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Atomic Mass and Formula Mass
Na: 22.990 amu
Cl: 35.453 amu
NaCl: 58.443 amu
Ca: 40.078 amu
Cl: 35.453 amu
CaCl
2
: 110.984 amu
Cl: 35.453 amu
Formula Mass
Average mass of a compound
also called molecular mass or molecular weight
Sum of the atomic masses of all atoms
CO
2
Formula mass = (1 x 12.011 amu)
Na
2
O
Formula mass = (2 x 22.990 amu)
= 44.009 amu
= 61.979 amu
Formula Mass
C
6
H
12
O
6
Formula mass =
Ca(NO
3
2
Formula mass =
(6 x 12.011 amu) + (12 x 1.008 amu) + (6 x 15.999 amu)
= 180.156 amu
= 164.086 amu
(1 x 40.078 amu) + (2 x 14.007 amu)+ (6 x 15.999 amu)
Molar Mass
Molar mass of an atom = mass(g) in one mole of atoms
Molar mass of carbon = 12.011 g/mol
Molar mass of a compound = mass(g) in one mole of
molecules or formula units
CO
2
molar mass = 44.009 g/mol
(= 12.011 amu + 2 x 15.999 amu) = 44.009 amu)
Mole (mol) is the amount of material containing
6.022 x 10
23
particles
Mole
Avogadro’s number
molar
mass
moles
atoms
molar
mass
moles
molecules
atomic mass
formula mass
Example: What is the mass of 12.0 million benzene (C 6
H
6
) molecules?
Example: Consider 4.49 g Ca 3
(PO
4
2
Its formula weight (FW) = 3(40.078) + 2(30.974) + 8(15.999) = 310.174 g/mol
How many mol of Ca
3
(PO
4
2
is this?
How many mol of Ca
2+ ions does this contain?
How many Ca
2+
ions are there?
Example: Consider 4.49 g Ca 3
(PO
4
2
Its formula weight (FW) = 3(40.078) + 2(30.974) + 8(15.999) = 310.174 g/mol
How many mol of P does that amount of Ca 3
(PO
4
2
contain?
How many mol of O does that amount contain?
How many grams of O are in that amount of salt?
Example: Consider 4.49 g Ca 3
(PO
4
2
Its formula weight (FW) = 3(40.078) + 2(30.974) + 8(15.999) = 310.174 g/mol
How many phosphate ions does that amount of Ca
3
(PO
4
2
contain?
Composition of Compounds
The chemical formula and the molar masses of the
elements in the compound indicates the relative quantities
of each element in a compound
In 1 mole CO
2
there is 1 mole C and 2 moles O
Example:
Calculate mass% composition of Cl in CCl
2
F
2
Mass percent Cl =
2 x molar mass Cl
molar mass CCl 2
F 2
x 100
= 58.64%
Molar mass of CCl 2
F 2
= 12.011 g/mol + 2(35.453 g/mol) + 2(18.998 g/mol) = 120.913 g/mol
Mass percent Cl =
2 x 35.453 g/mol
120.913 g/mol
x 100
Determining Empirical Formula from Experimental
Data
% A + %B + %C = 100
For the compound A
x
B
y
C
z
Example:
An unknown compound contains the following (by mass):
Determine its empirical formula
C
x
H
y
O
z
Simplest whole
number ratios of
atoms in compound
40.92% C
4.58% H
54.50% O
100 g
40.92 g C
4.58 g H
54.50 g O
Example:
1. Calculate the number of moles of C, H, and O
40.92 g C x = 3.407 mol C
12.011 g C
1 mol C
4.58 g H x = 4.54 mol H
1.008 g H
1 mol H
54.50 g O x = 3.406 mol O
15.999 g O
1 mol O
2. Write out the formula with the number of moles of each element
C
H
O
100 g
40.92 g C
4.58 g H
54.50 g O
Example:
C
H
O
3. Divide all by the smallest number
C
H
O
3.406 3.
C
H
O
4. Multiply through to get all whole numbers
1.33 x 3 = 4
1.0 x 3 = 3
Empirical Formula
C
3
H
4
O
3
Example: Determine the empirical formula of a compound that is 36.4% Mn,
21.2% S, and 42.4% O by mass
Example:
Organic Compounds
Organic compounds are composed of carbon and
hydrogen and a few other elements including nitrogen,
oxygen, and sulfur
Carbon always form four bonds
CH
4
methane
Organic Compounds
Carbon can bond to itself to form chain, branched, and ring structures
Organic Compounds
Carbon can also form double bonds and triple bonds with itself and other elements
Hydrocarbons
Hydrocarbons contain only carbon and hydrogen
examples: oil, gasoline, propane gas, natural gas
alkanes
(single bonds)
alkenes
(double bonds)
alkynes
(triple bonds)
Functionalized Hydrocarbons
Hydrocarbons that contain functional groups
Alcohols are organic compounds that have an −OH functional group
