Transcript Types of Formulas - SCH3U-CCVI

Empirical and Molecular
Formulas
SCH 3U
Types of Formulas
The formulas for compounds can be
expressed as an empirical formula and as a
molecular (true) formula.
Empirical
Formula
CH
Molecular/True
Formula
C2H2
Name
acetylene
CH
C6H6
benzene
CO2
CO2
Carbon dioxide
CH2O
C5H10O5
ribose
• 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.
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
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
Finding the Empirical Formula
A compound is Cl 71.65%, C 24.27%, and H
4.07%. What are the empirical and molecular
formulas? The molar mass is known to be
99.0 g/mol.
1. “Percent to Mass” - state mass percentages
as grams in a 100.00 g sample of the
compound.
Cl 71.65 g
C 24.27 g
H 4.07 g
2. “Mass to Moles”
71.65 g Cl x 1 mol Cl = 2.02 mol Cl
35.5 g Cl
24.27 g C x
1 mol C
12.0 g C
= 2.02 mol C
4.07 g H x
1 mol H
1.01 g H
=
4.04 mol H
3. “Divide by Small”
Find the smallest whole number ratio by
dividing each mole value by the smallest mole
values:
Cl: 2.02 mol
=
1 Cl
2.02 mol
C:
H:
2.02 mol
2.02 mol
=
1C
4.04 mol
= 2H
2.02 mol
4. Write the simplest or empirical formula
CH2Cl
Finding the Molecular Formula
Multiplier:
molar mass
= a whole number
empirical mass
To get Molecular Formula, first calculate
Empirical Formula, then multiply all
subscripts by the multiplier
Note: If your multiplier = 1, the molecular
formula = empirical formula
eg. if multiplier = 2, the MF = 2 x EF
Back to the problem….
We calculated Empirical Formula = CH2Cl & were given the
molar mass = 99.0 g/mol
5. Calculate EM (empirical mass)
= 1(C) + 2(H) + 1(Cl) = 49.5 g/mol
6. Calculate Multiplier:
Molar mass
=
99.0 g/mol = multiplier = 2
Empirical mass
49.5 g/mol
7. Molecular formula = Empirical Formula x multiplier
(CH2Cl) x 2
= C2H4Cl2
Learning Check
Aspirin is 60.0% C, 4.5 % H and 35.5%
O. Calculate its simplest formula. In
100 g of aspirin, there are 60.0 g C,
4.5 g H, and 35.5 g O.
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”
“Percent to Mass & Mass to Mole”
60.0 g C x
1 mol C
=
5.00 mol C
=
4.5 mol H
=
2.22 mol O
12.0 g C
4.5 g H
x
1 mol H
1.01 g H
35.5 g O x
1mol O
16.0 g O
“Divide by Small, Multiply ‘til Whole”
5.00 mol C
2.22
=
2.25
x 4 = 9 C
4.5 mol H
2.22
=
2.00
x 4 = 8 H
2.22 mol O
2.22
=
1.00
x 4 = 4 O
Therefore, the Molecular Formula (MF) = C9H8O4
Finding Subscripts for Decimals
A fraction between 0.1 and 0.9 must not
be rounded. Multiply all results by an
integer to give whole numbers for
subscripts.
Fraction
½
1/3
Decimal
0.5
0.333
Multiply by:
2
3
To Get:
1
1
¼
¾
0.25
0.76
4
4
1
3
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
S = 0.853 mol & divide by 0.853 = 1 S
N = 0.857 mol & divide by 0.853 = 1 N
Cl = 1.71 mol & divide by 0.853 = 2 Cl
Empirical Formula (EF) = SNCl2
Empirical Mass (EM) = 117.1 g/mol
Given Molar Mass = 351 g/mol
Multiplier =
351g/mol = 3,
117.1g/mol
 so MF = S3N3Cl6