Chemical Composition

Chapter 8

Nutrasweet Nutrasweet

Aspartic acidAspartic acidAnalysisAnalysis

Qualitative -- what elements -- C, H, Qualitative -- what elements -- C, H, N, & O.N, & O.

Quantitative -- how many of each Quantitative -- how many of each element --Celement --C44HH77NONO44..

Counting AtomsCounting Atoms

Atoms are too small to be seen or counted Atoms are too small to be seen or counted individually.individually.

Atoms can only be counted by weighing Atoms can only be counted by weighing them.them.

• all jelly beans are not identical.all jelly beans are not identical.• jelly beans have an average mass of 5 g.jelly beans have an average mass of 5 g.• How could 1000 jelly beans be counted?How could 1000 jelly beans be counted?

Jelly Beans & MintsJelly Beans & MintsMints have an average Mints have an average

mass of 15 g.mass of 15 g.

How would you count out How would you count out 1000 mints?1000 mints?

Why do 1000 mints have Why do 1000 mints have a mass greater than a mass greater than 1000 jelly beans?1000 jelly beans?

Atomic Mass UnitAtomic Mass Unit

Atoms are so tiny that the gram is much too Atoms are so tiny that the gram is much too large to be practical.large to be practical.

The mass of a single carbon atom is 1.99 The mass of a single carbon atom is 1.99 x 10x 10-23-23 g. g.

The atomic mass unit (amu) is used for atoms The atomic mass unit (amu) is used for atoms and molecules.and molecules.

AMU’s and GramsAMU’s and Grams

1 amu = 1.661 x 10 1 amu = 1.661 x 10 -24 -24 gg

Conversion FactorsConversion Factors1.661 x 101.661 x 10-24-24g/amug/amu6.022 x 106.022 x 102323amu/gamu/g

Calculating Mass Using Calculating Mass Using AMU’SAMU’S

1 N atom = 14.01 amu1 N atom = 14.01 amu

(23 N atoms)(14.01 amu/1N atom) = (23 N atoms)(14.01 amu/1N atom) = 322.2 amu322.2 amu

Calculating Number of Atoms Calculating Number of Atoms from Massfrom Mass

1 O atom = 16.00 amu1 O atom = 16.00 amu

(288 amu)(1 O atom/16.00 amu) = (288 amu)(1 O atom/16.00 amu) = 18 atoms O18 atoms O

Atomic MassesAtomic Masses

Elements occur in nature as mixtures of Elements occur in nature as mixtures of isotopesisotopes

Carbon =Carbon = 98.89% 98.89% 1212CC 1.11% 1.11% 1313CC<0.01% <0.01% 1414CC

Carbon atomic mass = 12.01 amuCarbon atomic mass = 12.01 amu

AMU’s & GramsAMU’s & Grams

1 atom C = 12.011 amu = 1.99 x 101 atom C = 12.011 amu = 1.99 x 10-23-23gg

1 mol C = 12.011 g1 mol C = 12.011 g

Use TI-83 or TI-83 Plus to store 6.022 x 10Use TI-83 or TI-83 Plus to store 6.022 x 1023 23

to A.to A.

MeasurementsMeasurements

dozen = 12dozen = 12gross = 12 dozen = 144gross = 12 dozen = 144

ream = 500ream = 500mole = 6.022 x 10mole = 6.022 x 102323

The MoleThe MoleThe number equal to the The number equal to the numbernumber of carbon atoms in of carbon atoms in exactly 12 grams of pure exactly 12 grams of pure 1212C.C.

1 mole 1 mole of anythingof anything = = 6.022 6.022 10 102323 units ofunits of that thingthat thing

Equal Equal molesmoles of substances have of substances have equal numbers of atoms, equal numbers of atoms, molecules, ions, formula units, etc.molecules, ions, formula units, etc.

Figure 8.1 (a): All Figure 8.1 (a): All these sample of these sample of pure elements pure elements contain the same contain the same number (a mole) of number (a mole) of atoms: 6.022 x 10atoms: 6.022 x 102323 atomsatoms

Cu – 63.55g Ag – 107.9g

Pb – 207.2g

Figure 8.2: One-Figure 8.2: One-mole samples of mole samples of iron (nails), iron (nails), iodine crystals, iodine crystals, liquid mercury, liquid mercury, and powdered and powdered sulfursulfur

How many atoms does each substance contain?

The MoleThe 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 6.022 x 1023

Al 26.98 1 6.022 x 1023

Avogadro’s number Avogadro’s number equalsequals

6.022 6.022 10 102323 units units

The MoleThe Mole

• One mole of rice grains is more than the One mole of rice grains is more than the number of grains of all rice grown since the number of grains of all rice grown since the beginning of time!beginning of time!

• A mole of marshmallows would cover the A mole of marshmallows would cover the U.S. to a depth of 600 miles!U.S. to a depth of 600 miles!

• A mole of hockey pucks would be equal in A mole of hockey pucks would be equal in mass to the moon.mass to the moon.

Unit CancellationUnit Cancellation

How many dozen eggs would 36 eggs be?How many dozen eggs would 36 eggs be?(36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs(36 eggs)(1 dozen eggs/12 eggs) = 3 dozen eggs

How many eggs in 5 dozen?How many eggs in 5 dozen?(5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs(5 dozen eggs)(12 eggs/1 dozen eggs) = 60 eggs

Calculating Moles & Number of AtomsCalculating Moles & Number of Atoms1 mol Co = 58.93 g1 mol Co = 58.93 g(5.00 x 10(5.00 x 102020 atoms Co)(1mol/6.022 x 10 atoms Co)(1mol/6.022 x 102323 atoms) atoms)

= = 8.30 x 108.30 x 10-4 -4 mol Comol Co

(8.30 x 10(8.30 x 10-4 -4 mol)(58.93g/1 mol) = mol)(58.93g/1 mol) = 0.0489 g Co0.0489 g Co

Moles are the doorway grams <---> moles <---> atoms

Molar MassMolar MassA substance’s A substance’s molar mass molar mass is the mass in is the mass in grams of one mole of the compound.grams of one mole of the compound.

