AP 9/21 TEST TODAY Turn in your review Pick up a periodic table As soon as we do good things, you will receive your test. You will have ONE hour to complete the test. Work must be shown in the space provide on your answer document. NO WORK=NO CREDIT. After the test, we will start equation writing. HW: lab report due Friday. NO EXCEPTIONS. review memorization stuff

BONUS #1: 2ZnS (s) + 3O 2(g)  2ZnO (s) + 2SO 2(g) If the typical yield is 86.78%, how much SO 2 should be expected if 4897 g of ZnS are used? BONUS #2

AP: 9/6 Today we will grade your quizzes in class (3 rd period will need to finish due to the assembly on Friday) You will need a calculator, a marker, your chapter 3 notes, and a blank piece of paper After we grade the quiz, we will cover section 3.1 and 3.2 Tomorrow we will cover sections 3.3, 3.4, and 3.7. Please bring your notes tomorrow. NO HW TONIGHT (YAY!). The remainder of the stoichiometry packet is due by Friday, I will give it back to you tomorrow.

Quick Assignment… Draw a mass spectrometer and label the parts What is this piece of equipment used for? Describe the five stages of this device. You will have no more than 15 minutes to finish this.

AP: 9/7 Today you will need the a blank piece of paper form the side table, a calculator, your chapter 3 notes Turn in your mass spectrometer HW We will cover sections 3.1, 3.2, 3.3, and 3.4 today HW: finish the NMSI packet 7-12 (due Friday) Lab on Friday (don’t get to excited, it’s probably the most boring lab you will perform in this class. It is meant to help you learn how to write a lab report)

Naming Practice On a piece of paper, answer the following: ▫1. Write the formula for the following compounds:  A. lead (II) sulfite  B. barium hydroxide  C. calcium chlorate  D. hydronitric acid ▫1. Name the following compounds:  A. HNO2  B. SnO2  C. Ag2CO3  D. N2O4

CHAPTER 3: STOICHIOMETRY AP CHEMISTRY

Chapter 3 Objective In this chapter, we will look at the quantities of materials consumed and produced in chemical reaction. The study of these quantities is called chemical stoichiometry.

3.1 Counting By Weighing If we were to take the individual masses of five nails, each nail would probably have a small deviation in mass, but would be relatively the same, let’s say an average mass of 2.5g per nail. If someone needed 1,000 nails for a building project, it is easier to take the mass of 1,000 nails, or 2,500g, than to count out all those nails. Objects do not have to have identical masses to be counted by weighing; they behave as if they are all identical. This is very lucky for chemists.

Average Atomic Mass % Isotope A (mass of A) + % Isotope B (mass of B) + …= avg. atomic mass Just like our nails deviated slightly, atoms of the same element may have different masses. We call these isotopes. Average Atomic Mass- an average of each isotope of an element, based on their percent abundance ▫Ex. Find the average atomic mass of element “X” if 1.40% of the sample is and isotope with a mass of amu, 24.10% is an isotope with a mass of amu, 22.10% is an isotope with a mass of amu and 52.40% is an isotope with a mass of amu ( ) ( ) ( ) ( ) = amu (probably lead)

3.2 Atomic Masses So if scientists can’t see atoms with their eyes, how do they know all these isotopes exist? They use a mass spectrometer. The mass spectrometer heats a sample of atoms into a gas, passes them through a beam of high- speed electrons, which knock off some of the atom’s electrons giving them positive charges, and then these positively charged atoms pass through an electric field. The atoms are deflected based on their size, and then detected on a computerized plate.

Figure 3.2 Neon Gas Note that these spikes on the readout reflect how much of each isotope is present in a sample.

Figure 3.3 (NMSI Exercise 1) When a sample of natural copper is vaporized and injected into a mass spectrometer, the results are shown. Use the data to calculate average atomic mass of natural copper. The mass values are Cu-63 (62.93) and Cu-65 (64.94).

Figure 3.3 Mass Spectrum of Natural Copper

Calculating the Mass of an Element Example. Use the mass spectrum shown on the left to calculate the average atomic mass of this element, then identify the element. Mass Percent Abundance (90) (91) (92) (94) (96) = or 90 Its zirconium!

3.3 The Mole The mole is the chemist’s way of counting atoms. Remember that the mole (or Avogadro’s number) is equal to x atoms of any element, which is then equal to the average atomic mass on the periodic table. The mole is standardized as the number of carbon atoms in 12.0g of pure carbon x amu = 1 g The mass of one mole of an element is equal to its atomic mass in grams.

