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Chemical Composition

Chemical Composition. Chapter 8. Nutrasweet . Aspartic acid Analysis Qualitative -- what elements -- C, H, N, & O. Quantitative -- how many of each element --C 4 H 7 NO 4 . Counting Atoms. Atoms are too small to be seen or counted individually.

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Chemical Composition

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  1. Chemical Composition • Chapter 8

  2. Nutrasweet • Aspartic acid • Analysis • Qualitative -- what elements -- C, H, N, & O. • Quantitative -- how many of each element --C4H7NO4.

  3. Counting Atoms • Atoms are too small to be seen or counted individually. • Atoms can only be counted by weighing them. • all jelly beans are not identical. • jelly beans have an average mass of 5 g. • How could 1000 jelly beans be counted?

  4. Jelly Beans & Mints • Mints have an average mass of 15 g. • How would you count out 1000 mints? • Why do 1000 mints have a mass greater than 1000 jelly beans?

  5. Atomic Mass Unit • Atoms are so tiny that the gram is much too large to be practical. • The mass of a single carbon atom is 1.99 • x 10-23 g. • The atomic mass unit (amu) is used for atoms and molecules.

  6. AMU’s and Grams • 1 amu = 1.661 x 10 -24 g • Conversion Factors • 1.661 x 10-24g/amu • 6.022 x 1023amu/g

  7. Calculating Mass Using AMU’S • 1 N atom = 14.01 amu • (23 N atoms)(14.01 amu/1N atom) = 322.2 amu

  8. Calculating Number of Atoms from Mass • 1 O atom = 16.00 amu • (288 amu)(1 O atom/16.00 amu) = 18 atoms O

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

  10. AMU’s & Grams • 1 atom C = 12.011 amu = 1.99 x 10-23g • 1 mol C = 12.011 g • Use TI-83 or TI-83 Plus to store 6.022 x 1023 to A.

  11. Measurements • dozen = 12 • gross = 12 dozen = 144 • ream = 500 • mole = 6.022 x 1023

  12. The Mole • The number equal to the number of carbon atoms in exactly 12 grams of pure 12C. • 1 mole of anything = 6.022  1023units of that thing Equal moles of substances have equal numbers of atoms, molecules, ions, formula units, etc.

  13. Figure 8.1 (a): All these sample of pure elements contain the same number (a mole) of atoms: 6.022 x 1023 atoms Pb – 207.2g Cu – 63.55g Ag – 107.9g

  14. Figure 8.2: One-mole samples of iron (nails), iodine crystals, liquid mercury, and powdered sulfur How many atoms does each substance contain?

  15. The Mole • Substance Average Atomic Mass # Moles# Atoms • (g) • Na22.9916.022 x 1023 • Cu63.55 16.022 x 1023 • S32.061 6.022 x 1023 • Al26.98 16.022 x 1023

  16. Avogadro’s number equals6.022  1023 units

  17. The Mole • One mole of rice grains is more than the number of grains of all rice grown since the beginning of time! • A mole of marshmallows would cover the U.S. to a depth of 600 miles! • A mole of hockey pucks would be equal in mass to the moon.

  18. Unit Cancellation • How many dozen eggs would 36 eggs be? • (36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs • How many eggs in 5 dozen? • (5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs

  19. Calculating Moles & Number of Atoms • 1 mol Co = 58.93 g • (5.00 x 1020 atoms Co)(1mol/6.022 x 1023 atoms) • = 8.30 x 10-4 mol Co • (8.30 x 10-4 mol)(58.93g/1 mol) = 0.0489 g Co • Moles are the doorway • grams <---> moles <---> atoms

  20. Molar Mass • A substance’s molar mass is the mass in grams of one mole of the compound. • CO2 = 44.01 grams per mole • 1 C = 1 (12.011 g) = 12.011 g • 2 O = 2 ( 16.00 g) = 32.00 g • 44.01 g

  21. Calculating Mass from Moles • CaCO3 • 1 Ca = 1 (40.08 g) = 40.08 g • 1 C = 1 (12.01 g) = 12.01 g • 3 O = 3 (16.00 g) = 48.00 g • 100.09 g • (4.86 molCaCO3)(100.09 g/1 mol) = 486 g CaCO3

  22. Calculating Moles from Mass • Juglone • 10 C = 10(12.01g) = 120.1 g • 6 H = 6(1.008 g) = 6.048 g • O = 3(16.00 g) = 48.00 g • 174.1 g • (1.56 g juglone)(1 mol/174.1 g) = 0.00896mol juglone

  23. Percent Composition • Mass percent of an element: • For iron in iron (III) oxide, (Fe2O3)

  24. % Composition • CuSO4. 5 H2O • 1 Cu = 1 (63.55 g) = 63.55 g • 1 S = 1 (32.06 g) = 32.06 g • 4 O = 4 (16.00 g) = 64.00 g • 5 H2O = 5 (18.02 g) = 90.10 g • 249.71 g

  25. % Composition (Continued) • % Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu • % S = 32.06 g/249.71 g (100 %) = 12.84 % S • % O = 64.00 g/249.71 g (100 %) = 25.63 % O • % H2O = 90.10 g/249.71 g (100 %) = 36.08 % H2O Check: Total percentages. Should be equal to 100 % plus or minus 0.01 %.

  26. Formulas • molecular formula = (empirical formula)x [x = integer] • molecular formula = C6H6 = (CH)6 • empirical formula = CH

  27. Formulas • Ionic compounds -- empirical formula • NaCl • CaCl2 • Covalent compounds -- molecular formula - • C6H12O6 • C2H6

  28. 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 (if necessary.)

  29. Calculating Empirical Formulas • 63.68 % C, 12.38 % N, 9.80 % H, & 14.14 % O • (63.68 g C)(1 mol/12.01g) = 5.302 mol C/.8837 mol = 6 • (12.38 g N)(1 mol/14.01g) = 0.8837 mol N/.8837 mol = 1 • (9.80 g H)(1 mol/1.01 g) = 9.70 mol H/.8837 mol = 11 • (14.14 g O)(1 mol/16.00g) = .8838 mol O/.8837 mol = 1 • C6NH11O

  30. Calculating Empirical Formulas • 4.151 g Al & 3.692 g O • (4.151 g Al)(1 mol/26.98 g) = 0.1539 mol Al/0.1539 mol = 1.000 • (3.692 g O)(1 mol/16.00 g) = 0.2308 mol O/0.1539 mol = 1.500 • 1 Al (2) = 2 Al • 1.5 O (2) = 3 O • Al2O3

  31. Molecular Formulas • 71.65 % Cl, 24.27 % C, & 4.07 % H • (71.65g Cl)(1 mol/35.45g) = 2.021 mol/2.021 mol = 1 • (24.27 g C)(1 mol/12.01g) = 2.021 mol/2.021 mol = 1 • (4.07 g H )(1 mol/1.01g) = 4.03 mol/2.021 mol = 2 • (EM)x = (MM) • (49.46)x = (98.96) • x = 2 (EF)x = (MF) (ClCH2)2 = Cl2C2H4

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