Part One: Mass and Moles of Substance. Molecular Mass = sum of the Atomic Masses in a molecule


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1 CHAPTER THREE: CALCULATIONS WITH CHEMICAL FORMULAS AND EQUATIONS Part One: Mass and Moles of Substance A. Molecular Mass and Formula Mass. (Section 3.1) 1. Just as we can talk about mass of one atom of an element as its Atomic Mass (or Atomic Weight), we can also extend that to mass of one molecule of a substance, the Molecular Mass (or Molecular Weight). Molecular Mass = sum of the Atomic Masses in a molecule Example: Molecular Mass of water is 18.0 amu 2. If the substance is ionic, we talk about its Formula Mass (or Formula Weight). Formula Mass = sum of the Atomic Masses in its formula Example: Formula Mass of CaF 2 is 78.1 amu B. The Mole Concept. (Sections 3.2) 1. Atoms combine together most fundamentally by number. (Their fixed masscombining ratios are simply a byproduct of this) When atoms interact, they COUNT each other, they don t WEIGH each other. 2. Problem: a laboratory sized sample of matter contains an enormously large number of atoms or molecules. 1 Liter of air contains ~ 3 x molecules 3. It is inconvenient to handle these large numbers in our chemical calculations, and yet we must continually keep track of numbers of molecules. 4. Solution: we have adopted the mole concept. a. Just like a baker uses the dozen to describe 12 of something, or a hardware store uses a gross to refer to 144 of something. Chapter 3 Page 1
2 b. Better yet the hardware store weighs out nails rather than counting them. Chemists do the same thing, weigh samples but know how to determine the number of atoms or molecules in that mass. c. Chemist uses the mole to refer to number of atoms in one gram of hydrogen. (original definition) Now many atoms is that? Turns out to be: 6.02 x of them. Since Carbon atoms are 12 times the mass of Hydrogen atoms, 12 grams of carbon would be one mole of Carbon atoms. 16 grams of Oxygen would be one mole of Oxygen atoms, and so forth. d. Modern definition of the mole is based on Carbon and not Hydrogen, because Carbon is easier to handle. 1 mole = 6.02 x particles of anything countable. D. The Mole. (Section 3.2) 1. One mole = amount of any substance that contains as many particles of that substance as there are atoms of C in 12 g pure Carbon mole = x particles = Avogadro s number One mole of Fe is x atoms of Fe One mole of H 2 O is x molecules of H 2 O One mole of NaCl is x formula units of NaCl 2. Abbreviation for mole is mol. 3. Why is mole defined this way? Because now: 1 mole of element X = Atomic Mass of X in grams Examples: 1 mole of C = grams C 1 mole of Na = grams Na Chapter 3 Page 2
3 4. Atomic Mass (Atomic Weight) of an element = number of grams in 1 mole of that element (55.85 = number of grams in mol of Fe): 26 Fe Abbreviate Atomic Mass as AM 5. So now: x atoms of Fe = 1 mol Fe = 55.85g Fe x x AM of X atoms of X moles of X grams of X x x AM of X 6. Problems: a. How many Na atoms in 10.00g Na? 1mol Na 10.00gNa atoms 22.99g Na 1mol = Na atoms b. A reaction used up 5.00g Ni. How many moles Ni were used up? 1 mol Ni 5.00gNi = mol Ni 58.69g Ni = mol Ni c. Which weighs more, a mole of C or a mole of H? By what factor? mol of C weighs 12 times a mol of H Chapter 3 Page 3
4 d. Calculate the mass of a Hydrogen atom in grams. 1 mol H 1atom H 1.008g H atoms H mol H = grams B. Molar Mass. (Section 3.2) 1. Molar Mass of a substance = mass of one mole of particles of that substance. 2. Molar Mass of a substance = the number of grams of that substance equal to the sum of Atomic Masses of elements in the formula. 3. Example: Molar Mass of CaF 2. = x = grams/mol 4. Example: Molar Mass of NaOH. = = grams/mol Therefore: 40.01g NaOH = 1 mol NaOH = x NaOH units = x Na + ion plus x OH  ions 5. Example: Molar Mass of H 2 O. = 2 x = grams/mol Thus, 18.02g H 2 O = 1 mole H 2 O = x H 2 O molecules. And contains x oxygen atoms x hydrogen atoms 6. Example: Molar Mass of H 2 = 2 x = grams/mol Thus, g H 2 = 1 mole H 2 = x H 2 molecules in which there are x H atoms present. Chapter 3 Page 4
