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Chapter 11

Chapter 11. Notes BB, Slide 1-6 Notes CC, Slides 7-9 Notes DD, Slides 10-11 Notes EE, Slides 12-15 Notes FF, Slides 16-17 Notes GG, Slides 18-19. Gram-atomic mass The atomic mass number of grams Contains 6.02X10 23 atoms 6.02X10 23 atoms is Avogadro's number.

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Chapter 11

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  1. Chapter 11 Notes BB, Slide 1-6 Notes CC, Slides 7-9 Notes DD, Slides 10-11 Notes EE, Slides 12-15 Notes FF, Slides 16-17 Notes GG, Slides 18-19

  2. Gram-atomic mass • The atomic mass number of grams • Contains 6.02X1023 atoms • 6.02X1023 atoms is Avogadro's number. • Mass in grams of one mole of atoms of an element • 1 mole = 6.02X1023 particles • Similar to: • 1 pair = 2 • 1 dozen = 12 Element Atomic Mass Gram-Atomic Mass C 12.0 12.0g Mg 24.3 24.3g O 16.0 16.0g Al 27.0 27.0g

  3. What are particles? • They are any representative units or chemical units like atoms, molecules, or formula units. • Remember: • Atoms – smallest particle of an element with all the properties of that element (Elements only, found on the periodic table) • Molecules – smallest chemical unit which can exist by itself in a stable form (Some elements and some compounds) • Two or more atoms are needed to make a molecule. • For any element, the gram-atomic mass equals the mass of one mole. • Carbon • 1mole = 12.0g = 6.02X1023 atoms

  4. Magnesium • 1mole = 24.3g = 6.02X1023 atoms • 2.00 moles C = 24.0g • 36.0g = _____ moles • Example Problems • Moles  Grams • Moles known X atomic mass = grams 1 mole • 6.00 moles K = _____ g • 1 mole K = 39.1g • 6.00 moles K X 39.1g = 235g K 1 mol • 7.40 moles Na = _____ g • 1 mole Na = 23.0g • 7.40 moles Na X 23.0g = 170g Na 1 mol

  5. Grams  Moles • Grams known X 1 mole = moles atomic mass • 160.5g S = _____ moles • 1 mole S = 32.1g • 160.5g S X 1 mole = 5.00 mol S 32.1g • 427g Ni = _____ moles • 1 mole Ni = 58.7g • 427g Ni X 1 mole = 7.27 mol Ni 58.7g

  6. Moles  atoms • Moles known X 6.02X1023 = atoms 1 mole • 6.00 moles K = __________ atoms • 1 mole K = 6.02X1023 atoms • 6.00 moles K X 6.02X1023 = 3.61X1024 atoms K 1 mol • 7.40 moles Na = __________ atoms • 1 mole Na = 6.02X1023 atoms • 7.40 moles Na X 6.02X1023 = 4.45X1024 atoms Na 1 mol

  7. What is a chemical formula? • 2 Types • Molecular – exact number of atoms • Empirical – simplest ratio of atoms • The formula H2S shows: • 2 atoms of H • I atom of S • Formula Mass • The sum of the atomic masses of all atoms shown in a formula • What is the formula mass of H2S? • Element# of atoms X at. Mass = contribution H 2 1.0 2.0 S 1 32.1 + 32.1 Formula mass = 34.1g

  8. For K2CrO4 • Element# of atoms X at. Mass = contribution K 2 39.1 78.2 Cr 1 52.0 52.0 O 4 16.0 + 64.0 Formula mass = 194.2g • Molecular mass • A formula mass of a molecular substance. • The term formula mass includes all molecular masses. • Formula mass is more general and can always be used.

  9. For Al2(SO4)3 • Element# of atoms X at. Mass = contribution Al 2 27.0 54.0 S 3 32.1 96.3 O 12 16.0 + 192.0 Formula mass = 342.3g

  10. Percent Composition • Mass percentage of the elements in a compound • Tells what percentage of the formula mass is due to each element present. • Steps: • 1. Find the formula mass. • 2. (contribution  formula mass) X 100 = percent composition Element# of atoms X at. Mass = cont.% comp. H 2 1.0 2.0 5.87% S 1 32.1 + 32.1 94.13% Formula mass = 34.1g (2/34.1)X100 (32.1/34.1)X100

