Chapter 9
The Mole concept and Chemical
Formulas
Law of definite proportions
• In a pure compound elements are always
present in the same definite proportion by
mass.
Formula Mass
• Sum of atomic masses of all the atoms
present in one formula unit of a substance
expressed in atomic mass units.
• Formula mass is also called as Molar
mass.
Problems
•
•
•
•
•
•
1.Calculate the molar mass of LiClO4.
2.Calculate the molar mass of (NH4)3PO4
3.Calculate the molar mass of C2H6O
Practice Ex: 9.2
Use sig figures like addition rules.
Practice Ex: 9.3
Percent composition:
• Percent composition of a compound is the mass
percent of each element in the compound.
• If the formula of a compound is known a twostep process is needed to calculate the
percent composition.
• Step 1 Calculate the molar mass of the formula.
• Step 2 Divide the total mass of each element in
the formula by the molar mass and multiply by
100.
Percent composition:
total mass of the element
x 100 = percent of the element
molar mass
Percent Composition From
Experimental Data
• Percent composition can be calculated from
experimental data without knowing the
composition of the compound.
• Step 1 Calculate the mass of the compound
formed.
• Step 2 Divide the mass of each element by the
total mass of the compound and multiply by 100.
• 5. A compound containing nitrogen and oxygen
is found to contain 1.52 g of nitrogen and 3.47 g
of oxygen. Determine its percent composition.
• Practice Ex: 9.4,9.5
The Mole: The Mole
• The mass of a single atom is too small to
measure on a balance.
• mass of hydrogen atom = 1.673 x 10-24 g
• 1 mole = 6.022 x 1023 objects
• This number is called as Avogadro’s Number
• 1 mole of any element contains 6.022 x
1023 particles of that substance.
• The atomic mass in grams of any
element2contains 1 mole of atoms.
• This is the same number of particles as there
are in exactly 12 grams of
1 mole = 6.022 x
23
10 objects
6.022 x
23
10
is a very
LARGE
number
6.022 x
23
10
is
Avogadro’s Number
number
If 10,000 people started to count
Avogardro’s number and counted at
the rate of 100 numbers per minute
each minute of the day, it would take
over 1 trillion years to count the total
number.
1 mole of any element contains
6.022 x 1023
particles of that substance.
The atomic mass in grams
of any element23
contains 1 mole of atoms.
This is the same number of particles
6.022 x 1023
as there are in exactly 12 grams of
12
6
C
Species
Quantity
Number of
H atoms
H
1 mole
6.022 x
23
10
Species
Quantity
Number of
Fe atoms
Fe
1 mole
6.022 x
23
10
Species C6H6
Quantity 1 mole
Number of
23
6.022 x 10
C6H6
molecules
• The molar mass of an element is its
atomic mass in grams.
• It contains 6.022 x 1023 atoms
(Avogadro’s number) of the element.
1 mol of atoms = 6.022 x 1023 atoms
1 mol of molecules = 6.022 x 1023 molecules
1 mol of ions = 6.022 x 1023 ions
Element
Atomic mass Molar mass
Number of
atoms
H
1.008 amu
1.008 g
6.022 x 1023
Mg
24.31 amu
24.31 g
6.022 x 1023
Na
22.99 amu
22.99 g
6.022 x 1023
Avogadro’s
Number of
Particles
6x
Particles
1 MOLE
Molar Mass
23
10
Avogadro’s
Number of
H2O molecules
23
10
6x
H2O
molecules
1 MOLE H2O
18.02 g H2O
These relationships are present when
hydrogen combines with chlorine.
H
Cl
HCl
6.022 x 1023 H
atoms
6.022 x 1023 Cl
atoms
6.022 x 1023 HCl
molecules
1 mol H atoms
1 mol Cl atoms
1 mol HCl
molecules
1.008 g H
35.45 g Cl
36.46 g HCl
1 molar mass H 1 molar mass Cl 1 molar mass HCl
atoms
atoms
molecules
In dealing with diatomic elements (H2,
O2, N2, F2, Cl2, Br2, and I2), distinguish
between one mole of atoms and one
mole of molecules.
Problems:
• 6. Convert 1.20 moles of CO to molecules.
• 7. Convert 2.53 moles of Ag to Ag atoms.
• 8. Convert 0.025 mole of magnesium
sulfate to formula units.
• Do Practice : Ex: 9.6
The Mass of a Mole:
•
•
•
•
•
•
•
•
•
The molar mass of an element is its atomic mass in grams.
•It contains 6.022 x 1023 atoms (Avogadro’s number) of the
element.
Moles=grams/molar mass.
