If you remember, we left off this part of the story with defining the mass of one 12

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1 Chapter 11 Chemical Calculations For the past several weeks we have been working on our qualitative understanding of first atoms, then molecules, and finally chemical reactions. In this chapter we enter a new phase: we begin to put numbers to everything and begin building up our quantitative tools so we can determine actual amounts of chemicals to be used in reactions The Mole To start building up on quantitative understanding of chemistry we have to go back to Chapter 2 where we discussed the mass of the atom. If you remember, we left off this part of the story with defining the mass of one 12 atom of 6C as being 12 atomic mass units or 12 u. The numbers associated with all the other atoms was then a relative number that told how much larger or smaller the mass of another atoms was. And then we added the complication that most elements in nature consist of a number of isotopes, and that the numbers we write on the periodic table represent a weighted average of all the isotopes. Now that we know how molecules are built, let s extend our concept of atomic masses to molecular mass 12 If the mass of one atoms of C is 12 units, then the molecule CH 4 which has one C atoms and 4 H atoms is goin to be 12+(4x1) = 16 u, etc. Key Concept: A molecular mass is the sum of the atomic masses of all the atoms in a molecule Let s generalize this one more step A formula mass is the sum of the atomic masses of all the atoms in an atom, molecule or ion. Again we are talking relative numbers one molecule of CH 4 is 16u and is 16u/12u more massive than one atom of C But dealing with ratios is tricky and abstract. I want something concrete and absolute to work with. So I can weigh a chemical out on a balance and know how much material I have. The key to this absolute chemical scale is the mole

2 2 Key concept: The quantity of a substance whose mass in grams is numerically equal to the formula mass of a substance is called a mole (abbreviated mol) -And- The molar mass of a compound is the number of grams need to make up one mole of a compound. Example: CH 4 Formula mass = units ( (1.008)) st 1 definition above: grams CH 4 = 1 mole CH4 nd 2 definition above Molar mass of CH = grams of CH 4 4 But I probably don t have to tell you this; you have probably had this a zillion times in high school, just maybe not presented in this manner. The above equations, linking the molar mass of a substance in grams with the mole is a key conversion factor that we will use over and over in chemical calculation and fits right in with our Dimensional Analysis that you learned in Chapter 1 Example 1: I have 20 grams of CH 4 How many moles is that Given want grams moles Unit conversion Grams X Moles = moles Grams grams CH 4 = 1 mole CH4 Plug & chug grams CH X 1 mole CH =1.25 moles CH g CH 4

3 3 Example 2: I need.450 moles of NaCl for an experiment. How many grams do I need to weigh out? 1 mole NaCl = g NaCl Given moles NaCl Conversion moles NaCl x Need Grams NaCl Grams NaCl = grams NaCl Moles NaCl Plug & Chug.450 moles NaCl x grams NaCl = 26.3 g NaCl 1 mol NaCl Clicker question: gram to mole conversion Ba(OH) 2 show 4 equations each with a different answer have them pick out correct equation 11-2 Avogadro s number 12 So our definition of a mole was that 12g of C is 1 mole of C But what is it really Now that we know that the C is composed of atoms of C, 12 how many atoms of C are in that 1 mole or 12 g of C? Key Concept: 23 1 mole of a thing is 6.022x10 things. 23 So g of CH 4 contains 6.022x10 molecules of CH and g of NaCl contains 6.022x10 g of NaCl(s) or 6.022x10 Na ions 23 - and 6.022x10 Cl ions And now we have another conversion factor 23 1 mole of stuff = 6.022x10 items of stuff A mole is a lot of stuff but I think you can handle it. Example calculations: I have a 10 pound lead weight that I use for an anchor on my boat. How many moles are in that anchor? How many atoms? 1 lbs= g 1 mole Pb = g

