2. Empirical Formulas
• Best way to identify an unknown
compound is to determine its chemical
formula.
• Empirical Formula: the smallest whole
number ratio of a compound. (Ionic
compounds are already in its empirical
formulas)
3. Empirical Vs.
Molecular Formulas
• Molecular Formula
• a formula that
shows the element
symbols & exact
number of each
type of atom in a
molecular
compound
• Empirical Formula
• a formula that shows the
simplest whole-number
ratio of elements in
compounds
4. CH4 Vs. Vs. C18H72
C6H24
Lowest ratio
In empirical
& molecular
formula
can be reduced
to lowest ratio
6 can go into 6
& 24--> CH4
same empirical formula.
But not the same
molecular formula
can be reduced
to lowest ratio
18 can go into
18 & 72--> CH4
5. • If you know the chemical formula you
can find percent composition.
CH4 C6H24
M= 12.01g/mol + (4 x 1.01g/mol)
= 16.00 g/mol
MC= 12.01g/mol
MH= 1.01g/mol
(12.00g/mol )x 1mol
16.00g
x 100% = 75%
(4.01g/mol) x 1mol
16.00g
x 100% = 25%
MC6H24= 96 g/mol
MC= 72 g/mol
(72 g/mol ) 1mol
x
96.00g
x100% = 75%
- same empirical formula
- should be ratios of each other
-helps determine percent composition
CH4 C6H24
&
6. Determining
Empirical Formula
• Find the empirical formula of a compound with
percentage composition 35.4% Na and the
remainder nitrogen
Step 1: Assume your conveniently working with a mass of 100.0
g.
100.0g - 35.4g = 64.6 g
Na= 35.4
g
N= 64.6 g
Step 2: Calculate the amount of each element.
nNa = (35.4 g) (1 mol)
22.99g
nN = (64.5 g) (1 mol)
14.01g
= 1.540 mol = 4.6110 mol
7. Determining
Empirical Formula
• Find the empirical formula of a compound with
percentage composition 35.4% Na and the
remainder nitrogen
Step 3: To determine the simplest ratio of the elements in the
compound, divide the amount of each element by the smallest
amount
nNa
= 1
nN
1.540 mol
4.6110 mol
nNa
=1.540 mol
nNa
= 1.540 mol
= 2.98
8. Determining
Empirical Formula
• Find the empirical formula of a compound with
percentage composition 35.4% Na and the
remainder nitrogen
Step 2: Calculate the amount of each element.
nNa
= 1
nN
1.540 mol
4.6110 mol
nNa
=
1.540 mol nNa
=
1.540 mol
= 2.98
The ratio for Na & N is1:2.98. Since we cannot
have a fraction of an element in a compound, the
value of N is rounded off to the nearest whole
number. As a result, the simplest whole-number
ratio is 1:3........NaN3
9. • If a number is within 0.05 of a whole
number, you can round it up or down to
the nearest whole number.
• But what do you do if one of the values
is not close to a whole number? You
multiply all the subscripts by the fraction
denominator gives whole numbers
• C1 x 3 H1.333 x 3 = C3H4
10. Practice
• Determine the empirical formula of a compound listed
that contains 52.2% carbon, 6.15% hydrogen and
41.7% oxygen
Step 1: Assume your conveniently working with a mass of 100.0
g.
nc = (52.2 g) (1 mol)
12.01g
= 4.3464 mol
nH = (6.15g) (1 mol)
1.01g
= 6.0891mol
n0 = (41.7g) (1 mol)
16.00g
= 2.6063mol
11. Practice
• Determine the empirical formula of a compound listed
that contains 52.2% carbon, 6.15% hydrogen and
41.7% oxygen
Step 2: Divide the amount of each element by the smallest
amount.
nc
=
= 1
nH =
= 1.67
n0 =
2.6063mol
n0
n0
n0
2.6063mol
2.6063mol
2.6063mol
6.0891mol
4.3464 mol
= 2.29
Step 3: These calculations
give an empirical formula of
C1.67H2.29O1
Multiplying each of the
subscripts by 3 give C5H7O3
12. Practice
• Write the empirical formula of compounds with the following
percentage
• 20.2% Al; 79.8% Cl
13. Practice
• Write the empirical formula of compounds with the following
percentage
• 50.85% carbon; 8.47% hydrogen; and 40.68% oxygen
14. Molecular Formula
• gives the exact number of atoms of each
element present
• “chemical formula” = molecular formula
• A compound’s molar mass is always a
whole number multiple of the molar mass
of the empirical formula. You ca find this
multiple x, by dividing the molar mass of
the compound by the molar mass of the
empirical formula
16. MCH2 = 1( 12.01 g/mol) + 1(1.01g/mol)
Find a Molecular Formula Given its
Empirical formula
Step 1: Find the empirical molar mass by adding the
molar mass of each element
Compound with empirical formula CH2 & the molar
mass of 84.18 g/mol
= 14.03 g/mol
17. x = 84.18 g/mol
14.03 g/mol
Find a Molecular Formula Given its
Empirical formula
Step 2: Solve for x,
Compound with empirical formula CH2 & the molar
mass of 84.18g/mol
x = 6
18. CH2= C6H12
Find a Molecular Formula Given its
Empirical formula
Step 3: therefore, the molar mass of the compound is
6 times the molar mass of the empirical formula
Compound with empirical formula CH2 & the molar
mass of 84.18g/mol
molecular formula
20. Using Percentage
Composition & Molar Mass
Data
• Determine the molecular formula of vitamin C
(ascorbic acid). This compound contains 40.5%
Carbon, 4.6% hydrogen, 54.5% oxygen. Its molar
mass is 176.14g/mol
