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Chemical Composition

Learn about chemical composition and analysis, including qualitative and quantitative methods for counting atoms, atomic masses, molar mass, and empirical formula determination. Understand the significance of Avogadro's number and unit cancellation in chemistry.

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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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