1. CHAPTER 3: STOICHIOMETRY AP CHEMISTRY

2. 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. 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.

4. 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)

5. 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.

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

8. Figure 3.3 Mass Spectrum of Natural Copper

9. 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 50 91 11 92 18 94 19 96 2 0.50(90) + 0.11(91) + 0.18(92) + 0.19(94) + 0.02(96) = 91.35 or 90 Its zirconium!

10. 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 10 23 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-12. 6.022 x 10 23 amu = 1 g The mass of one mole of an element is equal to its atomic mass in grams.

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

12. 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 453.59g 1 mol Au 16 oz 1 lb 196.97g Au = 0.1799 moles 0.1799 moles Au 6.022 x 10 23 atoms 1 mol Au = 1.083 x 10 23 atoms

13. 3.4 Molar Mass mass in grams of one mole of a substance may also be called molecular weight

14. Ex. Calculate the molecular mass of cisplantin, Pt(NH 3 ) 2 Cl 2. Pt 1 x 195.08 = 195.08 N 2 x 14.01 = 28.02 H 6 x 1.01 = 6.06 Cl 2 x 35.45 = 70.90 300.06 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.

15. Ex. How many grams are 3.25 moles of cisplantin? 3.25 mol cisplantin 300.06g 1 mol cisplantin = 975g

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

17. Ex. Find the percent composition of all elements in cisplantin, Pt(NH 3 ) 2 Cl 2. Pt 195.08 x 100 = 65.01% 300.06 N 28.02 x 100 = 9.34% 300.06 H 6.06 x 100 = 2.02% 300.06 Cl 70.90 x 100 = 23.63% 300.06

18. 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

19. Ex. A sample of a compound contains 11.66g of iron and 5.01g of oxygen. What is the empirical formula of this compound? 11.66g Fe 1 mol Fe = 0.2088 mol Fe 55.85g Fe 5.01g O 1 mol O = 0.3131 mol O 16.00g O Fe:O 0.2088 : 0.3131 1:1.5 or 2:3 0.2088 0.2088 Fe 2 O 3

20. 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 = 6.246 mol N 14.01g N 12.5g 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 6.246 NH 2

21. Combustion Analysis

22. 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 2 0.501g CO 2 1 mol CO 2 1 mol C 12.01gC 44.01g CO 2 1 mol CO 2 1 mol C = 0.1367 g C in compound

23. 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 = 0.01155 g H 0.513g cmpd - (0.137 + 0.012)= 0.364g O 0.364g O 1 mol O = 0.02275 mol O 16.00g O 0.137g C 1 mol C = 0.01141 mol C 12.01g C 0.01155g H 1 mol H = 0.01144 mol H 1.01 g H

24. C:H:O 0.01141: 0.01144 : 0.02275 0.01141: 0.01144 : 0.02275 0.01141 0.01141 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 C2H2O4C2H2O4

25. 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)

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

27. 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.

28. 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!

29. 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!

30. 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.

31. 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.

32. 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 3 17.04g NH 3 34.83g Li 3 N 1 mol Li 3 N 1 mol NH 3 = 2.63g NH 3

33. 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.

34. 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 3 2.0 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

35. 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 3 17.0g NH 3 32.0 g O 2 1 mol O 2 = 1 mol O 2 32.0g 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 = 24.0g NO 5 mol O 2 1 mol NO

36. 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

37. 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 2 1.00 mol CH 4 1 mol CCl 4 153.81g CCl 4 1 mol CH 4 1 mol CCl 4 = 154g CCl 4 (theoretical yield) 83.5 x 100 = 54.2% yield 154