1. Stoichiometry Chapter 3 -MW

2. What Is Stoich? Stoichiometry is the study of reactions:  Why do reactions occur?  How fast do they proceed?  What intermediary products if any are used?  How much of the reactants react? Stoichiometry - The study of quantities of materials consumed and produced in chemical reactions.

3. Mass Spectrometer Compares mass of atoms Atomic mass is defined by Carbon 12 = 12amu Parts of a mass spectrometer: Vaporizer Electron beams Ion- accelerating electric field Magnetic field Detector Plate

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5. Mass Spectrometer Process 1. A heater vaporizes a sample 2. A beam of high speed electrons knocks electrons off test atoms/molecules. 3. An electric field accelerates the sample ions 4. The accelerating ions have a magnetic field. They interacts with an applied magnetic field deflecting their path. The ions separate. 5. A detector plate measures the deflections -comparison of deflections gives ions’ masses -less massive particles deflect more

6. .6 Atomic Masses Elements occur in nature as mixtures of isotopes Carbon =98.89% 12 C 1.11% 13 C <0.01% 14 C Carbon atomic mass = 12.01 amu

7. Change % abundance into decimals & multiply by respective isotopic weights. Add together. If want % abundance; use “x” & “1 – x” to represent abundance. 7

8. Examples 1) There are two isotopes of carbon 12 C with a mass of 12.00000 amu(98.892%), and 13 C with a mass of 13.00335 amu (1.108%). 2) There are two isotopes of nitrogen, one with an atomic mass of 14.0031 amu and one with a mass of 15.0001 amu. What is the percent abundance of each if the weighted average is 14.01amu?

9. Answers 1) 0.98892(12.00000amu) + 0.01108(13.00335amu) = 11.86704amu +.1440771amu = 12.011117amu Matches P.table 14.0031(x) + 15.0001(1-x) = 14.01 14.0031x + 15.0001amu - 15.001x = 14.01 amu -0.997x = -0.9901 X =.9930192 or 99.3%, 1-x =.692% 9

10. 10 The Mole The number equal to the number of carbon atoms in exactly 12 grams of pure 12 C. 1 mole of anything = 6.022 x 10 23 units of that thing Avogadro’s number equals 6.022 x 10 23 units

11. 11 Molar Mass A substance’s molar mass (molecular weight) is the mass in grams of one mole of the compound. CO 2 = 44.01 grams per mole

12. Find the molar mass of CH 4 Mg 3 P 2 Ca(NO 3 ) 3 Al 2 (Cr 2 O 7 ) 3 CaSO 4 · 2H 2 O

13. CH 4 =12.0 + (4) 1.01 = 16.0g/mol Mg 3 P 2 = (3) 24.3 + (2) 30.974 = 135g/mol Ca(NO 3 ) 3 = 40.1 + (3) 14.0 + (9) 16.0 = 226g/mol Al 2 (Cr 2 O 7 ) 3 = (2) 27.0 + (6) 52.0 + (21) 16.0 = = 702g/mol CaSO 4 · 2H 2 O = 40.1 + 32.1 + (4) 16.0 + (2) 18.0 = 156g/mol 13

14. Percent Composition Mass percent of an element: For iron in iron (III) oxide, (Fe 2 O 3 )

15. Working backwards From percent composition, you can determine the empirical formula. Empirical Formula the lowest ratio of atoms in a molecule. Based on mole ratios.

16. 16 Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

17. A sample is 59.53% C, 5.38% H, 10.68% N, and 24.40% O. What is its empirical formula? 17 CHNO 59.535.3810.6824.40 12.01.0114.016.0 4.965.33.76281.525.7628.7628.7628.7628 6.5712(mult. By 2) C 13 H 14 N 2 O 4

18. A 0.2000 gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O 2. 0.2998 g of CO 2 and 0.0819 g of H 2 O are produced. What is the empirical formula?

