write a simple whole number ratio where sodium and chlorine are 1:1, oxygen will add a 4 to the end of that ratio NaClO4. This is the empirical formula. Now we have to name this compound. Remember back to naming ionic compounds. The anion is the perchlorate ion, so it should be sodium perchlorate.

Determining Empirical Formulas of Unknown Compounds Examples
Sodium dichromate is a bright orange, crystalline substance. An analysis of sodium dichromate gives the following mass percentages: 17.5 % Na, % Cr, and 42.8 % O. What is its empirical formula?

An oxide of osmium is a pale yellow solid. If g of the compound contains 2.16 g of osmium, what is its empirical formula?

An oxide of tungsten is a bright yellow solid. If g of the compound contains 4.23 g of tungsten, what is its empirical formula?

Determining the Molecular Formula of an Unknown Compound
For a compound in which the molar mass is known, the empirical formula can be used to obtain the molecular formula. Sometimes the empirical and molecular formulas are identical; other times the molecular formula is a whole number multiple of the empirical formula. Example: During physical activity, lactic acid (M = g/mol) forms in muscle tissue and is responsible for muscle soreness. Elemental analysis shows that it contains 40.0 mass % C, 6.71 mass %H and mass% O. Determine the empirical formula of lactic acid. Determine the molecular formula. We are going to work with an unknown compound again, but if we happen to know the molar mass of the compound, we can determine not only the empirical formula but the actual molecular formula. Remember, sometimes these two identities are the same, other times the molecular formula is a whole number multiple of the empirical formula. This example we are working with lactic acid, the stuff that gets in your muscles and makes you sore after a work out. We know the molar mass of lactic acid and what elements it is made of, but we need to determine in what ratio are each of these elements present. When we decompose it, we find that it is composed of carbon, hydrogen, and oxygen. This example, we are given mass% instead of exact masses. Remember back to mass percent that it is a ratio of the mass of one element and the mass of the entire substance. We do not have the mass of the entire substance, but to make it easy on ourselves, let’s assume that we have 100 g of lactic acid. How many grams of carbon would we have then g C (this assumption of 100 g won’t change anything with the chemical formula of the compound and it makes it easier on our calculations). There is also 6.71 g H and 53.3 g O. The next step is converting these masses into moles. From this, we find that there are the same number of moles of carbon and oxygen but twice as many moles of hydrogen. So our empirical formula will be CH2O. The second part of this example, however, requires us to determine the molecular formula. We have the empirical formula and we also have the molar mass of the compound. What we are going to do is see if the empirical formula will give us the same molecular mass or will it be some whole number multiple of it. Calculating the molar mass of the empirical formula we add 1 carbon, 2 hydrogen, and 1 oxygen and get g/mol. This is not the same as the molecular mass, but if you compare them, you see that it is 3 times greater. That must mean that we need three times more of everything in our recipe. So we need 3 carbon, 6 hydrogen, and 3 oxygen to make the molecular formula C3H6O3.

Determining the Molecular Formula of an Unknown Compound
A moth repellent, para-dichlorobenzene, has the composition 49.1 % C, 2.7 % H, and 48.2 % Cl. Its molecular weight is 147 amu. What is its molecular formula? We are going to work with an unknown compound again, but if we happen to know the molar mass of the compound, we can determine not only the empirical formula but the actual molecular formula. Remember, sometimes these two identities are the same, other times the molecular formula is a whole number multiple of the empirical formula. This example we are working with lactic acid, the stuff that gets in your muscles and makes you sore after a work out. We know the molar mass of lactic acid and what elements it is made of, but we need to determine in what ratio are each of these elements present. When we decompose it, we find that it is composed of carbon, hydrogen, and oxygen. This example, we are given mass% instead of exact masses. Remember back to mass percent that it is a ratio of the mass of one element and the mass of the entire substance. We do not have the mass of the entire substance, but to make it easy on ourselves, let’s assume that we have 100 g of lactic acid. How many grams of carbon would we have then g C (this assumption of 100 g won’t change anything with the chemical formula of the compound and it makes it easier on our calculations). There is also 6.71 g H and 53.3 g O. The next step is converting these masses into moles. From this, we find that there are the same number of moles of carbon and oxygen but twice as many moles of hydrogen. So our empirical formula will be CH2O. The second part of this example, however, requires us to determine the molecular formula. We have the empirical formula and we also have the molar mass of the compound. What we are going to do is see if the empirical formula will give us the same molecular mass or will it be some whole number multiple of it. Calculating the molar mass of the empirical formula we add 1 carbon, 2 hydrogen, and 1 oxygen and get g/mol. This is not the same as the molecular mass, but if you compare them, you see that it is 3 times greater. That must mean that we need three times more of everything in our recipe. So we need 3 carbon, 6 hydrogen, and 3 oxygen to make the molecular formula C3H6O3.