Taken by Paul Jung (AP ’09) at the Golden Gate Bridge in San Francisco

Take out the blank piece of paper We are going to review AMG by drawing a mole map and the AMG chart

Ex. Cody found a gold nugget that had a mass of oz. How many moles was this? How many atoms? oz Au 1 lb g 1 mol Au 16 oz 1 lb g Au = moles moles Au x atoms 1 mol Au = x atoms

3.4 Molar Mass mass in grams of one mole of a substance may also be called molecular weight

Ex. Calculate the molecular mass of cisplantin, Pt(NH 3 ) 2 Cl 2. Pt 1 x = N 2 x = H 6 x 1.01 = 6.06 Cl 2 x = g or amu The compound cis-PtCl 2 (NH 3 ) 2 cisplatinum is a platinum-based chemotherapy drug used to treat various types of cancers, including sarcomas, some carcinomas (e.g. small cell lung cancer, and ovarian cancer), lymphomas and germ cell tumors.

Ex. How many grams are 3.25 moles of cisplantin? 3.25 mol cisplantin g 1 mol cisplantin = 975g

AP 9/12 Pick up the paper from the side table Take out your lab, a blank piece of paper, and pick up a calculator Today we will perform part 1 of your lab, practice writing a hypothesis, perform more AMG problems, and possibly practice percent composition problems HW tonight (MAYBE): molar conversions and percent composition (paper you picked up from side table)

AP 9/13 Take out your lab, your calculation HW, the molar conversion problems, and a calculator We will finish the lab today and then practice problems over mole conversions We will also go over percent composition and empirical and molecular HW tonight: Molar conversions WS

Calculate the following: 1. The number of moles AND the mass in grams of 3.25 X atoms of sodium phosphate. 2. The mass in grams of 5.85 X units of iron (II) phosphide. 3. The number of moles of grams of dinitrogen pentoxide. 4. The mass in grams of 4.86 moles of calcium carbonate. 5. For question #4, calculate the mass of the carbonate ions present.

NMSI #7 (new type of question) Isopentyl acetate (C7H14O2), the compound responsible for the scent of bananas, can be produced commercially. Interestingly, bees release about 1 X 10-6 g of this compound when they sting. The resulting scent attracts other bees to join the attack. (a) calculate the number of molecules of isopentyl acetate released in a typical bee sting. (b) Calculate the number of carbon atoms present

AP 9/14 You will need a calculator, the paper from the side table, a white board, a marker, your chpt 3, and your chapter 3 notes. Today we will cover % composition, empirical/molecular formula, and stoichiometric calculations Before we get started, we will pass back papers HW: molar conversion paper AND whatever problems we don’t finish in class from the paper you picked up today

Percent Composition ▫ “mass percent” or “percent by mass”

Ex. Find the percent composition of all elements in cisplantin, Pt(NH 3 ) 2 Cl 2. Pt x 100 = 65.01% N x 100 = 9.34% H 6.06 x 100 = 2.02% Cl x 100 = 23.63%

Practice: Find the percent composition of the following: ▫A. nitric acid ▫B. glucose ▫C. sodium phosphite ▫D. percent of bromine in zinc (II) bromide

Practice: NMSI 8 and 9

Empirical Formula Poem Percent to mass, Mass to moles, Divide by smallest, And round (or multiply) ‘till whole 3.7 Determining the Formula of a Compound Empirical formula- simplest whole number ratio of the various types of atoms in a compound Molecular formula- the exact formula or (empirical formula) x

Remember: The empirical formula is the true formula for ionic compounds but may not be for molecular compounds. The formula to calculate molecular formula is: ▫MF formula=MF mass/EF mass ▫The MF mass will be given

Conceptual Practice: Which of the following have NO molecular formula: ▫A. calcium sulfate ▫B. methanol (CH3OH) ▫C. aluminum sulfide ▫D. dinitrogen pentoxide

Ex. A sample of a compound contains 11.66g of iron and 5.01g of oxygen. What is the empirical formula of this compound? (NOTE: you can check your answer BC we know it can only have two forms) 11.66g Fe 1 mol Fe = mol Fe 55.85g Fe 5.01g O 1 mol O = mol O 16.00g O Fe:O : :1.5 or 2: Fe 2 O 3