5 7. For a molecular substance S (e.g. S = H 2, H 2 O, CH 4, etc...) x x MolarMass molecules of S moles of S grams of S x x MolarMass 8. How many molecules of ozone in 24.0g ozone? 24.0g O 3 1 mol O molecules O g O 3 1 mole O 3 = molecules O 3 9. How many H atoms are present in 48.0g methane (CH 4 )? 48.0g CH 4 1 mol CH molecules CH g CH 4 1 mole CH 4 = molecules CH 4 4 H for every molecule CH 4, so (times 4) = 7.21 x H atoms. 10. How many moles of NaF are in 147 grams of it? 11. I want to measure out 2.5 moles of NiSO 4. How many grams do I measure out? Chapter 3 Page 5
6 Part Two: Determining Chemical Formulas A. Percent Composition from Compound Formulas. (Section 3.3) 1. If formula of compound is known, its percent composition by mass of each element can be found, and vice versa. compound formula Atomic Masses (AM) percent composition in terms of elements x Atomic Masses (AM) 2. First we ll go from formula to percent composition. Example: Calculate the percent composition of the elements in ammonium nitrate: given formula = NH 4 NO 3 Formula Mass (FM) = 80.0 g/mol 3 elements present; what is % N, % H, % O? % N = = = mass of N mass of compound 100% 2 AM of N FM of compound 100% g/mol 100% = 35.0% N 80.0 g/mol % H = = % O = = 4 AM of H FM of compound 100% g/mol 80.0 g/mol 3 AM of O FM of compound 100% g/mol 80.0 g/mol 100% = 5.04% H 100% = 60.0% O Chapter 3 Page 6
7 3. Note that the sum of percentages equals 100% % N % H % O = 100% This makes a good check. B. Chemical Formula from Percent Composition. 1. Knowing the Percent Composition from experiments, we can determine the Chemical Formula. 2. Example: A certain acidic substance has the following composition by mass: 2.45 % H, % S, and % O. Determine the compound formula. a. We want the mole ratio of the elements present, not their masses. b. Convert masses to moles by dividing by atomic mass (AM). c. For convenience, work with 100 g sample of compound and find moles of each element: 2.45g H 1 mol H = 2.43 mol H 1.008g H 39.0g S 1 mol S =1.218 mol S 32.06g S 58.48g O 1 mol O = mol O 16.00g O d. Now take this mole ratio and convert it to simplest wholenumber ratio; easiest to first try dividing all 3 numbers by the smallest: 2.43 mol H =1.99 mol H 2H mol S =1 mol S 1S mol O = mol O 3O formula = H 2 SO 3 (sulfurous acid) Chapter 3 Page 7
8 3. This method produces only the simplest or empirical formula of a compound. Illustration: 4. Example: Analysis shows that a compound of Hydrogen and Oxygen is 5.93 % H and % O. Calculate it s formula: 5.93g H 1 mol H 1.008g H 94.07g O 1 mol O 16.00g O = mol H = mol O Now divide through by smallest number: mol H mol O formula = HO =1.001 mol H =1 mol O See, this predicts simplest or empirical formula. Actual chemical formula is H 2 O 2, hydrogen peroxide. 5. Percent compositions alone give only empirical formula for a compound. 6. This is generally all that is needed for ionic compounds. 7. For molecular substances we also need to know the Molar Mass. 8. Example: In the preceding problem, if we were also given that the Molar Mass of the compound is 34.0 amu, then we could find the true molecular formula: empirical formula is HO (Formula Mass = 17.0 g/mol) convert to formula H 2 O 2 with Molar mass = 34.0 g/mol Chapter 3 Page 8
9 9. Summary: percent Atomic Mass mole smallest wholecomposition ratios of number elements mole ratio 10. One more problem: empirical true Molar Mass true formula emp Formula Mass molecular formula A hydrocarbon has a percent composition of % H and % C and a MW = What is it s molecular formula? a. First get empirical formula: 7.744g H 1 mol H = mol H 1.008g H 92.27g C 1 mol C = mol C 12.01g C So empirical formula = CH b. Convert to molecular formula using Molar Mass. Formula Mass (FM) of CH = amu 1:1 ratio x 78.1/13.02 = 6:6 molecular formula = C 6 H 6 (benzene) 11. Molecular Formula from Combustion Data: A gram sample of an organic compound containing C, H and O is analyzed by combustion analysis and grams of CO 2 and grams of H 2 O are produced. In a separate experiment, the molar mass is found to be g/mol. Determine the empirical formula and the molecular formula of the organic compound. Chapter 3 Page 9
