Download Empirical and Molecular revised

(4.6/4.7) Empirical and
Molecular Formulas
SCH 3U
• An empirical formula represents the
simplest whole number ratio of the
atoms in a compound.
• The molecular formula is the true or
actual ratio of the atoms in a
compound.
Types of Formulas
The formulas for compounds can be
expressed as an empirical formula and as a
molecular (true) formula.
eg. empirical formula = CH2O
Learning Check
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
3) C3H6O3
Finding the Empirical Formula
a) A compound is 71.65% Cl, 24.27% C, and
4.07% H. What is the empirical formula?
1. Assume a 100 g sample. Determine the
mass, in grams, of each element present.
Cl 71.65 g
C 24.27 g
H 4.07 g
2. Calculate the number of moles of each
element.
nCl = 71.65 g
= 2.021 mol
35.45 g/mol
nC = 24.27 g
= 2.021 mol
12.01 g/mol
nH = 4.07 g
1.01 g/mol
= 4.03 mol
3. Divide each by the smallest number of moles
to obtain the simplest whole number ratio.
**If whole numbers are not obtained, multiply
subscripts by the smallest number that will give
whole numbers**
Cl: 2.021 mol
= 1.000 Cl (1)
2.021 mol
C:
H:
2.021 mol
2.021 mol
=
1.000 C (1)
4.04 mol
= 2.00 H (2)
2.021 mol
4. Write the simplest or empirical formula.
CH2Cl
Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5%
O. Calculate the empirical formula.
nC = 60.0 g
=
5.00 mol
=
4.5 mol
=
2.22 mol
12.01 g/mol
nH = 4.5 g
1.01 g/mol
nO = 35.5 g
16.00 g/mol
5.00 mol C
2.22
=
2.25 mol X 4 = 9 mol C
4.5 mol H
2.22
=
2.0 mol
X 4 = 8 mol H
2.22 mol O
2.22
=
1.00 mol
X 4 = 4 mol O
Therefore, the Empirical Formula (EF) = C9H8O4
Finding the molecular formula
a) A compound is Cl 71.65%, C 24.27%, and H 4.07%. What
is the empirical formula? (from yesterday, CH2Cl)
b) The molar mass is known to be 99.0 g/mol. What is the
molecular formula?
1. Calculate EFM (empirical formula mass)
1(12.01g/mol) + 2(1.01 g/mol) + 1(35.45g/mol) = 49.48 g/mol
2. Calculate Multiplier:
Molar mass (M)
=
EFM
99.0 g/mol
49.48 g/mol
=
2.00
3. Multiply the empirical formula subscripts by the multiplier
(CH2Cl) x 2
= C2H4Cl2
Learning Check
A compound is 27.4% S, 12.0% N and
60.6 % Cl. If the compound has a molar
mass of 351 g/mol, what is the molecular
formula?
Solution
nS = 27.4 g
=
0.855 mol
=
0.857 mol
=
1.71 mol
32.06 g/mol
nN = 12.0 g
14.01 g/mol
nCl = 60.6 g
35.45 g/mol
Solution
0.855 mol S
0.855
=
1.00 mol
0.857 mol N
0.855
=
1.00 mol
1.71 mol Cl
0.855
=
2.00 mol
Therefore, the Empirical Formula (EF) = SNCl2
Solution
empirical formula mass (EFM)
= 32.06 g/mol + 14.01 g/mol + (2)(35.45 g/mol)
= 116.97 g/mol
Molar Mass = 351 g/mol
Multiplier = 351 g/mol = 3.00
116.97 g/mol
 so MF is S3N3Cl6