19. Get C from CO2, H from H2O and O from subtracting C + O from original amount. C0.2998g x 12g = 0.0817636g C CO 2 44g H0.0819g x 2.02g = 0.00919g H H 2 018.0g 0.0909546g 0.2000g - 0.0909546g = 0.1090454g O 19

20. 0.0817636g C 0.00919g H 0.1090454g O 12.0 1.0116.0 1 1.331 (x 3) C 3 H 4 O 3 20

21. 21 Molecular Formula Molar mass = (empirical formula) n [n = integer] empirical formula = CH, Molar mass = 78.0g (CH) x = 78.0g,(12 + 1.01) x = 78.0g, x = 6 molecular formula = (CH) 6 = C 6 H 6

22. Example A compound is made of only sulfur and nitrogen. It is 69.6% S by mass. Its molar mass is 184 g/mol. What is its formula? 69.6/32.1 = 2.16830.4/14.0 = 2.171 46.1x = 184, x = 3.99 S4N4S4N4

23. 23 Chemical Equations Chemical change involves a reorganization of the atoms in one or more substances.

24. 24 Chemical Equation A representation of a chemical reaction: C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O reactants products

25. Chemical Equations Are sentences. Describe what happens in a chemical reaction. Reactants  Products Equations should be balanced. Have the same number of each kind of atoms on both sides because...

26. Abbreviations (s)  (g)  (aq) heat  catalyst

27. 27 Chemical Equation C 2 H 5 OH + 3O 2  2CO 2 + 3H 2 O The equation is balanced. 1 mole of ethanol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

28. Practice Ca(OH) 2 + H 3 PO 4  H 2 O + Ca 3 (PO 4 ) 2 Cr + S 8  Cr 2 S 3 KClO 3 (s)  Cl 2 (g) + O 2 (g) Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s)

29. 3Ca(OH) 2 + 2H 3 PO 4  6H 2 O + Ca 3 (PO 4 ) 2 16Cr + 3S 8  8Cr 2 S 3 2KClO 3 (s)  Cl(s) + 3O 2 (g) Fe 2 S 3 (s) + 6HCl(g)  FeCl 3 (s) + 3H 2 S(g) Fe 2 O 3 (s) + 2Al(s)  2Fe(s) + Al 2 O 3 (s) 29

30. All chemical reactions can be placed into one of six categories. Here they are, in no particular order: 1) Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An example of this kind of reaction is the burning of napthalene: C 10 H 8 + 12 O 2 ---> 10 CO 2 + 4 H 2 O.30

31. 2) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a more complicated one. These reactions come in the general form of: A + B ---> AB One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide: 8 Fe + S 8 ---> 8 FeS 31

32. 3) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions come in the general form: AB ---> A + B One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen gas: 2 H 2 O ---> 2 H 2 + O 2 32

33. 4) Single displacement: This is when one element trades places with another element in a compound. These reactions come in the general form of: A + BC ---> AC + B One example of a single displacement reaction is when magnesium replaces hydrogen in water to make magnesium hydroxide and hydrogen gas: Mg + 2 H 2 O ---> Mg(OH) 2 + H 2 33

34. 5) Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form: AB + CD ---> AD + CB One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate: Pb(NO 3 ) 2 + 2 KI ---> PbI 2 + 2 KNO 3 6) Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H + ion in the acid reacts with the OH - ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O 34

35. 6) Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H + ion in the acid reacts with the OH - ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water: HA + BOH -> H 2 O + BA One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH -> NaBr + H 2 O 35

36. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 Calculating Masses of Reactants and Products 1.Balance the equation. 2.Convert mass to moles. 3.Set up mole ratios. 4.Use mole ratios to calculate moles of desired substituent. 5.Convert moles to grams, if necessary.

37. Examples One way of producing O 2 (g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. 1) How many moles of O 2 (g) are produced? 2) How many grams of potassium chloride? 3) How many grams of oxygen?

38. 2KClO 3 (s)  Cl(s) + 3O 2 (g) 25.5g KClO 3 x 1 mol x 3 mol = 0.3109756 mol O 2 123g2 mol 25.5g KClO 3 x 1 mol x 2 mol x 74.6g = 15.5g  Cl 123g2 mol 1 mol 25.5g KClO 3 x 1 mol x 3 mol x 32.0g = 9.95g O 2 123g2 mol 1 mol 38

39. Examples 1) A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H 2 (g) are produced? 2) How many grams of each reactant are needed to produce 15 grams of iron from the following reaction? Fe 2 O 3 (s) + Al(s)  Fe(s) + Al 2 O 3 (s)

