Chapter 6 Chemical Reactions: Mole and Mass Relationships

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1 Chapter 6 Chemical Reactions: Mole and Mass Relationships 6.1 The Mole and Avogadro s What is a Mole? - A Chemist s way of counting! - Cooks don t count out individual grains of sugar or rice when they are cooking. Instead they measure out bulk quantities in more convenient units, like cups or tablespoons. - Atoms are much, much smaller than grains of sugar, so chemists take a similar approach. - Masses of individual atoms are very small, in the amu range - We want a group size of atoms or molecules that would be of an easily measured mass and would be convenient to use. - group sizes used in other industries include: shoes pair eggs dozen soda case paper ream It is common to give a name to a number. Usually, the smaller the item measured, the larger the group size. - Chemist use a group size named a. A mole is officially defined to be equal to the number of C atoms in exactly The number of particles in a mole is N, Avogadro's Number Ch 06 Page 1

2 How big is Avogadro s Number? If you had Avogadro's number of grains of sand, you could spread them out over the entire state of California and have a layer tall! If you had Avogadro s number of sheets of paper and divided them into a million equal piles, each would stretch from here to! If Avogadro s number of were distributed equally among all the people on earth (~4 billion), each would have enough money to spend a million dollars every hour, day and night, throughout his lifetime and they could not spend even half of it! - How big is a mole of Cu? (i.e x10 23 atoms of Cu?) - Does it surprise you that you can hold Avogadro s number of atoms in the palm of your hand? - Why would chemists pick such a big oddball number? A mole (abbreviated mol) is also the amount of any substance whose mass, in grams, (known as its molar mass) is numerically equal to its formula weight. If one molecule of water weighs 18.0 amu then one mole of water molecules weight We say that the molar mass = Ch 06 Page 2

3 - Example: 1 mole of Aluminum weighs 1 mole of NaCl weighs - The weight of one molecule is known as the. - The weight of one formula unit of an ionic substance is a. - The weight of one mole of a substance is known as its. Problem: Determine the molar mass of Cu(NO 3 ) 2. Ch 06 Page 3

4 6.2 Gram-Mole Conversions Conversions: Molar mass serves as a conversion factor between and. If two things are equal to each other, we can divide one by the other to get a factor equal to 1. (Anything divided by itself = ) For water we can develop two conversion factors: 18.0g H 2 O and 1 mol H 2 O 1 mol H 2 O and 18.0 g H 2 O At any given time, we will use the one that makes the units cancel! Problem: How many moles of NaCl are present in 48.2 g NaCl? Problem: What is the mass of 1.50 mol of sodium carbonate? How many formula units are present? Ch 06 Page 4

5 Problem How many moles of cytosine are in 253 g of cytosine? O N NH The structure for cytosine is: H 2 N Problem: How many molecules of oxygen are there in 10.8 g oxygen? How many individual atoms are present? How many oxygen atoms are in the same sample? Ch 06 Page 5

6 6.3 Mole Relationships and Chemical Equations (aka Stoichiometry) For the reaction: 3 H 2 + N 2 2 NH 3 Up until now we interpreted this as molecules of H 2 react with molecule of N 2 to produced molecules of NH 3. Now we can also interpret this as moles of H 2 will react with mole of N 2 to produce moles of NH 3. Stoichiometric coefficients and reaction quantities ONLY WORK IN MOLES!! (not grams) THE REACTION ARROW SHOULD BE CONSIDERED A MOLE CROSSING!!! These coefficients can be used to set up ratios of moles of reactants and/or products. There are 6 conversion factors that can be made using the stoichiometric factors of this equation. 3 mol H2 or 3 mol H2 or 1 mol N2. 1 mol N2 2 mol NH3 2 mol NH3 Ch 06 Page 6