Ex. What is the empirical formula of hydrazine, which contains 87.5% N & 12.5%H? What is the molecular formula if the gram formula mass is 32.07g? 87.5gN 1 mol N = mol N 14.01g N 12.5g H 1 mol H = mol H 1.01 g H N:H = : : 1.98 = 1 : NH 2

NMSI # 10, 11, 12

Combustion Analysis

Suppose you isolate an acid from clover leaves and know that it contains only the elements C, H, and O. Heating 0.513g of the acid in oxygen produces 0.501g of CO 2 and 0.103g of H 2 O. What is the empirical formula of the acid? Given that another experiment has shown that the molar mass of the acid is 90.04g/mol, what is its molecular formula? C + O 2  CO g CO 2 1 mol CO 2 1 mol C 12.01gC 44.01g CO 2 1 mol CO 2 1 mol C = g C in compound

2H 2 + O 2  2H 2 O 0.103g H 2 O 1 mol H 2 O 2 mol H 1.01g H 18.02g H 2 O 1 mol H 2 O 1 mol H = g H 0.513g cmpd - ( )= 0.364g O 0.364g O 1 mol O = mol O 16.00g O 0.137g C 1 mol C = mol C 12.01g C g H 1 mol H = mol H 1.01 g H

C:H:O : : : : : 1: :1:2 CHO 2 efm = = /45.02 = 2 C2H2O4C2H2O4

3.8 Chemical Equations During a chemical reaction, atoms are rearranged when existing bonds are broken and reformed in a different arrangement. Since atoms are not lost or gained during a reaction, a balanced equation must have the same number of atoms of each element on each side. Symbols representing physical states: (s) (l) (g) (aq)

Observe the diagram to the right. Balance the equation and then draw in the correct number of expected products.

3.9 Balancing Chemical Equations We balance equations by adding coefficients, never by changing formulas. Most equations can be balanced by inspection. Some redox reactions require a different method.

Ex. Al(s) + Cl 2 (g)  AlCl 3 (s) 2Al(s) + 3 Cl 2 (g)  2AlCl 3 (s) N 2 O 5 (s) + H 2 O(l)  HNO 3 (l) N 2 O 5 (s) + H 2 O(l)  2 HNO 3 (l) Pb(NO 3 ) 2 (aq) + NaCl(aq)  PbCl 2 (s) + NaNO 3 (aq) Pb(NO 3 ) 2 (aq) + 2 NaCl(aq)  PbCl 2 (s) +2 NaNO 3 (aq) Let’s Practice! Try these on your own!

PH 3 (g) + O 2 (g)  H 2 O(g) + P 2 O 5 (s) 2PH 3 (g) + 4O 2 (g)  3H 2 O(g) + P 2 O 5 (s) You should be able to write and balance an equation when given the reactants and products in words. Ex. Phosphine, PH 3 (g) is combusted in air to form gaseous water and solid diphosphorus pentoxide. Try this one!

NH 3 (g) + Na(l)  H 2 (g) + NaNH 2 (s) 2NH 3 (g) + 2Na(l)  H 2 (g) + 2NaNH 2 (s) When ammonia gas is passed over hot liquid sodium metal, hydrogen is released and sodium amide, NaNH 2, is formed as a solid product.

AP 9/16 Turn in the molar conversion homework BUT NOT the other 3 stoichiometry problems. You will need your chapter 3 notes, a calculator, and something to write with We are practicing stoichiometric calculations today. Grab a marker and whiteboard if you prefer to do your work on that. HW: Stoichiometry practice you picked up today. Only do # 1,2,3,5,7,8,10,12,13 NEXT WEEK: Monday we will go over limiting and excess reactants. Tuesday we will go over combustion analysis. Wednesday you will have your first test.

3.10 Stoichiometric Calculations Although coefficients in balanced chemical equations tell us how many atoms will react to form products, they do not tell use the actual masses of reactants and products we will use or expect to produce. For this, we need to relate the reactants and products in terms of their mole ratios. ▫The mole ratio = moles required/moles given.