Combustion Analysis Combustion analysis is a method used to measure the amounts of carbon and hydrogen in a combustible organic compound. Unknown compound is burned in pure O2. All of the H in the compound is converted into H2O and all of the C is converted into CO2. Products are absorbed, and their mass is measured. From this information, the empirical formula may be determined. If compound contains another element other than C and H, its mass may be calculated by subtracting the masses of C and H from the original compound mass. There is a third experimental method for determining the molecular formula for unknown compounds that is used for combustible organic compounds (means they contain carbon). The method for this is given here. The unknown compound is burned in O2, a process called combustion when the hydrogen is converted to water, and the carbon is converted to carbon dioxide. The experimental apparatus measures the amounts of these different gases by using special absorbers, each of which only absorb one of the product compounds, either carbon dioxide or water. The masses of these absorbers are measured before and after the experiment and the difference is attributed to the amount of product that was produced. If there is another element in the compound other than carbon and hydrogen, its mass can be calculated after finding the masses of hydrogen and carbon and subtracting them from the original substance mass. Again, this is much more easily understood in example form. Vitamin C contains carbon, hydrogen, and oxygen. We are given the molecular mass. A g sample is burned, changing into carbon dioxide and water. We are given the data of the masses of the absorbers. To find out the mass of carbon dioxide that was formed, we simply subtract the “before” from the “after”. We repeat this process for the water. Now, we have masses of carbon dioxide and water, but in our previous examples, we were given the masses of the individual elements and asked to turn that into moles. How can we determine the moles of carbon in a specific mass of carbon dioxide? First, we are going to convert the mass of carbon dioxide into moles. Now, how many moles of carbon are in one mole of carbon dioxide? 1 – use the chemical formula as a molar ratio. Thus we can determine the moles of carbon. Likewise, we convert the mass of water into moles of water. How many moles of hydrogen are in 1 mole of water? 2! Now we have the moles of each. This would be great and ready for comparison, except we have one more element that we have to consider – oxygen. We have to find the moles of oxygen in the original compound. The oxygen that was converted into carbon dioxide and water does not count. That was the oxygen used for combustion. In order to find this, we have to know the masses of carbon and hydrogen that came out and subtract that from the original g. How would you find the mass of carbon if you know the number of moles? Molar mass!! Do that with both carbon and hydrogen and subtract the sum of those values from g. Now we have a mass of oxygen, but to compare it for our molecular formula, we have to find the moles of oxygen. How do we do that? Molar mass! Ok, now we can see that we have the same amounts of carbon and oxygen again, but a different amount of hydrogen. We have to compare the moles of hydrogen to the moles of carbon and oxygen and we find that it is 1.3 times that. We have to get that value to a whole number and the easiest way is to multiply it by 3. When we multiply it by three we find we need 3 carbon, 4 hydrogen, and 3 oxygen. This gives us the empirical formula for vitamin C. Now we have to compare this to the molecular formula and adjust our formula. The mass of one of our empirical formula units gives us g/mol. This is half the molecular mass of the compound, so we have to multiply our subscripts by 2. Thus, our molecular formula is C6H8O6.