COCO22 = 44.01 grams per mole = 44.01 grams per mole

1 C = 1 (12.011 g) = 12.011 g1 C = 1 (12.011 g) = 12.011 g

2 O = 2 ( 16.00 g) = 2 O = 2 ( 16.00 g) = 32.00 g32.00 g

44.01 g44.01 g

Calculating Mass from MolesCalculating Mass from Moles

CaCOCaCO33

1 Ca = 1 (40.08 g) = 40.08 g1 Ca = 1 (40.08 g) = 40.08 g1 C = 1 (12.01 g) = 12.01 g 1 C = 1 (12.01 g) = 12.01 g 3 O = 3 (16.00 g) = 3 O = 3 (16.00 g) = 48.00 g48.00 g 100.09 g100.09 g(4.86 molCaCO(4.86 molCaCO33)(100.09 g/1 mol) = )(100.09 g/1 mol) = 486 g CaCO486 g CaCO33

Calculating Moles from MassCalculating Moles from Mass

JugloneJuglone10 C = 10(12.01g) = 120.1 g10 C = 10(12.01g) = 120.1 g6 H = 6(1.008 g) = 6.048 g6 H = 6(1.008 g) = 6.048 g33 O = 3(16.00 g) = O = 3(16.00 g) = 48.00 g48.00 g 174.1 g174.1 g(1.56 g juglone)(1 mol/174.1 g) = (1.56 g juglone)(1 mol/174.1 g) = 0.008960.00896 mol mol

juglonejuglone

Percent CompositionPercent Composition

Mass percent of an element:Mass percent of an element:

For iron in iron (III) oxide, (FeFor iron in iron (III) oxide, (Fe22OO33))

mass Fegg

%..

( . 11 1 6 91 5 9 69

10 0% ) 69 9 4 %

mass m ass o f elem en t in com poundmass o f com pound

% ( 10 0 % )

% Composition% Composition

CuSOCuSO44. . 5 H5 H22OO

1 Cu = 1 (63.55 g) = 63.55 g1 Cu = 1 (63.55 g) = 63.55 g1 S = 1 (32.06 g) = 32.06 g1 S = 1 (32.06 g) = 32.06 g4 O = 4 (16.00 g) = 64.00 g4 O = 4 (16.00 g) = 64.00 g5 H5 H22O = 5 (18.02 g) = O = 5 (18.02 g) = 90.10 90.10 gg

249.71 g249.71 g

% Composition % Composition (Continued)(Continued)

% Cu = 63.55 g/249.71g (100 %) = % Cu = 63.55 g/249.71g (100 %) = 25.45 % Cu25.45 % Cu% S = 32.06 g/249.71 g (100 %) = % S = 32.06 g/249.71 g (100 %) = 12.84 % S12.84 % S% O = 64.00 g/249.71 g (100 %) = % O = 64.00 g/249.71 g (100 %) = 25.63 % O25.63 % O% H% H22O = 90.10 g/249.71 g (100 %) = O = 90.10 g/249.71 g (100 %) = 36.08 % H36.08 % H22OO

Check: Total percentages. Should be equal to 100 % plus or minus 0.01 %.

FormulasFormulas

molecular formula = (empirical formula)molecular formula = (empirical formula)xx

[x = integer][x = integer]

molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

empirical formula = CHempirical formula = CH

FormulasFormulas

Ionic compounds -- empirical formula Ionic compounds -- empirical formula NaClNaCl CaClCaCl22

Covalent compounds -- molecular formula - Covalent compounds -- molecular formula - CC66HH1212OO66

CC22HH66

Empirical Formula DeterminationEmpirical Formula Determination

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound.compound.

2.2. Determine moles of each element in 100 Determine moles of each element in 100 grams of compound.grams of compound.

3.3. Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

4.4. Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers (if necessary.)obtain all whole numbers (if necessary.)

Calculating Empirical FormulasCalculating Empirical Formulas63.68 % C, 12.38 % N, 9.80 % H, & 14.14 % O63.68 % 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(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(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(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(14.14 g O)(1 mol/16.00g) = .8838 mol O/.8837 mol = 1

CC66NHNH1111OO

Calculating Empirical FormulasCalculating Empirical Formulas4.151 g Al & 3.692 g O4.151 g Al & 3.692 g O(4.151 g Al)(1 mol/26.98 g) = 0.1539 mol Al/0.1539 mol = 1.000(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(3.692 g O)(1 mol/16.00 g) = 0.2308 mol O/0.1539 mol = 1.500

1 Al (2) = 2 Al1 Al (2) = 2 Al1.5 O (2) = 3 O1.5 O (2) = 3 O

AlAl22OO33

Molecular FormulasMolecular Formulas

71.65 % Cl, 24.27 % C, & 4.07 % H71.65 % Cl, 24.27 % C, & 4.07 % H(71.65g Cl)(1 mol/35.45g) = 2.021 mol/2.021 mol = 1(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(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(4.07 g H )(1 mol/1.01g) = 4.03 mol/2.021 mol = 2

(EM)(EM)x x = (MM)= (MM)

(49.46)(49.46)x x = (98.96)= (98.96)

x = 2x = 2

(EF)x = (MF)(ClCH2)2 = Cl2C2H4