10 a. All C ends up producing CO 2 so every C atom gives rise to a CO 2 molecule or every mol of C gives rise to a mole of CO 2. b. Same for H and H 2 O, except every mol of H gives rise to only half a mole of H 2 O g CO g H 2 O 1 mol CO 2 44 g CO 2 = mol CO 2 = mol C 1 mol H 2O 18 g H 2 O = mol H 2O = mol H Therefore: C and H are present in 1:1 ratio by atoms or moles. c. Now what about oxygen? Must find first what mass oxygen in original sample by subtracting out masses of C and H mol C 12 g/mol = g C mol H g/mol = g H g = g oxygen g O 1 mol O 16 g O = mol O mole ratio: mol C; mol H; mol O divide by smallest 4 : 4 : 1 simplest formula: C 4 H 4 O Formula Mass = 68 True Molar Mass = 136 (2x as much) So molecular formula is C 8 H 8 O The Law of Multiple Proportions (revisited) = when two elements, A and B, form more than one compound together, the amount of B combining with a given amount of A in each compound forms a whole number ratio: amt B combining w/a in cmpd 1 = ratio of whole numbers amt B combining w/a in cmpd 2 Chapter 3 Page 10
11 13. This supplements the Law of Definite Proportions. 14. Example: Remember finding two oxides of hydrogen, with different mass combining ratios: 1.0 g H for every 8.0 g O in one compound 1.0 g H for every 16.0 g O in the other Confirms the Law of Multiple Proportions: for every gram present of hydrogen: 8.0 g O in H 2 O = 1 (whole number ratio) 16.0 g O in H 2 O 2 2 Part Three: Stoichiometry A. What is Stoichiometry? (Sec ) 1. Stoichiometry = description of quantitative relationships: a. among elements within a compound. b. among substances within a chemical reaction. 2. Concerned with: a. mole ratios which combine together. (balancing chemical equations) b. mass ratios which combine together. B. Calculations Based on Chemical Equations. (Section 3.6) 1. Here we use mole interpretation of the balanced equation. 2. Balance exists on two levels, mole level and mass level. Chapter 3 Page 11
12 CH O 2 CO H 2 O mole 1 mol 2 mol 1 mol 2 mol level x MM x MM x MM x MM CH 4 O 2 CO 2 H 2 O 16 g/mol 32 g/mol 44 g/mol 18 g/ mol mass 16 g 64 g 44 g 36 g level 3. Tells us that 16 g CH 4 requires exactly 64 g O 2 to completely burn, or a 1:4 ratio by mass. 4. Problem: What mass of oxygen is required to react completely with 100 g methane? How many grams CO 2 will form? CH O 2 CO H 2 O 5. Schematic: Chapter 3 Page 12
13 6. Another problem: How many grams of CH 4 and O 2 would produce 54 g of H 2 O? C. The Limiting Reagent Concept. (Section 3.8) 1. Limiting reagent = reactant that is used up first when reactants are not mixed in their exact stoichiometric ratios. 2. Example: initially have 4 slices of ham, 6 slices of bread, and wish to make sandwiches. 2 slices bread + 1 slice ham 1 sandwich a. How many sandwiches can you make? Only 3, until the bread runs out. b. What is limiting reagent? Bread. c. How much ham is left over? 1 slice. d. Which reagent is present in excess? Ham. 3. Problem of Limiting Reagent: How many grams of MgO could be formed by the reaction of 30 g Mg with 16 g O 2 in the following reaction? 2 Mg + O 2 2 MgO Strategy: a. Recognize it as a limiting reagent problem. How? b. Convert to moles of reactants first. Chapter 3 Page 13
14 c. Pick one of the reactants. d. Calculate how many moles of the other reactant would be required to completely use up the one you picked. e. Not enough of the other one? Other must be the limiting reagent. f. The moles of limiting reagent determines how much product could form. How many grams of MgO could be formed by the reaction of 30 g Mg with 16 g O 2 in the following reaction? 2 Mg + O 2 2 MgO Let s work it: O 2 is limiting!! 4. Another example problem: 7.0 grams of nitrogen and 3.0 grams of hydrogen are mixed, and the following reaction takes place. What is the limiting reagent, and how many grams of ammonia would be produced if the reaction goes to completion? N H 2 2 NH 3 Chapter 3 Page 14
15 D. Percent Yields. (Section 3.8) 1. Definition: Percent yield = 2. Theoretical yield is calculated by assuming the reaction goes to completion (all of limiting reagent used up), no side reactions, and nothing lost in handling. 3. Actual yield is amount of desired product we actually isolated in our lab experiment. 4. Example problem: In preparing Ni(OH) 2, 5.0 g of product are isolated after combining 10.0 g NiCl 2 and 10.0 g NaOH. What is % yield of Ni(OH) 2? % yield = = 70% Chapter 3 Page 15
16 NOTES: Chapter 3 Page 16
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