Relationship between mass moles and molecules in a

Relationship between mass, moles and molecules in a compound Mass (g) X molar mass (__g__ mole) Amount (moles) Grams moles = moles gram X 6. 022 x 1023 (units mole) # molecules or Formula units moles units = units or molecules moles grams = grams mole Mullis 1

Molar mass • Molar mass of a substance = mass in grams of one mole of the substance. • A compound’s molar mass is NUMERICALLY equal to its formula mass. 2 mol H x 1. 01 g H = 2. 02 g H 1 mol O x 16. 00 g O = 16. 00 g O 1 mol O molar mass H 2 O = 18. 02 g/mol • Formula mass H 2 O = 18. 02 amu • Molar mass H 2 O = 18. 02 g/mol Mullis 2

Molar Mass Example What is the molar mass of K 2 SO 4? 2 mol K x 39. 10 g K = 78. 20 g K 1 mol S x 32. 10 g S = 32. 07 g S 1 mol S 4 mol O x 16. 00 g O = 64. 00 g O 1 mol O molar mass K 2 SO 4 = 174. 27 g/mol How many moles of each element are present in this compound? 2 mol K, 1 mol S, 4 mol O Mullis 3

What is the molar mass of C 6 H 12 O 6? 6 mol C x 12. 01 g C = 72. 06 g C 1 mol C 12 mol H x 1. 01 g H = 12. 12 g H 1 mol H 6 mol O x 16. 00 g O = 96. 00 g O 1 mol O molar mass C 6 H 12 O 6 = 180. 18 g/mol How many moles of each element are present in this compound? 6 mol C, 12 mol H, 6 mol O Mullis 4

Converting to grams from moles How many moles of glucose are in 4. 15 x 10 -3 g C 6 H 12 O 6? 4. 15 x 10 -3 g x 1 mol C 6 H 12 O 6 = 2. 30 x 10 -5 mol C 6 H 12 O 6 180. 18 g How many molecules of glucose are in 4. 15 x 10 -3 g C 6 H 12 O 6? 2. 30 x 10 -5 mol C 6 H 12 O 6 x 6. 022 x 10 23 molecules = 1 mol (2. 30 x 6. 022)(10(-5+23)) = 13. 90 x 10 – 18 molecules = 1. 39 x 10 – 19 molecules Mullis 5

What is the mass in grams of 6. 25 moles copper (II) nitrate? Cu 2+ NO 3 - : formula is Cu(NO 3)2 Find molar mass of Cu(NO 3)2 first. 1 mol Cu x 63. 55 g Cu = 63. 55 g Cu 1 mol Cu 2 mol N x 14. 01 g N = 28. 02 g N 1 mol N 6 mol O x 16. 00 g O = 96. 00 g O 1 mol O molar mass Cu(NO 3)2 = 187. 57 g/mol Now find mass in grams of 6. 25 moles: 6. 25 moles x 187. 57 g = 1172 g Ans. 1170 g Cu(NO 3)2 1 mol Mullis 6

Atoms and Ions Within Compounds • How many carbon atoms are in one mole of C 2 H 6? 1 mole C 2 H 6 | 2 moles C | 6. 022 x 1023 atoms = 1. 204 x 1024 atoms |1 mole C 2 H 6 | 1 mole C • How many MOLES of carbon atoms are in one mole of C 2 H 6? 1 mole C 2 H 6 | 2 moles C = 2 moles C atoms |1 mole C 2 H 6 • How many moles of hydroxide ions are in one mole of calcium hydroxide? How many moles of Ca 2+? 1 mole Ca(OH)2 | 2 moles OH-- =2 moles hydroxide ions |1 mole Ca(OH)2 | 1 mole Ca 2+ =1 mole calcium ions |1 mole Ca(OH)2 Mullis 7

Percentage Composition % Composition is the % by mass of each element in a compound. Find the percentage composition of sodium chloride. Na+ Cl - : formula is Na. Cl 1 mol Na x 22. 99 g Na = 22. 99 g Na 1 mol Cl x 35. 45 g Cl = 35. 45 g Cl 1 mol Cl molar mass Na. Cl = 58. 44 g/mol 22. 99 g Na x 100 = 39. 34 % Na 58. 44 g Na. Cl 35. 45 g Cl x 100 = 60. 66 % Cl Mullis 58. 44 g Na. Cl 8

Find the percentage composition of sodium nitrate. Na+ NO 3 - : formula is Na. NO 3 1 mol Na x 22. 99 g Na = 22. 99 g Na 1 mol N x 14. 01 g N = 14. 01 g N 1 mol N 3 mol O x 16. 00 g O = 48. 00 g O 1 mol O molar mass Na. NO 3 = 85. 00 g/mol 22. 99 g Na x 100 = 27. 05 % Na 85. 00 g Na. NO 3 14. 01 g N x 100 = 16. 48 % N 85. 00 g Na. NO 3 48. 00 g O x 100 = 56. 47% O 85. 00 g Na. NO 3 Mullis 9

Percentage Composition • Why mass instead of moles? • Isn’t 2/3 of the water molecule hydrogen? • Moles indicate the amounts of each atom needed to make the molecule stable from an electron standpoint. 2. 02 g H x 100 18. 02 g H 2 O = H O 11. 21 % H H Mullis 10

Empirical Formula Use % composition to convert to original formula: 1. Assume 100 g sample, so % = grams 2. Convert grams to moles for each element 3. Divide the number of moles for each element by the smallest number of moles 4. The result for each type of element is its subscript in the empirical formula. 5. The order of elements is usually: Organics: C, H, O, N Inorganics: Metal, nonmetal, oxygen 6. Keep 4 decimal places when dividing numbers. If the result has a decimal between. 2 and. 8, may need to multiply all numbers by the number needed to get a whole number. Ex: 3. 5 should be multiplied by 2 to get 4. Then multiply all other elements by the same number. Mullis 11

Example: Empirical and Molecular formula What is the empirical formula of a compound with 54. 82% C, 5. 624% H, 32. 45% O, 7. 104% N? 54. 82 g C| 1 mole C = 4. 568 mole C /. 5074 = 9 C | 12 g C 5. 624 g H| 1 mole H =5. 624 mole H /. 5074 = 11 H | 1 g. H C 9 H 11 O 4 N 32. 45 g O| 1 mole O = 2. 028 mole O /. 5074 = 4 O | 16 g O 7. 104 g N| 1 mole N =0. 5074 mole N /. 5074 = 1 N | 14 g N If a compound has this same composition but its molecular weight is 394 g/mol, what is its molecular formula? MW. C 9 H 11 O 4 N = 197 g/mol 394/197 = 2 so molecular formula is C 18 H 22 O 8 N 2 Mullis 12

Oxidation Numbers • Used to indicate the general distribution of electrons among the bonded atoms in molecular compounds or polyatomic ions. • Analogous to charges in ionic compounds. • An oxidation number is assigned to each element. • Assign the ones you know 1 st. • Find the others based on the numbers it takes to make the charge equal to the charge of the ion or compound. (A compound has a charge of zero. ) Mullis 13

Oxidation Numbers: Rules 1. 2. 3. 4. 5. Pure element = 0 F = -1 O = -2 (except in peroxides and bonds with halogens) H = +1 (except in bonds with metals) The more electronegative element = same (-) charge as its anion 6. The less electonegative element = same (+) charge as its cation 7. The sum of a compound or polyatomic ion’s oxidation numbers is equal to its charge. Mullis 14