Remember: THE MOLE RATIO IS USED IN STOICHIOMETRY. There are four types of stoichiometric problems: ▫MOLE TO MOLE ▫MOLE TO MASS ▫MASS TO MOLE ▫MASS TO MASS

PRACTICE 1.For the following reaction write the mole ratio for lead (II) nitrate and sodium nitrate Pb(NO 3 ) 2 (aq) + 2 NaCl(aq)  PbCl 2 (s) +2 NaNO 3 (aq) 2. Write the mole ratio of aluminum chloride to chlorine gas 2Al(s) + 3 Cl 2 (g)  2AlCl 3 (s)

Mole to mole Mass given  mole given  mole unknown  mass unknown Given quantity is in moles Unknown quantity is in moles Need 1 conversion factor to solve ▫Mole ratio to convert between mole given & mole unknown

Practice: Ammonia is formed from the synthesis of nitrogen gas and hydrogen gas in the Haber process. How many moles of hydrogen are needed to react with 2.00 mol of nitrogen?

Practice: Hydrogen gas reacts with oxygen to synthesize water. If how many moles of water will be produced from 5.52 moles of hydrogen?

Mass to mole Mass given  mole given  mole unknown  mass unknown Given quantity is a mass (g, kg, etc.) Unknown quantity is in moles Need 2 conversion factors to solve: ▫Molar mass to convert from mass given to mole given ▫Mole ratio to convert from mole given to mole unknown

Practice: Sodium hydroxide neutralized sulfuric acid to form a salt and water. How many moles of sodium hydroxide are needed to form 23.0 g of the salt?

Mole to mass Mass given  mole given  mole unknown  mass unknown Given quantity is in moles Unknown quantity is mass (g, kg, etc.) Need 2 conversion factors to solve: ▫Mole ratio to convert from mole given to mole unknown ▫Molar mass to convert from mole unknown to mass unknown

Practice: Hydrogen peroxide decomposes to form water and oxygen gas. This reaction proceeds faster when a catalyst is added. How many grams of water will be produced from 2.25 moles of H 2 O 2 ?

Mass to mass Mass given  mole given  mole unknown  mass unknown  Given quantity is a mass (g, kg, etc.)  Unknown quantity is a mass  Need 3 conversion factors to solve: Molar mass to convert from mass given to mole given Mole ratio to convert from mole given to mole unknown Molar mass to covert from mole unknown to mass unknown

Practice: What mass of NH 3 is formed when 5.38g of Li 3 N reacts with water according to the equation: Li 3 N(s) + 3H 2 O  3LiOH(s) + NH 3 (g)

Ex. What mass of NH 3 is formed when 5.38g of Li 3 N reacts with water according to the equation: Li 3 N(s) + 3H 2 O  3LiOH(s) + NH 3 (g)? 5.38g Li 3 N 1 mol Li 3 N 1 mol NH g NH g Li 3 N 1 mol Li 3 N 1 mol NH 3 = 2.63g NH 3

AP 9/19 Pick up the packet from the side table, turn in your stoichiometry HW. You will need a calculator, something to write with, and possibly your chapter 3 notes. Today we will review a few stoichiometry problems and then go over limiting/excess and theoretical yield.

On a piece of paper, work the following: 1.If excess sulfuric acid reacts with 30.0 g of sodium chloride in a double replacement reaction, how many grams of hydrochloric acid are produced? 2. How many moles are required to replace 4.00 g of silver (I) nitrate which is dissolved in water. 3. C 6 H 6(l) + Cl 2(g)  C 6 H 5 Cl (s) + HCl (g) When 36.8g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8g. What is the percent yield?

3.11 The Concept of Limiting Reagent Limiting reagent- reagent that restricts or determines the amount of product that can be formed If in a stoichiometry problem you are given amounts of two or more reactants and asked to determine how much product forms, you must determine which reactant is limiting and use it to work the problem.

Which is limiting? Excess? To decide which reactant is limiting in a rxn, use one of your givens to solve for the other. Then compare how much you have (given) to how much you need under ideal conditions (solved for). ▫If you have more than you need, the 1 st substance is limiting. ▫If you have less than you need, the 2 nd substance is limiting.

Limiting reactant example CO (g) + 2H 2(g)  CH 3 OH If 500. mol of CO and 750. mol of H 2 are present, which is the limiting reactant? ▫Solve to determine how much H 2 would be needed to completely react 500. mol CO. ▫Do you have 1000 mol H 2 ? No, you only have 750. mol. ▫H 2 is the limiting reactant.

Limiting reactant example #1 3ZnCO 3(s) + 2C 6 H 8 O 7(aq)  Zn 3 (C 6 H 5 O 7 ) 2(aq) + 3H 2 O (l) + 3CO 2(g) If there is 1 mol of ZnCO 3 & 1 mol of C 6 H 8 O 7, which is the limiting reactant? Answer: ▫1 mol ZnCO 3 could react with 0.67 mol C 6 H 8 O 7, which is less than is available. ▫ZnCO 3 is limiting.