Combustion Analysis Examples
Acetic acid contains only C, H, and O. A 4.24 g sample of acetic acid is completely burned. It gives 6.21 g of carbon dioxide and 2.54 g of water. What is the empirical formula for acetic acid?

Combustion Analysis Examples
Example: Vitamin C (M = g/mol) is a compound of C, H, and O found in many natural sources. A g sample of vitamin C is placed in a combustion chamber and burned. Using the following data, find the molecular formula of vitamin C. Mass of CO2 absorber before combustions 83.85 g Mass of CO2 absorber after combustion 85.35 g Mass of H2O absorber before combustion 37.55 g Mass of H2O absorber after combustion 37.96 g

Stoichiometry of Chemical Equations
4 Al(s) O2(g) Al2O3(s) 4 g Al g O2 yield 2 g Al2O3 This equation means: 4 Al atoms O2 molecules yield 2 molecules of Al2O3 or 4 Al moles O2 moles yield 2 moles of Al2O3 4 mol 3 mol 2 mol 108 g g = g 45

Calculating Amounts of Reactant and Product
A balanced chemical equation relates amounts of involved substances in relationship to one another; thus, if you know the amount of one substance, all other quantities may be calculated from the chemical equation. In a balanced equation, the number of moles of one substance is stoichiometrically equivalent to the number of moles of any other substance. Stoichiometrically equivalent means there is a definite amount of one substance is formed from, produces, or reacts with a definite amount of the other. Stoichiometric coefficients express the molar ratios of involved components. Once you have a balanced chemical reaction, you have the different substances involved related to each other by their amounts. If you know the chemical equation and one of the amounts, you can calculate all the other amounts. You do this through the molar ratio of the components. For instance, in our first example with Mg and O, there were 2 mol of Mg for every 1 mole of O2, and 2 mol of Mg for every 2 mol of MgO. The stoichiometric coefficients are used to produce molar ratios!!! Let’s see this in action.

Using Mole Ratios 2Na + Cl2  2NaCl
How many moles of chlorine are required to react with 15.0 moles of sodium? How many moles of sodium chloride are produced from 6.0 moles of chlorine? How many moles of sodium are required to produce 25.2 moles of sodium chloride?

Using Mole Ratios 2H2O  O2 + 2H2
How many moles of oxygen and hydrogen are produced from 8.6 moles of water? How many moles of water are required to produce 46.1 moles of oxygen?

Stoichiometry Steps Core step in all stoichiometry problems!!
1. Write a balanced equation. 2. Identify known & unknown. 3. Line up conversion factors. Mole ratio - moles  moles Molar mass - moles  grams Molarity - moles  liters solution Molar volume - moles  liters gas Mole ratio - moles  moles Core step in all stoichiometry problems!! 4. Check answer. 49

Welcome to Mole Island 1 mol = molar mass 1 mol =
1 mole = 22.4 L @ STP 1 mol = 6.02 x 1023 particles 50

Stoichiometry Island Diagram
Known Unknown Substance A Substance B Mass Mass 1 mole = molar mass (g) 1 mole = molar mass (g) Use coefficients from balanced chemical equation Volume Mole Mole Volume 1 mole = STP 1 mole = STP (gases) (gases) 1 mole = x 1023 particles (atoms or molecules) 1 mole = x 1023 particles (atoms or molecules) Particles Particles 51