7 6.4 Mass Relationships and Chemical Equations This is sometimes hard to grasp when you start right off with a chemical application. Let s try it first with a non-chemical application Assume that the following relationship exists: 1 steering wheel + 4 tires + 2 headlights 1 car Problem: If you have 8 tires, how many cars can you make? Problem: But what if you got a delivery of 1000 lb container of tires? How many cars can you make? You need to know: If each tire weighs 50 lb then Now let s relate this to chemistry. Mole-to-Mole Conversions This is like conversions between tires and cars. We need to know the relationship between the items. These are carried out using mole ratios as conversion factors. Ch 06 Page 7

8 Problem: Rusting involves the reaction of iron with oxygen to form iron(iii) oxide. 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) (a) What are the mole ratios of the product to each reactant and of the reactants to each other? (b) How many moles of iron(iii) oxide are formed by the complete oxidation of 6.2 mol of iron? Start by writing down the known information (6.2 mol of iron), and select the mole ratio that allows the quantities to cancel, leaving the desired quantity. Problem: Using the equation 3 H 2 + N 2 2 NH 3 how many moles of hydrogen does it take to fully react with 3.0 moles of nitrogen? How many moles of ammonia are produced? Ch 06 Page 8

Mass to Mole and Mole-to-Mass Conversions This is like conversions between lb of tires and cars. We need to know the relationship between the items.

9 Mass to Mole and Mole-to-Mass Conversions This is like conversions between lb of tires and cars. We need to know the relationship between the items. We needed to add the knowledge of how much a tire weighed. Now we need to add the knowledge of how much a mole weighs. Problem: How many moles of ammonia can be produced from the reaction of 58g of N 2 with excess H 2? We need 2 pieces of information: a) the mole ratio between b) the molar mass of Start by listing the known value on the left, then set up ratios that allow units to cancel. Problem: How many grams of ammonia can be produced from the reaction of 2.00 moles of H 2 with excess N 2? We need 2 pieces of information: a) the mole ratio between b) the molar mass of Start by listing the known value on the left, then set up ratios that allow units to cancel. Ch 06 Page 9

Mass-to-Mass Conversions This is the type of information that is what we usually want to know, but we can't get it directly from the chemical equation because EQUATIONS ONLY WORK IN MOLES, NOT IN

10 Mass-to-Mass Conversions This is the type of information that is what we usually want to know, but we can't get it directly from the chemical equation because EQUATIONS ONLY WORK IN MOLES, NOT IN GRAMS (mass)!!!! Our equations only work in specified units of measure. If we get other units given, we need to convert first. Imagine in our problem with the cars that we needed to know what the final weight of the cars produced so that we could figure out how we could ship them. Now in addition to knowing the weight of each reactant (tires) we also need to know the weight of each product (car). Assuming each car weighs 3200 lb 1000 lb tire x 1 tire = 20 tires x 1 car = 5 cars x 3200 lb = 16,000 lb 50 lb tire 4 tires 1 car We can work out the same problem without calculating the intermediate results lb tire x 1 tire x 1 car x 3200 lb = 16,000 lb 50 lb tire 4 tires 1 car The same is true for chemical equations. They only work in moles. If we are given weights (grams) then we must first convert to moles. Show example using arrows with grams above and moles below. Ch 06 Page 10

11 Problem: All alkali metals react with water to produce hydrogen gas and the corresponding alkali metal hydroxide. A typical reaction is that between lithium and water: The unbalanced equation is Li (s) + H 2 O (l) LiOH (aq) + H 2(g) The balanced equation is Li (s) + H 2 O (l) LiOH (aq) + H 2(g) How many grams of H 2 will be formed by the complete reaction of 80.57g of Li with water? Ch 06 Page 11

There is also the possibility that the reactant or product may be given in terms of molecules or formula units. In that case we need to add to our concept map. 6.

One is usually present in excess. In the previous problem, it was obvious that water was in excess, and the reaction would stop when we ran out of lithium.