4 4 Given lbs Want Moles Set up lbs x grams x moles =Mole lbs grams Plug & Chug 10.0 lbs Pb x g Pb x 1 mole Pb = 21.9 moles Pb 1 lbs Pb g Pb If I wanted to go all the way to atoms Given lbs Want Atoms Set up lbs x grams x moles x atoms = atoms lbs gram Mole Plug & Chug lbs Pb x g Pb x 1 mole Pb x 6.022x10 atoms = 1.32x10 atoms Pb 1 lbs Pb g Pb 1 mole Pb Clicker question Start with Giga molecules of Calcium Chloride (written not formula) Give 3 wrong and 1 correct equation to convert to grams 11-3 Simplest (Empirical) Formulas Back in chapter 2 we talked about analyzing a compound using mass percent where mass % = (mass of element A)/(total mas of all elements in a compound) X100% This is one of the simplest and oldest ways that chemists use to analyze compounds. In this section you will learn how to go from a mass % to an empirical formula But first, what is an empirical formula? Key Concept: An Empirical formula is the simplest possible formula for a compound. It expresses the ratio of one atom to another, but it does not convey the total amount of each element in a compound

5 5 Example: Benzene and acetylene have the same empirical formula CH. This says that there is 1 C atom for every H atom in both of these compound The actual molecular formula of Benzene is C6H 6 and that of acetylene is C2H2 so you can see how the actual molecular formula is a multiple of the empirical formula The way McQuarrie explains how to go from a mass % to an empirical formula makes sense, but he hasn t really boiled it down to a simple step-by-step procedure for you to follow. So let s try doing it this way: Key Procedure: Calculating Empirical Formulas from mass % 1. Calculate the numbers of grams of each element that would occur if the sample was 100 g. We do this to convert from % to g in one easy step. 2. Calculate the Number of moles for each element in a 100 gram sample 3. Find the element with the smallest number of moles, and then divide the number of moles for each element by that smallest number. - This should result in small whole numbers that will represent the subscripts for the elements in the empirical formula - If you do not have whole numbers, multiply all the numbers in the equation by integers until you get whole numbers for all coefficient in the equation Example calculation: Let s try an example, hydrazine is 87.42% nitrogen and 12.58% hydrogen, what is its empirical formula? 1. The first step is to convert % to grams assuming you have a 100 gram sample 100g X.8742 = g N 100 g X = g H 2. Second step - convert from grams to moles g N/14.01 g/mol = 6.24 moles N g H / g/mol = moles H 3. Divide through by the smallest number of moles moles H/ 6.24 moles of N So 2 moles of H 1 mole of N means is stoichiometically equivalent to

6 Three things to notice. A. Always put the largest number over the smallest. If you have more than 2 elements choose the smallest overall and use that as the denominator for all calculations B. You will usually get nice whole numbers, but sometimes may end up with ratios like 1.5/1. In this case you need to multiply both the numerator and the denominator by a single number that will give you all nice whole numbers. In this case 2(1.5)=3 and 2(1) =2, so the actual ratio is 3/2 C. The resulting formula is called the empirical formula. It is not the molecular formula you have the ratio right but the overall numbers can be factors of this NH, (N H, N H etc.) A second example (if you want it) An unknown liquid has 12.5% H, 27.5% C and 50% O, what is the empirical formula of this compound? H : 100g x 12.5% = 12.5g C: 100g x 37.5% = 37.5g O: 100g x 50% = 50.0g H: 12.5g x 1 mole/1.008 g = 12.5 mole C: 37.5g x 1 mole/12.01g = mol O: 50 g x 1 mole/16.00 g = /3.125 = 4 units of H 3.125/3.125 = 1 unit of C 3.125/3.125 = 1 unit of O 4 moles of H 1 mole of C 1 mole of O Empirical formula = CH O Clicker question? Long an complicated break into steps? 4 The book does some additional problems where you burn a compound in a gas and look at the resulting oxide or nitride. I think I will skip this Using the empirical formula to determine the atomic mass of an element in a compound While this may be a fairly standard lab technique, its not one that we do so I will skip this section 6

7 11-5 Molecular Formulas So if we can only get an empirical formula from a mass %, how do we get an actual molar formula? We need one more piece of information, the molar mass. You have seen that the empirical formula is the simplest possible molecular formula. You can think of it as a lowest common denominator. The actual molecular formula is some multiple of this: Key Equation molecular formula = empirical formula x N molecular mass = empirical mass x N Where N is some whole number Example: A compound is 85.7% C and 14.3% H and has a molar of 42. What is the molecular formula of this compound? There are a couple of ways to do this. First the long way gc in 100g sample = 85.7g gh in 100g sample = 14.3 g mole C = 85.7/12.01 = 7.14 moles C mole H = 14.3/1.008 = moles H Dividing by mole C 2 moles of H Empirical formula = CH 2 ; Empirical mass = 14 Molecular mass = empirical mass x N 42 = 14x N 42/14=N N=3 Molecular formula = empirical formula x N = (CH 2) x 3 =C H 3 6 I like the long way because you see that there are three empirical units in the molecular unit. 7

