CHAPTER 3 STOICHIOMETRY AP CHEMISTRY Chapter 3 Objective

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. 5 g 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, 500 g, 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 203. 973 amu, 24. 10% is an isotope with a mass of 205. 9745 amu, 22. 10% is an isotope with a mass of 206. 9759 amu and 52. 40% is an isotope with a mass of 207. 9766 amu. 0. 0140(203. 973) + 0. 2410(205. 9745) + 0. 2210(206. 9759) + 0. 5240(207. 9766) = 207. 21 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 highspeed 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 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 90 91 92 94 96 Percent Abundance 50 11 18 19 2 0. 50(90) + 0. 11(91) + 0. 18(92) + 0. 19(94) + 0. 02(96) = 91. 35 or 90 Its zirconium!

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Things of note… Good things! Less Good Things • Great job of reading and following instructions! • Staying on task and working productively with partners! • Paying attention to what is needed and what you have available. • Working safely!! • Cleaning up. • Leaving supplies out, and glassware and tables dirty.

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 6. 022 x 1023 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. 01 g of pure carbon-12. • 6. 022 x 1023 amu = 1 g • The mass of one mole of an element is equal to its atomic mass in grams.

Ex. Cody found a gold nugget that had a mass of 1. 250 oz. How many moles was this? How many atoms? 1. 250 oz Au 1 lb 453. 59 g 1 mol Au 16 oz 1 lb 196. 97 g Au = 0. 1799 moles Au 6. 022 x 1023 atoms 1 mol Au = 1. 083 x 1023 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 N H Cl 1 x 195. 08 = 195. 08 2 x 14. 01 = 28. 02 6 x 1. 01 = 6. 06 2 x 35. 45 = 70. 90 300. 06 g or amu The compound cis-Pt. Cl 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 = 975 g 300. 06 g 1 mol cisplantin

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 195. 08 x 100 = 65. 014% 300. 06 N 28. 02 x 100 = 9. 338% 300. 06 H 6. 06 x 100 = 2. 02% 300. 06 Cl 70. 90 x 100 = 23. 63% 300. 06

3. 7 Determining the Formula of a Compound Empirical Formula Poem • Empirical formulasimplest whole number ratio of the various types of atoms in a compound • Molecular formula- the exact formula or (empirical formula)x Percent to mass, Mass to moles, Divide by smallest, And round (or multiply) ‘till whole

Ex. A sample of a compound contains 11. 66 g of iron and 5. 01 g of oxygen. What is the empirical formula of this compound? 11. 66 g Fe 1 mol Fe = 0. 2088 mol Fe 55. 85 g Fe 5. 01 g O 1 mol O = 0. 3131 mol O 16. 00 g O Fe: O 0. 2088 : 0. 3131 1: 1. 5 or 2: 3 0. 2088 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. 07 g? 87. 5 g. N 1 mol N = 6. 246 mol N 14. 01 g N 12. 5 g H 1 mol H = 12. 38 mol H 1. 01 g H N: H = 6. 246 : 12. 38 1 : 1. 98 = 1 : 2 6. 246 NH 2 EFM = 14. 01 + 2(1. 01) = 16. 03 g MFM/EFM = 32. 07 g/16. 03 g = 2 MF = N 2 H 4

Combustion Analysis

Suppose you isolate an acid from clover leaves and know that it contains only the elements C, H, and O. Heating 0. 513 g of the acid in oxygen produces 0. 501 g of CO 2 and 0. 103 g 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. 04 g/mol, what is its molecular formula? C + O 2 CO 2 0. 501 g CO 2 1 mol C 12. 01 g. C 44. 01 g CO 2 1 mol C = 0. 1367 g C in compound 2 H 2 + O 2 2 H 2 O 0. 103 g H 2 O 1 mol H 2 O 2 mol H 1. 01 g H 18. 02 g H 2 O 1 mol H = 0. 01155 g H

0. 513 g cmpd - (0. 137 + 0. 012)= 0. 364 g O 1 mol O = 0. 02275 mol O 16. 00 g O 0. 137 g C 1 mol C = 0. 01141 mol C 12. 01 g C 0. 01155 g H 1 mol H = 0. 01144 mol H 1. 01 g H C: H: O 0. 01141: 0. 01144 : 0. 02275 0. 01141 1: 1: 1. 98 = 1: 1: 2 CHO 2 efm = 12. 01 + 1. 01 + 32. 00 = 45. 02 90. 04/45. 02 = 2 C 2 H 2 O 4

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.

