Mass Relationships in Chemical Reactions Copyright The Mc

Mass Relationships in Chemical Reactions Copyright © The Mc. Graw-Hill Companies, Inc. Permission required for reproduction or display.

Chapter 3 3. 1 Atomic mass. 3. 2 Avogadro’s number and the molar mass of an element. 3. 3 Molecular mass. 3. 5 Percent composition of compounds. 3. 6 Experimental determination of empirical formulas. 3. 7 Chemical reactions and chemical equations. 3. 8 Amounts of reactions and products. 3. 9 Limiting reagents. 3. 10 Reaction yield.

Micro World atoms & molecules Macro World grams Atomic mass is the mass of an atom in atomic mass units (amu) By definition: 1 atom 12 C “weighs” 12 amu On this scale 1 H = 1. 008 amu 16 O = 16. 00 amu 3. 1

Natural lithium is: 7. 42% 6 Li (6. 015 amu) 92. 58% 7 Li (7. 016 amu) Average atomic mass of lithium: 7. 42 x 6. 015 + 92. 58 x 7. 016 = 6. 941 amu 100 3. 1

Average atomic mass (6. 941)

The mole (mol) is the amount of a substance that contains as many elementary entities as there atoms in exactly 12. 00 grams of 12 C 1 mol = NA = 6. 0221367 x 1023 Avogadro’s number (NA) 3. 2

eggs shoes Molar mass is the mass of 1 mole of in grams marbles atoms 1 mole 12 C atoms = 6. 022 x 1023 atoms = 12. 00 g 1 12 C atom = 12. 00 amu 1 mole 12 C atoms = 12. 00 g 12 C 1 mole lithium atoms = 6. 941 g of Li For any element atomic mass (amu) = molar mass (grams) 3. 2

One Mole of: S C Hg Cu Fe 3. 2

1 12 C atom 12. 00 g 1. 66 x 10 -24 g x = 23 12 12. 00 amu 6. 022 x 10 C atoms 1 amu = 1. 66 x 10 -24 g or 1 g = 6. 022 x 1023 amu M = molar mass in g/mol NA = Avogadro’s number 3. 2

Do You Understand Molar Mass? How many atoms are in 0. 551 g of potassium (K) ? 1 mol K = 39. 10 g K 1 mol K = 6. 022 x 1023 atoms K 1 mol K 6. 022 x 1023 atoms K 0. 551 g K x x = 1 mol K 39. 10 g K 8. 49 x 1021 atoms K 3. 2

Worked Example 3. 4

Worked Example 3. 6

Worked Example 3. 7

Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule. 1 S SO 2 2 O SO 2 32. 07 amu + 2 x 16. 00 amu 64. 07 amu For any molecule molecular mass (amu) = molar mass (grams) 1 molecule SO 2 = 64. 07 amu 1 mole SO 2 = 64. 07 g SO 2 3. 3

Do You Understand Molecular Mass? How many H atoms are in 72. 5 g of C 3 H 8 O ? 1 mol C 3 H 8 O = (3 x 12) + (8 x 1) + 16 = 60 g C 3 H 8 O 1 mol C 3 H 8 O molecules = 8 mol H atoms 1 mol H = 6. 022 x 1023 atoms H 1 mol C 3 H 8 O 8 mol H atoms 6. 022 x 1023 H atoms 72. 5 g C 3 H 8 O x x x = 1 mol C 3 H 8 O 1 mol H atoms 60 g C 3 H 8 O 5. 82 x 1024 atoms H 3. 3

Formula mass is the sum of the atomic masses (in amu) in a formula unit of an ionic compound. 1 Na Na. Cl 22. 99 amu 1 Cl + 35. 45 amu Na. Cl 58. 44 amu For any ionic compound formula mass (amu) = molar mass (grams) 1 formula unit Na. Cl = 58. 44 amu 1 mole Na. Cl = 58. 44 g Na. Cl 3. 3

Do You Understand Formula Mass? What is the formula mass of Ca 3(PO 4)2 ? 1 formula unit of Ca 3(PO 4)2 3 Ca 3 x 40. 08 2 P 8 O 2 x 30. 97 + 8 x 16. 00 310. 18 amu 3. 3

Percent composition of an element in a compound = n x molar mass of element x 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound 2 x (12. 01 g) x 100% = 52. 14% 46. 07 g 6 x (1. 008 g) %H = x 100% = 13. 13% 46. 07 g 1 x (16. 00 g) %O = x 100% = 34. 73% 46. 07 g %C = C 2 H 6 O 52. 14% + 13. 13% + 34. 73% = 100. 0% 3. 5

Heavy Light KE = 1/2 x m x v 2 v = (2 x KE/m)1/2 F = q x v x B 3. 4

Percent Composition and Empirical Formulas Determine the empirical formula of a compound that has the following percent composition by mass: K 24. 75, Mn 34. 77, O 40. 51 percent. 1 mol K n. K = 24. 75 g K x = 0. 6330 mol K 39. 10 g K 1 mol Mn n. Mn = 34. 77 g Mn x = 0. 6329 mol Mn 54. 94 g Mn n. O = 40. 51 g O x 1 mol O = 2. 532 mol O 16. 00 g O 3. 5

Percent Composition and Empirical Formulas n. K = 0. 6330, n. Mn = 0. 6329, n. O = 2. 532 0. 6330 ~ K : ~ 1. 0 0. 6329 Mn : 0. 6329 = 1. 0 0. 6329 2. 532 ~ O : ~ 4. 0 0. 6329 KMn. O 4 3. 5