12 There is also the possibility that the reactant or product may be given in terms of molecules or formula units. In that case we need to add to our concept map. 6.5 Limiting Reagent and Percent Yield When reagents are mixed to start a reaction, we almost never have exactly the right amounts to exactly use up all of the reactants. One is usually present in excess. In the previous problem, it was obvious that water was in excess, and the reaction would stop when we ran out of lithium. The limiting reagent is the that runs out first. Sometimes the Limiting Reagent (L.R.) is not so obvious. To find the L.R., find how much of any one, you could form from each. Ch 06 Page 12

13 The one that can produce the, product is the limiting reagent. The actual amount of product that could be formed is based on the. Problem: Imagine we were making cheese sandwiches. Each sandwich needs 2 slices of bread and 1 slice of cheese. If we have 4 loafs of bread (with 22 slices each) and 3 packages of cheese (with 16 slices each), A) Which will run out first (limiting reagent) B) How many sandwiches can I make? Problem: If 2.0 g of H 2 is reacted with 8.0 g N 2 according to the unbalanced reaction H 2 (g) + N 2 (g) NH 3 (g) A) which is the limiting reagent and B) how many grams of ammonia can be formed? Ch 06 Page 13

14 Theoretical & Percent Yield In the real world, most reactions do not occur with 100% efficiency. The amount of product produced is usually somewhat less than that predicted by limited reagent stoichiometric calculations. Theoretical yield = predicted amount of product. Percent yield = the percent of theoretical actually obtained Non-chemical example Let s think of this again of a non-chemical application to start with. Assume I pass out a survey to 50 students. The theoretical yield would be 50 returned surveys. In reality, I am likely to get less than 50 returned, (not everybody participates). If I get 45 returned surveys, the percent yield on my survey would be: 45/50 x 100 = 90% Problem: In the previous problem we predicted that 9.7g of NH 3 should form. The product produced was measured and only 8.3 g was obtained. Calculate the percent yield for the reaction. Ch 06 Page 14

15 Problem: The reaction of ethylene (H 2 C=CH 2 ) with water to give ethyl alcohol (CH 3 CH 2 OH) occurs at 78.5% actual yield. How many grams of ethyl alcohol are formed by reaction of 25.0 g of ethylene? ethylene, MM = 28.0 g/mol, ethyl alcohol, MM = 46.0 g/mol. Plan of attack: Write a balanced equation, and then treat this as a typical mass-mass relationship problem to find the amount of ethyl alcohol that can theoretically be formed from 25.0 g of ethylene, and then multiply the answer by 78.5% to find the amount actually formed. H 2 C=CH 2 + H 2 O CH 3 CH 2 OH Solution: The theoretical yield of ethyl alcohol is So, the actual yield is as follows. Ch 06 Page 15

16 Problem A 0.270g sample of "Fool's Gold" (iron pyrite) was heated in the presence of excess oxygen. The unbalanced equation is FeS2(s) + O2(g) Fe3O4 + SO2(g) a) What mass of each product was formed, assuming that all of the iron sulfide reacted? Start by writing a balanced equation. FeS2(s) + O2(g) Fe3O4(s) + SO2(g) Since equations only work in moles, we will need MM conversion factors. FeS 2 ( g/mol) O 2 (32.00 g/mol) Fe 3 O 4 ( g/mol) SO 2 (64.07 g/mol) Mass of SO 2 produced: Mass of Fe 3 O 4 produced. b) If you performed the above experiment but only obtained 0.200g of sulfur dioxide, what was the percent yield for the reaction? % Yield = Actual Yield x 100 Theoretical Yield 1 c) If you started with 0.40 moles of FeS 2 and there is a 69.4% yield, how much sulfur dioxide will be produced? Ch 06 Page 16

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction.

Stoichiometry is the relationship between the amount of reactants used and the amount of products produced in a chemical reaction. Unit 7 STOICHIOMETRY 1. Introduction to Stoichiometry 2. Mole Mole Stoichiometry 3. Mass Mole Stoichiometry 4. Mass Mass Stoichiometry 5. Mass Volume & Volume Volume Stoichiometry 6. Excess & Limiting