Let’s Practice! Try these on your own! Ex. Al(s) + Cl 2 (g) Al. Cl 3(s) 2 Al(s) + 3 Cl 2 (g) 2 Al. Cl 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) + Na. Cl(aq) Pb. Cl 2(s) + Na. NO 3(aq) Pb(NO 3)2(aq) + 2 Na. Cl(aq) Pb. Cl 2(s) +2 Na. NO 3(aq)

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! PH 3(g) + O 2(g) H 2 O(g) + P 2 O 5(s) 2 PH 3(g) + 4 O 2(g) 3 H 2 O(g) + P 2 O 5(s)

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

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.

Ex. What mass of NH 3 is formed when 5. 38 g of Li 3 N reacts with water according to the equation: Li 3 N(s) + 3 H 2 O 3 Li. OH(s) + NH 3(g)? 5. 38 g Li 3 N 1 mol NH 3 17. 04 g NH 3 34. 83 g Li 3 N 1 mol NH 3 = 2. 63 g NH 3

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.

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? 2 Fe 2 S 3(s) + 6 H 2 O(l) + 3 O 2(g) 4 Fe(OH)3(s) +6 S(s) 1. 0 mol 2. 0 mol limiting 3. 0 mol 1. 0 mol Fe 2 S 3 4 mol Fe(OH)3 2 mol Fe 2 S 3 2. 0 mol H 2 O 4 mol Fe(OH)3 6 mol H 2 O 3. 0 mol O 2 4 mol Fe(OH)3 3 mol O 2 = 2. 0 mol Fe(OH)3 = 1. 3 mol Fe(OH)3 = 4. 0 mol Fe(OH)3 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.

Ex. If 17. 0 g of NH 3(g) were reacted with 32. 0 g of oxygen in the following reaction, how many grams of NO(g) would be formed? 4 NH 3(g) + 5 O 2(g) 4 NO(g) + 6 H 2 O(l) 17. 0 g 32. 0 g Xg 17. 0 g NH 3 1 mol NH 3 = 1 mol NH 3 17. 0 g NH 3 32. 0 g O 2 1 mol O 2 = 1 mol O 2 32. 0 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 30. 01 g NO = 24. 0 g NO 5 mol O 2 1 mol NO

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 • Theoretical yield • • Actual yield- experimental yield x 100

Ex. In the reaction of 1. 00 mol of CH 4 with an excess of Cl 2, 83. 5 g of CCl 4 is obtained. What is theoretical yield, actual yield and % yield? Actual yield = 83. 5 g CCl 4 CH 4 + 2 Cl 2 CCl 4 + 2 H 2 1. 00 mol CH 4 1 mol CCl 4 153. 81 g CCl 4 1 mol CH 4 1 mol CCl 4 = 154 g CCl 4 (theoretical yield) 83. 5 x 100 = 54. 2% yield 154

Let’s Talk About Monday’s Lab! • Lab reports will be written up in bound laboratory notebooks that produce a carbonless copy of each page. • A copy will be torn out and turned in for grading, leaving you with a copy of each lab in your book. Some colleges and universities want to see your lab notebook before granting credit for chemistry lab classes. Therefore, it is important that you keep a neat, accurate representation of your laboratory work. • When labs are returned, you should make any major corrections on your copy of the original so that your lab notebook is as accurate as possible.

LAB FORMAT • All lab reports must be handwritten in blue or black ink. All errors should be drawn through with a single line. No whiteout! • Each lab should include the following numbered parts: ▫ ▫ ▫ ▫ ▫ Title, your name, date Objective (purpose of the lab) Procedure (Do not copy the lab handout. Simplify!) Pre-lab Questions (These are usually found at the end of the lab. (Write the question and the answer. ) Data Tables, if applicable (Any tables needed should be drawn before entering the lab area. ) Observations (Record all observations directly into your lab book at the time of observation) Calculations, if applicable (show all work!) Questions (Write question and answer to any end of experiment questions. ) Conclusion (This is the most important section! It includes your interpretation of what took place in the lab. Discuss your results (including any unknowns). Include errors and how they affected your results. ) • Depending on the lab, there will often be sections that must be written up before you can enter the lab! If this is not done, you will receive a point deduction from the overall lab grade and will not be able to do the lab during the class.