Calculating Reactants and Products Examples
Copper (I) sulfide is heated with oxygen gas to form powdered copper (I) oxide and gaseous sulfur dioxide. How many moles of oxygen are required to react with mol of copper (I) sulfide? How many grams of sulfur dioxide are formed when mol of copper (I) sulfide is reacted? How many kilograms of oxygen are required to form kg of copper (I) oxide? Ok, here we are given a reaction that we must translate into a chemical equation and then balance. It is very important that you be able to do these steps because if you are wrong in any part of the equation, the remainder of your problem will be incorrect because it will be based on incorrect information. Now that we have a balanced chemical equation, we look at the first question. We have to determine the moles of oxygen that react with 10.0 mol of copper sulfide. Let’s get our railroad tracks and molar ratios ready. If we have 10.0 mol of copper sulfide, how can we convert that into moles of oxygen? Using the molar ratio. From the balanced chemical equation, we see that we have 3 mol of O2 for every 2 mole of Cu2S. Line those units up and calculate!! Next, we are asked for grams of sulfur dioxide (a product ) if we react the same 10.0 mole of copper sulfide. Don’t get overwhelmed by the question. Work the units. We know that we can get from mol of copper sulfide to mol of sulfur dioxide if we use the molar ratio. Can we get from mol of sulfur dioxide to grams of sulfur dioxide? Yes, using the molar mass!!

Calculating Reactants and Products Examples
Thermite is a mixture of iron (III) oxide and aluminum powders that was once used to weld railroad tracks. It undergoes a reaction to yield solid aluminum oxide and molten iron. How many grams of iron form when 135 g of aluminum reacts? How many atoms of aluminum react for every 1.00 g of aluminum oxide that is formed? Ok, here we are given a reaction that we must translate into a chemical equation and then balance. It is very important that you be able to do these steps because if you are wrong in any part of the equation, the remainder of your problem will be incorrect because it will be based on incorrect information. Now that we have a balanced chemical equation, we look at the first question. We have to determine the moles of oxygen that react with 10.0 mol of copper sulfide. Let’s get our railroad tracks and molar ratios ready. If we have 10.0 mol of copper sulfide, how can we convert that into moles of oxygen? Using the molar ratio. From the balanced chemical equation, we see that we have 3 mol of O2 for every 2 mole of Cu2S. Line those units up and calculate!! Next, we are asked for grams of sulfur dioxide (a product ) if we react the same 10.0 mole of copper sulfide. Don’t get overwhelmed by the question. Work the units. We know that we can get from mol of copper sulfide to mol of sulfur dioxide if we use the molar ratio. Can we get from mol of sulfur dioxide to grams of sulfur dioxide? Yes, using the molar mass!!

Stoichiometry Examples (1)
Answer the following questions for the reaction given: 2KClO3  2KCl + 3O2 How many grams of KClO3 are required to produce 9.00 L of O2 at STP? How many molecules of O2 will be produced from 8.00 moles of KClO3? How many grams of KCl are produced from grams of KClO3? 54

Answer the following questions for the reaction given: Cu AgNO3  2 Ag Cu(NO3)2 How many grams of silver will be formed from grams of copper? How many atoms of silver will be formed from 8.42 moles of silver nitrate? How many atoms of copper are required to produce 87.6 grams of copper (II) nitrate? 55

Answer the following questions for the reaction given: 2 Na + Cl2  2 NaCl How many grams of NaCl will be produced from L of Cl2 at STP? How many grams of NaCl will be produced from grams of Na? 56

Answer the following questions for the reaction given: 2TiO2 + 4Cl2 + 3C  CO2 + 2 CO + 2TiCl4 How many moles of chlorine will react with 4.55 moles of carbon? How many grams of titanium (IV) oxide will react with 4.55 moles of carbon? How many molecules of TiCl4 will form from 115 g TiO2? 57

Water from a Camel Camels store the fat tristearin (C57H110O6) in the hump. As well as being a source of energy, the fat is a source of water, because when it is used the following reaction takes place: 2 C57H110O6(s) O2(g)  114 CO2(g)+ 110 H2O(l) What mass of water can be made from kg of fat? 58

Rocket Fuel The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10.0 kg of diborane completely? (The products are B2O3 and H2O). Balanced chemical equation 3 B2H O B2O H2O 59