9. Convert 1.50 moles of CH4molecule to mass in grams.
10. Convert 2.50 moles of NaCl formula units to mass in grams.
11. Convert 1.68 moles of N2 molecule to mass in grams.
12. Convert 1.68 moles of N atoms to mass in grams.
Do Practice Ex: 9.7.
13. What is the mass in grams of a molecule whose mass on the
amu scale is 104.00 amu?
Problems
•
•
•
•
Do Practice Ex: 9.8
Do Practice Ex: 9.9
Do Practice Ex: 9.10
14. 56.04 g of N2 contains how many N2
molecules?
• 15. How many moles of benzene, C6H6, are
present in 390.0 grams of benzene?
• 16.How many grams of (NH4)3PO4 are
contained in 2.52 moles of (NH4)3PO4?
• Do Practice Ex: 9.11,9.12.
Problems
• 17. Calculate the mass in grams for :
•
a) a single atom of Cu.
•
b) a single molecule of CO2
• 18. Caffeine has the formula C8H10N4O2
How many grams of N are present in a
50.0g sample of caffeine.
• Do Practice Ex: 9.14.
Problems
• 19. The compound cholesterol has a formula C27H46O. How many
H atoms are present in 2.000 g sample of cholesterol.
• Percent purity: Percent by mass of a specified substance in an
impure sample of the substance.
• 20. A 32.00 g sample of HNO3 has a purity of 96.20 % by mass.
Calculate the following for this sample of nitric acid.
• a) The mass in grams of HNO3 present,b) the mass in grams of
impurities present.
• Do Practice Ex: 9.16.
• 21. How many Fe atoms are present in 1 25.00 g sample of Fe-ore
that has a purity of 87.70 % by mass Fe? Assume that there are no
Fe containing impurities present.
Empirical Formula versus
Molecular Formula
•
•
•
The empirical formula or simplest
formula gives the smallest wholenumber ratio of the atoms present in a
compound.
The molecular formula is the true
formula of a compound.
Two compounds can have identical
empirical formulas and different
molecular formulas.
Molecular Formula
C2H4
Empirical Formula
CH2
Smallest Whole
Number Ratio
C:H 1:2
Molecular Formula
C6H6
Empirical Formula
CH
Smallest Whole
Number Ratio
C:H 1:1
Calculating Empirical Formulas
• Step 1 Assume a definite starting
quantity (usually 100.0 g) of the
compound, if if the actual amount is not
given, and express the mass of each
element in grams.
• Step 2 Convert the grams of each
element into moles of each element using
each element’s molar mass.
Calculating Empirical Formulas
• Step 3 Divide the moles of atoms of each element by
the moles of atoms of the element that had the smallest
value.
• – If the numbers obtained are whole numbers, use them
as subscripts and write the empirical formula.
• – If the numbers obtained are not whole numbers, go on
to step 4.
• Step 4 Multiply the values obtained in step 3 by the
smallest numbers that will convert them to whole
numbers
• Use these whole numbers as the subscripts in the
empirical formula.
Some Common Fractions and Their Decimal Equivalents
Decimal
Resulting Whole
Equivalent
Number
Common Fraction
1
4
0.25
1
1
3
0.333…
1
0.666…
2
0.5
1
0.75
3
2
3
1
2
3
4
Multiply the
decimal equivalent
by the number in
the denominator of
the fraction to get
a whole number.
Problems:
• 22.The analysis of a salt shows that it contains
56.58% potassium (K); 8.68% carbon (C); and
34.73% oxygen (O). Calculate the empirical
formula for this substance.
• 23. The percent composition of a compound is
25.94% nitrogen (N), and 74.06% oxygen (O).
Calculate the empirical formula for this
substance.
• Do Practice Ex: 9.18, 9.19, 9.20
Calculating the Molecular Formula
from the Empirical Formula
•The molecular formula can be calculated
from the empirical formula if the molar
mass is known.
•The molecular formula will be equal to the
empirical formula or some multiple, n, of it.
•To determine the molecular formula
evaluate n.
•n is the number of units of the empirical
formula contained in the molecular formula
Problems:
• 24. What is the molecular formula of a
compound which has an empirical formula
of CH2 and a molar mass of 126.2 g?
• Do Practice EX: 9.23, 9.24
Combustion analysis
• A method used to measure the amounts of
C and H present in a combustible
compound that contains these two
elements when burned in O2.
• 25. Ethylene is burned in combustion
analysis apparatus and 3.14 g of CO2 and
1.29 g of H2O are produced. What is the
formula of ethylene?
• Do Practice Ex: 9.21.