  11. (78.2/194.2)X100 For K2CrO4 Element # of atoms X at. Mass = cont.% Comp. K 2 39.1 78.2 40.3% Cr 1 52.0 52.0 26.8% O 4 16.0 + 64.0 33.0% Formula mass = 194.2g For Al2(SO4)3 Element# of atoms X at. Mass = cont. % Comp. Al 2 27.0 54.0 15.8% S 3 32.1 96.3 28.1% O 12 16.0 + 192.0 56.1% Formula mass = 342.3g (52.0/194.2)X100 (64.0/194.2)X100 (54.0/342.3)X100 (96.3/342.3)X100 (192.0/342.3)X100

  12. Finding an Empirical Formula • Steps: • 1. Assume the given percentages are grams and you have 100 grams total. • 2. Convert the assumed grams to moles (See Notes BB). • 3. Divide all of the mole values by the smallest number of moles. This answer is the number of atoms in the formula. • 4. Write the formula.

  13. Example 1 • A compound is 50.1% sulfur and 49.9% oxygen by mass. What is its empirical formula? • Step 1: 50.1g S 49.9g O 1 mol S = 32.1g 1 mol O = 16.0g • Step 2: • 50.1g S X 1 mole = 1.56 mol S 32.1g • 49.9g O X 1 mole = 3.12 mol O 16.0g • Step 3: • 1.56 = 1 3.12 = 2 1.56 1.56 • Step 4: • SO2

  14. Example 2 • A compound is 40.1% sulfur and 59.9% oxygen by mass. What is its empirical formula? • Step 1: 40.1g S 59.9g O 1 mol S = 32.1g 1 mol O = 16.0g • Step 2: • 40.1g S X 1 mole = 1.25 mol S 32.1g • 59.9g O X 1 mole = 3.74 mol O 16.0g • Step 3: • 1.25 = 1 3.74 = 2.99  3 1.25 1.25 • Step 4: • SO3

  15. Example 3 • A compound is 39.8% copper, 20.1% sulfur, and 40.1% oxygen by mass. What is its empirical formula? • Step 1: 39.8g Cu 20.1g S 40.1g O 1mol Cu = 63.5g 1 mol S = 32.1g 1 mol O = 16.0g • Step 2: • 39.8g Cu X 1 mole = .6269 mol Cu 63.5g • 20.1g S X 1 mole = .6262 mol S 32.1g • 40.1g O X 1 mole = 2.506 mol O 16.0g • Step 3: • .6269 = 1Cu .6262 = 1 S 2.506 = 4 O .6262 .6262 .6262 • Step 4: • CUSO4

  16. Finding Empirical Formulas • (See Notes EE for the steps.) • 94.9 grams of iron and 120.7 grams of chlorine combine to form a compound. What is its empirical formula? • Step 1: • Skip this step. The numbers were given in grams. • Step 2: • 94.9g Fe X 1 mole = 1.7 mol Fe 55.8g • 120.7g Cl X 1 mole = 3.4 mol Cl 35.5g • Step 3: • 1.7 = 1 3.4 = 2 1.7 1.7 • Step 4: • FeCl2

  17. A compound is 52.94% aluminum and 47.06% oxygen by mass. What is its empirical formula? • Step 1: 52.94g Al 47.06g O 1 mol Al = 27.0g 1 mol O = 16.0g • Step 2: • 52.94g Al X 1 mole = 1.96 mol Al 27.0g • 47.06g O X 1 mole = 2.94 mol O 16.0g • Step 3: • The numbers do not come out to be whole numbers. So multiply by 2. • 1.96 = 1 X 2 = 2 Al 2.94 = 1.5 X 2 = 3 O 1.96 1.96 • Step 4: • Al2O3

  18. Hydrates • A hydrate is a compound that contains some water in its crystal structure. • The water is called water of hydration or water of crystallization. • The water can usually be removed by heating the compound to a temperature greater than the boiling point of water, 100ºC. • The water is shown following a raised dot in the formula. • If there is more than one water molecule, a coefficient follows the dot. • Example • Ba(OH)2 • 8 H20

  19. Finding the formula mass of water (H2O). • Element# of atoms X at. Mass = contribution H 2 1.0 2.0 O 1 16.0 + 16.0 Formula mass = 18.0g • When you find the formula mass of a hydrate, leave water together like a single element. Ba(OH)2 • 8 H20 Element# of atoms X at. Mass = cont. % Comp. Ba 1 137.3 137.3 43.55% O 2 16.0 32.0 10.15% H 2 1.0 2.0 0.63% H2O 8 18.0 + 144.0 45.67% Formula mass = 315.3g (137.3/315.3)X100 (32.0/315.3)X100 (2.0/315.3)X100 (144.0/315.3)X100

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