40. 5.11in=12.98, 3.23in=8.20cm, 0.0381in=0.0968cm vol = 10.3cm 3 D = 2.7g/cm 3 27.8g Al x 1mol x 3 mol H 2 x 27.0g = 3.12g H 2 27g 2 mol Al 1 mol ------------------------------------------------------------------ 15g Fe x 1 mol x 2 mol Al x 27.0g = 7.25g Al 55.85g 2 mol Fe 1 mol 15g Fe x 1 mol x 1 mol Fe2O3 x 159.7g = 21.5g Al 55.85g 2 mol Fe 1 mol 40

41. Examples K 2 PtCl 4 (aq) + NH 3 (aq)  Pt(NH 3 ) 2 Cl 2 (s)+ KCl(aq) What mass of Pt(NH 3 ) 2 Cl 2 can be produced from 65 g of K 2 PtCl 4 ? How much KCl will be produced? How much from 65 grams of NH 3 ?

42. K 2 PtCl 4 (aq) + 2NH 3 (aq)  Pt(NH 3 ) 2 Cl 2 (s)+ 2KCl(aq) 65 g K 2 PtCl 4  x 1mol x 1mol x 300g = 47g Pt(NH 3 ) 2 Cl 2 415g1mol 1mol How much KCl will be produced? 65 g K 2 PtCl 4  x 1mol x 2mol x 74.6g = 23g KCl 415g1mol 1mol How much from 65 grams of NH 3 ? 65g NH 3 x 1mol x 1mol x 300g = 574g Pt(NH 3 ) 2 Cl 2 17.0g2mol 1mol 65g NH 3  x 1mol x 2mol x 74.6g = 285g KCl 17.0g 2mol 1mol. 42

43. Limiting Reagent Reactant that determines the amount of product formed. The one you run out of first. Makes the least product. Book shows you a ratio method. It works. So does mine

44. Example Ammonia is produced by the following reaction N 2 + H 2  NH 3 What mass of ammonia can be produced from a mixture of 100. g N 2 and 500. g H 2 ? How much unreacted material remains?

45. 100g N 2 x 1mol x 2mol x 17.0g = 121g NH 3 28.0g 1mol 1mol 500g H 2 x 1mol x 2mol x 17.0g = 2805g NH 3 2.02g 3mol 1mol 121g NH 3 x 1mol x 3mol x 2.02g = 21.6g H 2 17.0g 2mol 1mol 500g – 21.6g = 478g H 2 unreacted 45

46. Excess Reagent The reactant you don’t run out of. The amount of stuff you make is the yield. The theoretical yield is the amount you would make if everything went perfect. The actual yield is what you make in the lab.

47. Percent Yield % yield = Actual x 100% Theoretical % yield = what you got x 100% what you could have got

48. Examples Aluminum burns in bromine producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

49. 2Al + 3Br 2  2AlBr 3 6.0g Al x 1mol x 2mol x 267g = 59.3g AlBr 3 27.0g 2mol 1mol 50.3 x 100 = 84.8% 59.3 Copyright©2000 by Houghton Mifflin Company. All rights reserved. 49

50. Examples Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br 2   HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? If the reaction did go to completion, how much excess reagent would be left?

51. Hg + Br 2  HgBr 2 If 10.0 g of Hg and 9.00 g of Br 2 are reacted, how much HgBr 2 will be produced? 10.0 g Hg x 1mol x 1mol x 361g = 17.96g 201g 1mol 1mol 9.00 g Br 2 x 1mol x 1mol x 361g = 20.31g 160g 1mol 1mol how much excess reagent would be left?.743 x 17.96g = 13.34 g HgBr 2 13.34 g HgBr 2 x 1mol x 1mol x 160g = 5.91gBr 2 361g 1mol 1mol 9.00-5.91=3.09g excess

52. Examples Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2HCl (aq)  ZnCl 2 (aq) + H 2 (g) Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl 2 are eventually isolated. What is the composition of the brass?

53. Zn(s) + 2HCl(aq)  ZnCl 2 (aq) + H 2 (g) 0.5065 g of brass is reacted with excess HCl, 0.0985g ZnCl 2 x 1mol x 1mol x 65.39g =.0473597g 136g 1mol 1mol 0.5065 - 0.0473597g = 0.4591 x 100 = 90.6% Cu 9.35% zn 53