Limiting Reactants The limiting reactant (limiting reagent) is the compound that determines the extent of the reaction. The limiting reactant gets used up first, not allowing the reaction to continue. The other reactant(s) are considered to be in excess. To determine the limiting reactant, choose the reactant that yields the lower amount of product. Limiting Reactant used up in a reaction determines the amount of product Excess Reactant added to ensure that the other reactant is completely used up cheaper & easier to recycle If one or more of the reactants is not used up completely but is left over when the reaction is completed, then the amount of product that can be obtained is limited by the amount of only one of the reactants A limiting reactant is the reactant that restricts the amount of product obtained. The reactant that remains after a reaction has gone to completion is present in excess. 60

+ Building Cars Cb + 4 T CbT4 plus 16 tires excess 8 car bodies
8 car bodies 48 tires 8 cars Cb T CbT4 61

Grilled Cheese Sandwiches
Bread Cheese  ‘Cheese Melt’ 2 B C  B2C 100 bread 30 slices ? sandwiches Multiple Choice: a) 130 sandwiches b) 100 sandwiches c) 90 sandwiches d) 60 sandwiches e) 30 sandwiches f) Not enough information given 62

Limiting Reactants Making Methane
Methane, CH4 + plus 8 hydrogen molecules excess 8 carbon atoms 24 hydrogen molecules 8 methane molecules C H CH4 63

Limiting Reactants 1. Write a balanced equation. 2. Either:
For each reactant, calculate the amount of product formed. Convert one reactant to the other and compare. 3. Smaller answer indicates: limiting reactant amount of product 65

Limiting Reactant Examples
Consider the following reaction: 2Na + Cl2  2NaCl If 4 moles of sodium and 8 moles of chlorine react, how many moles of sodium chloride are produced? If 16 moles of sodium and 4 moles of chlorine react, how many moles of sodium chloride are produced? For the reaction in part b, how many moles of the excess reactant remain?

If 100. g of aluminum metal reacts with 100. g of chlorine gas, how many grams of aluminum chloride will form? A g B g C g D. 494 g 67

A fuel mixture is composed of two liquids, hydrazine (N2H4) and dinitrogen tetraoxide (N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 100. g of N2H4 and 200. g of N2O4 are mixed? 68

How many grams of solid aluminum sulfide can be prepared by the reaction of 10.0 g of aluminum and 15.0 g of sulfur? How much of the nonlimiting reactant is in excess? 69

According to the balanced chemical equation, how many atoms of silver will be produced from combining 100. g of copper with g of silver nitrate? Cu(s) AgNO3(aq)  Cu(NO3)2(aq) Ag(s) 70

At STP, what volume of “laughing gas” (dinitrogen monoxide) will be produced from 50. g of nitrogen gas and 75 g of oxygen gas? How many grams of water will be produced from 50. g of hydrogen and 100. g of oxygen? 71

8. An unbalanced chemical equation is given as __N2H4(l) + __N2O4(l) __N2(g) __ H2O(g). If you begin with 400 g of N2H4 and 900 g of N2O4… Find the number of liters of nitrogen produced at STP, assuming the reaction goes to completion. Find the mass of excess reactant left over at the conclusion of the reaction. Number 5A is not a reasonable question. Water at STP is NOT a gas. I also have a slide (next slide) that shows how the problem should be phrased and worked out. 72

An unbalanced chemical equation is given as __Na(s) + __O2(g) __Na2O (s) If you have 100 g of sodium and 60 g of oxygen… Find the number of moles of sodium oxide produced. Find the mass of excess reactant left over at the conclusion of the reaction. Number 5A is not a reasonable question. Water at STP is NOT a gas. I also have a slide (next slide) that shows how the problem should be phrased and worked out. 73

Generic Stoichiometry
Z (aq) Y (aq)  M (s) T2 (g) Given the following information: Z = 20 g/mol M = 6 g/mol Y = 10 g/mol T = 5 g/mol If you combine 100 g of solution Z with 1.8 x 1024 molecules of solution Y: How many moles of M will precipitate out of the solution? What volume of T2 gas will be produced at STP? 74

