Chemical Composition Chapter 8

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

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?

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?

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.

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

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

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

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

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.

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

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

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

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

The Mole Substance Average Atomic Mass # Moles # Atoms (g) Na 22.99 1 6.022 x 1023 Cu 63.55 1 6.022 x 1023 S 32.06 1 6.022 x 1023 Al 26.98 1 6.022 x 1023

Avogadro’s number equals 6.022  1023 units

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.

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

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

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

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

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) = 0.00896 mol juglone

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

% 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

% 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 %.

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

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

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.

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

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

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