Pre-AP 9/20 Turn in your HW Pick up the test review Take out a piece of paper Today you will review for the test by performing a gallery walk on the back tables. You will answer those questions on the blank piece of paper. Further instructions will be given once class starts. HW: answer the questions on the test review. You will take the atomic structure test on Thursday.

9/20 AP Pick up the review from the side table and take out the limiting and excess problem you were asked to calculate for HW. Today we will go over that problems and then you will have time in class to review. HW: The stoichiometry “story packet” is due Monday. You only need to do three problems of your choice. However, I would read the information in the packet. LAB REPORT IS DUE FRIDAY. NO EXCPETIONS

Limiting reactant example #2 Aspirin, C 9 H 8 O 4, is synthesized by the rxn of salicylic acid, C 7 H 6 O 3, with acetic anhydride, C 4 H 6 O 3. 2C 7 H 6 O 3 + C 4 H 6 O 3  2C 9 H 8 O 4 + H 2 O When 20.0g of C 7 H 6 O 3 and 20.0g of C 4 H 6 O 3 react, which is the limiting reactant? How many moles of the excess reactant are used when the rxn is complete? What mass in grams of aspirin is formed?  C 7 H 6 O 3, mol, 26.1 g

Once we know how much of a product if formed in a reaction, we can find the percent yield of the reaction. Theoretical yield- amount of product that should form according to stoichiometric calculations Percent yield = Actual yield x 100 Theoretical yield Actual yield- experimental yield

Percent Yield example C 6 H 6(l) + Cl 2(g)  C 6 H 5 Cl (s) + HCl (g) When 36.8g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8g. What is the percent yield? Given: 36.8g C 6 H 6, excess Cl 2, actual yield=38.8g C 6 H 5 Cl Unknown: ? g C 6 H 5 Cl, percent yield

Percent Yield example Methanol can be produced through the rxn of CO and H 2 in the presence of a catalyst. If 75.0 g of CO reacts to produce 68.4 g CH 3 OH, what is the percent yield? Answer: 79.7%

Percent Yield example 2ZnS (s) + 3O 2(g)  2ZnO (s) + 2SO 2(g) If the typical yield is 86.78%, how much SO 2 should be expected if 4897 g of ZnS are used?

Ex. How many moles of Fe(OH) 3 (s) can be produced by allowing 1.0 mol Fe 2 S 3, 2.0 mol H 2 O and 3.0 mol O 2 to react? 2Fe 2 S 3 (s) + 6H 2 O(l) + 3O 2 (g)  4Fe(OH) 3 (s) +6S(s) 1.0 mol 2.0 mol 3.0 mol limiting In this case, we can see that H 2 O gave use the smallest amount of product, so it is the limiting reagent. Thus, we use that answer as our overall answer. 1.0 mol Fe 2 S 3 4 mol Fe(OH) 3 = 2.0 mol Fe(OH) 3 2 mol Fe 2 S mol H 2 O 4 mol Fe(OH) 3 = 1.3 mol Fe(OH) 3 6 mol H 2 O 3.0 mol O 2 4 mol Fe(OH) 3 = 4.0 mol Fe(OH) 3 3 mol O 2

Ex. If 17.0g of NH 3 (g) were reacted with 32.0g of oxygen in the following reaction, how many grams of NO(g) would be formed? 4NH 3 (g) + 5O 2 (g)  4NO(g) + 6H 2 O(l) 17.0g 32.0g Xg 17.0g NH 3 1 mol NH 3 = 1 mol NH g NH g O 2 1 mol O 2 = 1 mol O g O 2 We need 5 mol O 2 to every 4 mol NH 3, so O 2 is limiting. 1 mol O 2 4 mol NO g NO = 24.0g NO 5 mol O 2 1 mol NO

Ex. In the reaction of 1.00 mol of CH 4 with an excess of Cl 2, 83.5g of CCl 4 is obtained. What is the theoretical yield, actual yield and % yield? Actual yield = 83.5g CCl 4 CH 4 + 2Cl 2  CCl 4 + 2H mol CH 4 1 mol CCl g CCl 4 1 mol CH 4 1 mol CCl 4 = 154g CCl 4 (theoretical yield) 83.5 x 100 = 54.2% yield 154