CHEM 1405 - Chapter 8
THE MOLE CONCEPT
The Mole:
The amount of substance that contains 6.022 x 1023 elementary entities is said to be one mole.
This number is called the Avogadro’s number
One mole contains 6.022 x 1023 elementary entities
1 mole atoms
= 6.022 x 1023 atoms
1 mole molecules
= 6.022 x 1023 molecules
1 mole ions
= 6.022 x 1023 ions
1 mole oranges
= 6.022 x 1023 oranges
1 mole of element = Atomic mass in grams
1 mole of compound = Molecular mass in grams
1 mole of ions = Ionic mass in grams
Number of moles = Mass in grams / molar mass
The Molar mass
Mass of one mole of a substance expressed in grams is the molar mass
Molar Volume
Volume occupied by 1 mole of a gas at STP is the molar volume. This is equal to 22.4 L at STP
Substance Formula mass or
molar mass
particles per mole
moles
Molecular mass
H2
2.016 amu
2.016 g
6.022 x 1023 molecules of H2
= 2 (6.022 x 1023) atoms of H
1 mole molecules
2 mole atoms
H2O
18.02 amu
18.02 g
6.022 x 1023 molecules of H2O
= 2 (6.022 x 1023) atoms of H
= 1 (6.022 x 1023) atoms of O
1 mole molecules
2 mole atoms
1 mole atoms
NH3
17.03amu
17.03 g
6.022 x 1023 molecules of NH3
= 1 (6.022 x 1023) atoms of N
= 3(6.022 x 1023) atoms of H
1 mole molecules
1 mole atoms
3 mole atoms
Ex.: How many atoms are there in a penny that weighs 3 g of Cu atoms
= (3g Cu) (1mol Cu /63.5g Cu)(6.022 x 1023 Cu atoms/1 mol Cu)
= 3 x 1022 Cu atoms
Percentage Composition of Elements
Percentage composition of Element
=
n x Molar mass of element x 100
Molar mass of Compound
‘n’ is the number of moles of the element present in one mole of the compound.
Calculate the percentage composition of elements in Sulfuric acid.
Empirical Formula
The formula that gives the simplest whole number ratio of the constituting atoms in a molecule
of the compound is called the empirical formula.
Molecular Formula:
The formula that shows the actual number of each kind of atoms present in a molecule
Finding the empirical formula
Q: A compound contains 44.66% Potassium (K), 18.13% Phosphorous (P), 36.63% Oxygen
(O) and 0.5785% Hydrogen (H). Find the empirical formula.
Element
Percentage by mass
# of moles
K
44.66
(44.66/39.10) = 1.142
# of moles
Least value
(1.142/. 5739) = 1.990
P
18.13
(18.13/30.97) = 0.5854
(.5854/. 5739) = 1.020
1
O
36.63
(36.63/16.00) = 2.289
(2.289/ .5739) = 3.989
4
H
0.5785
(.5785/1.008) = 0.5739
(.5739/.5739) = 1.000
1
Thus, the formula of the compound is K2HPO4.
Molecular Formula = n x Empirical Formula
Molecular Formula mass
n =
Empirical Formula mass
Ratio
2