8 8 Now the short way Multiply the molar mass by the % 42 g/mole x 85.7%/100% = 36.0 g/mol C 42 g/mol x 14.3%/100% = 6.00 g/mol H Divide the g/mol fo reach element by the atomic mass g/mol C /12 g/atom = 3 atoms/mole = C 3 6 g/molh / 1g/atom H = 6 atoms/mol = H 6 Molecular formula = C H 3 6 Key calculation Given % composition and molar mass be able to determine the molecular formula. This can be done by two different procedures choose the one that you like best Combustion Analysis This is another fairly standard technique that is used in many labs, but I think we will skip for now Coefficients in Chemical Equations Back in chapter 3 you learned how to balance chemical equations. In this process you came up with a set of balancing coefficients, or numbers in front of every molecule that told you how many of each molecule you needed so that all the atoms that appeared in the reactant molecules were also present in the product molecules so that the mass of every atom was conserved These coefficients are also the key to our stoichiometry problems. For example let s start with the balanced equation for the burning butane: 2C4H 10(l) + 13O 2(g) 8CO 2(g) + 10H2O(g) The equation says that 2 molecules of C H 13 molecules O 8 molecules CO 10 molecules H O Which the same as 2 moles of C H 13 moles O 8 moles CO 10 moles H O This gives you a wealth of conversion factors. A cigarette lighter might have 5 grams of butane in it. How much O 2 is required to burn this amount of butane? First, how many moles of butane do I have What is the Molecular mass of butane? 4(12.01) + 10(1.01) = 58.24

9 9 5 g x 1 mole/58.24 g = moles Now from the above reaction we know that 2 moles 13 moles of Oxygen (or we also say is equivalent to). Now just as we did unit conversions, lets set this up as a proportionality 2 moles butane = 13 moles oxygen 2 moles butane/13 moles oxygen = 1 -or- 13 moles oxygen/2 mole butane =1 Now with these conversions, how do we convert moles of butane to moles of oxygen? moles butane X 13 moles oxygen=.5580 moles of oxygen 2 moles of butane Since O 2 has a molecular mass of 32.0, our weight of O 2 is.5580 moles O 2 x 32g O 2 = grams O2 1 mol O 2 Out of curiosity, lets continue this to calculate the amount of water made in this reaction. From our chemical reaction we know that 2 moles of butane 10 moles of water, so we have the proportion 2 mol butane/10 mol water = 1 -or- 10 mol water/2 mol Butane =1 If we have mol butane x 10 mol water =.429 moles of water 2 mol butane Molar mass of water= so.429 mole water x g water = grams of water 1 mole water Key Procedure: Stiochiometric calculations 1. Balance the equation 2. Convert the mass of given substance to moles 3. Use balanced equation to set up molar conversion factors 4. Use appropriate conversion to calculate the moles of substance you want 5. Use molecular mass of substance to convert moles into grams of substance.

10 Stoichiometry Figure 11.7 Puts the above procedure into a nice flow chart. But I can t seem to grab it from the computer Let s try this in another example; the thermite reaction. Used in rockets, underwater welding and incendiary bombs Fe2O 3 (s)+al(s) Fe(l) + Al2O 3(s) (Unbalanced) (review names iron(iii) oxide and Aluminum oxide (why no (III)?) If I want to generate 15 grams of Fe, how much Al and how much Fe2O 3 is needed Step 1. balance reaction Fe2O 3 (s)+2al(s) 2 Fe(l) + Al2O 3(s) Fe 2 2 Al 2 2 O 3 3 Fe2O 3 (s)+2al(s) 2 Fe(l) + Al2O 3(s) Step 2. Convert 15 g Fe to moles 15 g Fe X 1 mole Fe = moles Fe g Fe Step 3 Convert moles Fe to moles Al mol Fe X 2 mol Al = mole Al; 2 mol Fe Step 4 Convert mole Al to grams Al mole Al x g Al = 7.12 g Al 1 mol Al Step 3 for Fe O : Converting moles Fe to moles Fe O mol Fe X 1 mol Fe O = mol Fe O 2 mol Fe Step 4 Convert moles Fe2O 3 to g Fe2O mol Fe O X g Fe O = 21.5 g Fe O Mole Fe O