Combust 11. 5 g ethanol Collect 22. 0 g CO 2 and 13. 5 g H 2 O g CO 2 mol C g C 6. 0 g C = 0. 5 mol C g H 2 O mol H g H 1. 5 g H = 1. 5 mol H g of O = g of sample – (g of C + g of H) 4. 0 g O = 0. 25 mol O Empirical formula C 0. 5 H 1. 5 O 0. 25 Divide by smallest subscript (0. 25) Empirical formula C 2 H 6 O 3. 6

A process in which one or more substances is changed into one or more new substances is a chemical reaction A chemical equation uses chemical symbols to show what happens during a chemical reaction 3 ways of representing the reaction of H 2 with O 2 to form H 2 O reactants products 3. 7

How to “Read” Chemical Equations 2 Mg + O 2 2 Mg. O 2 atoms Mg + 1 molecule O 2 makes 2 formula units Mg. O 2 moles Mg + 1 mole O 2 makes 2 moles Mg. O 48. 6 grams Mg + 32. 0 grams O 2 makes 80. 6 g Mg. O IS NOT 2 grams Mg + 1 gram O 2 makes 2 g Mg. O 3. 7

Balancing Chemical Equations 1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation. Ethane reacts with oxygen to form carbon dioxide and water C 2 H 6 + O 2 CO 2 + H 2 O 2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts. 2 C 2 H 6 NOT C 4 H 12 3. 7

Balancing Chemical Equations 3. Start by balancing those elements that appear in only one reactant and one product. C 2 H 6 + O 2 2 carbon on left C 2 H 6 + O 2 6 hydrogen on left C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 1 carbon on right multiply CO 2 by 2 2 CO 2 + H 2 O 2 hydrogen on right 2 CO 2 + 3 H 2 O multiply H 2 O by 3 3. 7

Balancing Chemical Equations 4. Balance those elements that appear in two or more reactants or products. C 2 H 6 + O 2 2 oxygen on left 2 CO 2 + 3 H 2 O multiply O 2 by 7 2 4 oxygen + 3 oxygen = 7 oxygen (3 x 1) on right (2 x 2) 7 C 2 H 6 + O 2 2 2 CO 2 + 3 H 2 O 2 C 2 H 6 + 7 O 2 4 CO 2 + 6 H 2 O remove fraction multiply both sides by 2 3. 7

Balancing Chemical Equations 5. Check to make sure that you have the same number of each type of atom on both sides of the equation. 2 C 2 H 6 + 7 O 2 4 CO 2 + 6 H 2 O 4 C (2 x 2) 4 C 12 H (2 x 6) 12 H (6 x 2) 14 O (7 x 2) 14 O (4 x 2 + 6) Reactants 4 C 12 H 14 O Products 4 C 12 H 14 O 3. 7

Amounts of Reactants and Products 1. Write balanced chemical equation 2. Convert quantities of known substances into moles 3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity 4. Convert moles of sought quantity into desired units 3. 8

Methanol burns in air according to the equation 2 CH 3 OH + 3 O 2 2 CO 2 + 4 H 2 O If 209 g of methanol are used up in the combustion, what mass of water is produced? grams CH 3 OH moles CH 3 OH molar mass CH 3 OH 209 g CH 3 OH x moles H 2 O grams H 2 O molar mass coefficients H 2 O chemical equation 4 mol H 2 O 18. 0 g H 2 O 1 mol CH 3 OH = x x 32. 0 g CH 3 OH 1 mol H 2 O 2 mol CH 3 OH 235 g H 2 O 3. 8

Limiting Reagents 2 NO + 2 O 2 2 NO 2 NO is the limiting reagent O 2 is the excess reagent 3. 9

Do You Understand Limiting Reagents? In one process, 124 g of Al are reacted with 601 g of Fe 2 O 3 2 Al + Fe 2 O 3 Al 2 O 3 + 2 Fe Calculate the mass of Al 2 O 3 formed. g Al mol Al g Fe 2 O 3 124 g Al x mol Fe 2 O 3 needed OR mol Al needed mol Fe 2 O 3 1 mol Al 27. 0 g Al x g Fe 2 O 3 needed 1 mol Fe 2 O 3 2 mol Al Start with 124 g Al 160. g Fe 2 O 3 = x 1 mol Fe 2 O 3 g Al needed 367 g Fe 2 O 3 need 367 g Fe 2 O 3 Have more Fe 2 O 3 (601 g) so Al is limiting reagent 3. 9

Use limiting reagent (Al) to calculate amount of product that can be formed. g Al mol Al 2 O 3 g Al 2 O 3 2 Al + Fe 2 O 3 Al 2 O 3 + 2 Fe 124 g Al x 1 mol Al 27. 0 g Al x 1 mol Al 2 O 3 2 mol Al 102. g Al 2 O 3 = x 1 mol Al 2 O 3 234 g Al 2 O 3 3. 9

Reaction Yield Theoretical Yield is the amount of product that would result if all the limiting reagent reacted. Actual Yield is the amount of product actually obtained from a reaction. % Yield = Actual Yield Theoretical Yield x 100 3. 10

Chemistry In Action: Chemical Fertilizers Plants need: N, P, K, Ca, S, & Mg 3 H 2 (g) + N 2 (g) 2 NH 3 (g) NH 3 (aq) + HNO 3 (aq) NH 4 NO 3 (aq) fluorapatite 2 Ca 5(PO 4)3 F (s) + 7 H 2 SO 4 (aq) 3 Ca(H 2 PO 4)2 (aq) + 7 Ca. SO 4 (aq) + 2 HF (g)

Problems • 4 , 5 , 6 , 7 , 9 , 10 , 14 , 16.