Theoretical/Actual/Percent Yields
Theoretical Yield – the amount of product indicated by the stoichiometrically equivalent molar ratio in the balanced equation Never obtained exactly in reality Actual Yield – the amount of product that you actually obtain Less than the theoretical yield Percent Yield – the actual yield expressed as a percent of the theoretical yield Since actual yield is less than theoretical yield, this value is always less than 100% In multi-step reaction sequences, the percent yields of the steps are expressed as fractions and multiplied together to find the overall percent yield. Yields have to do with laboratory work. What we have been calculating so far with our chemical equations is the theoretical yield. This is the amount of product that the math and chemistry tells us that we are going to achieve. However, in reality, there is always some error in experimentation. Thus, the theoretical yield is never actually obtained. The actual yield represents the amount of product that is actually obtained in the laboratory. The closer the actual yield is to the theoretical yield, the better you have done on the experiment. To quantify this value, we calculate percent yield, which is simply the actual yield / theoretical yield and changed to a percent by multiplying by Since the actual yield should always be less than the theoretical yield, the percent yield should always be less than 100%. If you have a mult-istep reaction, you must determine the percent yields from each of the individual steps and multiply those together to find the overall percent yield. We will do a couple problems to get the hang of this.

Percent Yield Examples
When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are actually formed. Water and carbon dioxide are also formed. Calculate the theoretical and % yields of KCl. In order to calculate percent yield, we must first find the theoretical yield – what we were supposed to get. We do this exactly how we have been doing it. We just have a different resulting number that we got in actuality to compare it to.

Silicon carbide (SiC) is an important ceramic material that is made by allowing sand (silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When g of sand are processed, 51.4 g of SiC are recovered. What is the percent yield of SiC in this process? In order to calculate percent yield, we must first find the theoretical yield – what we were supposed to get. We do this exactly how we have been doing it. We just have a different resulting number that we got in actuality to compare it to.

Marble (calcium carbonate) reacts with hydrochloric acid solution to form calcium chloride solution, water, and carbon dioxide. What is the % yield of carbon dioxide if 3.65 g of the gas is collected when 10.0 g of marble reacts? In order to calculate percent yield, we must first find the theoretical yield – what we were supposed to get. We do this exactly how we have been doing it. We just have a different resulting number that we got in actuality to compare it to.

Consider the reaction: W X  Y. A scientist needs to obtain 500. g of Y in a reaction that has an 80.% yield. How many atoms of W are needed to produce the desired result? How many liters of X at STP are needed? In order to calculate percent yield, we must first find the theoretical yield – what we were supposed to get. We do this exactly how we have been doing it. We just have a different resulting number that we got in actuality to compare it to.

Lithium Hydroxide Scrubber Modified by Apollo 13 Mission
Interior view of the Apollo 13 Lunar Module (LM) as the astronauts jerry-rig a system to use the Command Module lithium hydroxide canisters to purge carbon dioxide from the LM. Photo: NASA/JSC Astronaut John L. Swigert holds the jury-rigged lithium hydroxide scrubber used to remove excess carbon dioxide from the damaged Apollo 13 spacecraft. 80

Water in Space CO2(g) + 2 LiOH(s)  Li2CO3(aq) + H2O(l)
In the space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO2 daily. What volume of liquid water will be produced when this amount of CO2 reacts with an excess of LiOH? (Hint: The density of water is about 1.00 g/mL.) CO2(g) LiOH(s)  Li2CO3(aq) + H2O(l) 81

Lithium Hydroxide Scrubbers w/ % Yield
Determine the mass of LiOH required for a seven-day mission in space for four astronauts. Assume each passenger expels 20 mol of CO2 per day. The lithium hydroxide scrubbers are only 85 % efficient (percent yield is 85%). Make sure to plan for a slight delay to allow the astronauts some room for error. 82

Unit 3: Stoichiometry Pre-AP Chemistry

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