Unit IIIChemical

Composition(i.e. The Mole)

Atomic MassesAtomic masses use Carbon 12 (12C) as the standardCalculated with the aide of a mass

spectrometer

This process is also known as gas chromatography

As gas chromatograph counts the number of particles present in a given sample

Produces a graph like the one aboveWith this information percentages can be

calculated and formulas can be determined

Atomic mass is the average of all the naturally occurring isotopes of that elementCarbon = 12.011

Isotope Symbol Composition of the nucleus

% in nature

Carbon-12

12C 6 protons

6 neutrons

98.89%

Carbon-13

13C 6 protons

7 neutrons

1.11%

Carbon-14

14C 6 protons

8 neutrons

<0.01%

The MoleAbbrev. mol1 mol = # C atoms in 12 g of pure

12C

Avogadro’s numberEqual to 6.022 x 1023 atoms in

1 mol CNamed in honor of the Italian

chemist Amadeo Avogradro (1776-1855)

I didn’t discover it. Its just named

after me!

1 mol = 6.022 x 1023 atoms = molar mass (g) = 22.4 L

To covert amongst mol, atoms, grams, & liters, use the following equivalencies:1 mol = 6.022 x 1023 = molar mass = 22.4 L

atoms (grams)Or use ½ of the following chart:

Keys to Use:1. Only use 1 side of the chart

2. Proceed from one shaded box to another shaded box

3. When following the arrows, perform the indicated function

4. When going against the arrows, perform the opposite function

Molar MassThe molar mass is determined by summing the masses of the component atomsExample: What is the molar mass of MgCO3

24.31 g + 12.01 g + 3(16.00 g) =24.31 g + 12.01 g + 3(16.00 g) =84.32 84.32 gg

Molar CalculationsExample 1: How many grams of lithium are in 3.50 mol of lithium?3.50 mol Li → g Li

(3.50 • 6.94) / 1 = 24.29(3 SF’s in original problem) → Round24.3 → SSN → 2.43 x 101

Don’t forget the units → g Li

3.50 mol Li

1 mol Li

6.94 g Li= 2.43 x 101 g Li

Example 2: How many mol are in 98.2 g of NaCl?98.2 g NaCl → mol NaCl

(98.2 • 1) / 58.44 = 1.680355921(3 SF’s in original problem) → Round1.68 → SSN → 1.68 x 100

Don’t forget the units → mol NaCl

98.2 g NaCl

58.44 g NaCl

1 mol NaCl= 1.68 x 100 mol

NaCl

Example 3: How many atoms are in 411 g of calcium phosphate?411 g Ca3(PO4)2 → atoms Ca3(PO4)2

(411 • 6.022 x 1023) / 310.18 = 7.9793 x 1023

(3 SF’s in original problem) → Round7.98 x 1023 → SSN → 7.98 x 1023

Don’t forget the units → atoms Ca3(PO4)2

411 g Ca3(PO4)2

310.18 g Ca3(PO4)2

6.022 x 1023 atoms Ca3(PO4)2 =7.98 x 1023 atoms

Ca3(PO4)2

Example 4: How many liters of dioxygen heptioide gas does 7.5 x1024 atoms occupy?7.5 x 1024 atoms O2I7 → L O2I7

(7.5 x 1024 • 22.4) / 6.022 x 1023 = 278.977084(2 SF’s in original problem) → Round280 → SSN → 2.8 x 102

Don’t forget the units → L I2O7

7.5 x 1024 atoms O2I7

6.022 x 1023 atoms O2I7

22.4 L I2O7

= 2.8 x 102 L O2I7

Percent Composition of Compounds

Mass Percent = (Part / Whole) 100%Calculate the percent composition of

magnesium carbonate (MgCO3)Molar Mass = 84.32 g

24.31 g + 12.01 g + 3 (16.00)

Mg = (24.31 / 84.32) 100 = 28.83 %

C = (12.01 / 84.32) 100 = 14.24 %

O = (48.00 / 84.32) 100 = 56.93 %

100 %

Determining the Formula of a Compound

There are 2 basic formulas:1. Empirical

Simple – Can not be simplifiedExamples: H20, MgCl2

2. MolecularComplex – Can be simplifiedExamples: H8O4, Mg3Cl6Molecular Formula = (Empirical Formula )n

Example: (H20)4 = H8O4

n = molar mass / Empirical Formula Mass

Note: % are grams (if based on 100)Adipic acid contains 49.32% C, 6.85% H, &

43.84% O by mass. What is the E.F.?

C: (49.32/12.01)= 4.10 / 2.74 = 1.50 2 = 3

H: (6.85 /1.01) = 6.78 / 2.74 = 2.47 2 = 5

O: (43.84 /16.00)= 2.74 / 2.74 = 1 2 = 2

Answer:

C3H5O2

Moles Divide by smallest #

of mol

Mole Ratio X by

integer to obtain a whole #

# of atoms

Empirical Formula Determination

Empirical Formula Determination

Sample ProblemsSample Problem 1:.6884 g of lead combined with .2356 g of Cl

to form a binary compound. Calculate the empirical formula.

Sample Problem 2:A compound’s percent by mass is as follows:

Copper = 33.88%, Nitrogen = 14.94%, and Oxygen = 51.18%. Determine the empirical formula of the compound.

Problem:The E.F. for adipic acid is C3H5O2

The molar mass is 146 g/ molSteps:

1. Determine the E.F. mass of C3H5O2

(3 12.01) + (5 1.01) + (2 16.00) = 73.08 g

2. Find the n factorn = molar mass / E.F. massn = 146 / 73.08 = 2

3. M.F. = (E.F.)n = (C3H5O2)2

= C6H10O4

Molecular Formula Determination

Molecular Formula Determination

Sample ProblemsSample Problem 1:Calculate the M.F. of a compound with the

E.F. CH2O and a molar mass of 150 g/mol.

Sample Problem 2:A gas is composed of 71.75% Cl, 24.27% C,

and 4.07% H (by mass) & its molar mass is 98.96 g/mol. What is the M.F.?

Formulas of HydratesA hydrate is a compound that has a specific number of water molecules bound to its atomsWritten with each formula unit following a dotExamples:

1. Calcium chloride dihydrate → CaCl2 • 2 H2O

2. Magnesium sulfate heptahydrate→ MgSO4 • 7 H2O

Samples:1. Sodium cabonate decahydrate (molar mass ?)

2. Barium hydroxide octahydrate (molar mass ?)