11 Stoichiometry without chemical equations Using the principal of conservation of mass, it is sometimes possible to do a stoichiometric calculation without a chemical equation! For example say I am going to react 10 g of sodium in water to make NaOH If NaOH is the only product of this reaction that contains sodium, then the principle of conservation of mass says that all the sodium I start off with has to appear in the NaOH product. So Na NaOH So, not even knowing the reaction I can do my stoichiometric calculation: 10g Na x 1 mole Na 23 g Na =.435 mole Na.435 mole Na x 1 mole NaOH =.435 mol NaOH 1 mole NaOH.435 mol NaOH x 40 g NaOH = 17.4 g NaOH 1 mol NaOH Limiting Reactant When two or more substances react the mass of the product is determined by the limiting reactant. Let s go back to the thermite reaction. When we left that problem we had determined that 7.12 g of Al and 21.5 g of Fe2O 3 were required to make 15 of Fe using this reaction What would happen if we used 15 g of Al and 21.5 g of Fe2O 3, would we make any more product? NO. Adding excess of one reactant doesn t give you more product, because the other reactant limits the amount of product that can be made. Key Concept: In a chemical reacion with more that one reactant, usually one reactant, called the limiting reactant, will limit the amount of product made in the reaction. Other reactants, that are present in excess are called excess reactants. You will often be given problems like this, where you are give the gram amounts of two reactants and you need to figure out which one is the limiting reactant and which one is the excess reactant.

12 12 So let me give you a procedure to figure out which is which Key Procedure and definition: Finding the limiting reactant 1. Pick a product and stick with it for the rest of the problem 2. Calculate how many moles of this product you can get from reactant A alone 3. Calculate how many moles of this product you can get from reactant B alone C, D, E??? The reactant that give you the SMALLEST product is the limiting reactant. The other reactants are all excess reactants. The amount of product obtained with this limiting reactant is the Theoretical Yield Let s try this with a variation of our butane lighter problem. If I start the butane lighter with 0.5 grams of butane, and place it in a 5 gallon jug and seal the jug, which will run out first, the air in the jug or the lighter fuel? We already calculated that 0.5 grams of butane is moles of butane. And we know from the balanced equation 2C H (l) + 13O (g) 8CO (g) + 10H O(g) That 2 moles butane 13 moles O 8 moles CO 10 moles H O Now I will skip ahead a few steps on how to go from 5 gallons of air to moles of O, and just give you the final answer that 5 gallons of air contains.187 moles of 2 O 2 Step 1 pick a product I will pick CO 2 Step 2 calculate amount for product given by reactant A.0086 m butane x 8 mol CO =.0344 moles CO 2 mol butane Step 3 calculate amount for product given by reactant B.187 m O 2 x 8 mol CO 2 =.115 mole CO2 13 m O Butane is the limiting reagent because it makes the smallest amount of product. O is the excess reagent because you have an excess amount of it. 2

13 Which product do you choose? It doesn t matter, and the result will be the same. However, on a test, the next question is usually how much is your expected yield of X so if you chose the reagent that answers that question, you are one step closer to answering that question Percentage Yield In the above section I talked about the theoretical yield. That the amount of yield you calculate from the limiting reactant. You actual yield almost never matches this theoretical value. Why? Reaction fails Reaction not complete Side reaction that gives other products You may lose some product by accident Chemists like to keep track of how close their yield is to the theoretical with a calculation called the % Yield Key Calculation: % yield = (actual yield/theoretical yield) x 100 % Example: Going back to the previous example. If our actual yield was.0301 moles of CO and our theoretical yield was.0344 mole of CO, what is our % yield?.0301/.0344 x 100 = 87.5% Note 1: You can calculate % yield based on grams of product or moles of product. It doesn t matter which one you use they should give you the same number of you do the calculation correctly. Note 2: You might think that the % yield should always be less than 100% but guess again. Sometimes it is >100%. How can that be? Sometimes when you purify your product, you accidentally purify some Kaka with it. Now the weight you record is higher than it should be because of the added Kaka