PP_18_Empirical_Molecular_Formulas

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Empirical and Molecular
Formulas
Empirical vs Molecular
Formula
• The Molecular Formula (MF) gives the actual
number of each type of atom present.
• The Empirical Formula (EF) gives the lowest wholenumber ratio of the atoms present.
• Example:
C2H6
and
C3H9
– they have the same EF CH3 yet have very different MF
Types of Formulas
The formulas for compounds can be
expressed as an empirical formula and as a
molecular(true) formula.
Empirical
Molecular (true)
Name
CH
C 2H 2
acetylene
CH
C 6H 6
benzene
CO2
CO2
CH2O
C5H10O5
Timberlake LecturePLUS
carbon dioxide
ribose
3
• 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.
Timberlake LecturePLUS
4
EF vs. MF
When you have an ionic compound:
the EF = MF
For some molecular compounds:
the EF = MF, but not always
Determining the EF
Remember: The EF is the lowest whole-number
ratio of the moles of each atom present.
Example: CH4 has 1 mol C atoms
4 mol H atoms
Determining the EF
If a compound consists of:
62.1% C
13.8% H
24.1% N
The percentages are based on MASS not
MOLES
We can compare moles, not masses
Determining the EF
In order to determine the ratio of C:H:N, we
need to know the mole ratio
Step 1: Convert % of each into grams
Make it easy on yourself, assume a sample
size of 100.00g
62.1g C
13.8g H
24.1g N
Determining the EF
Now that you know how many grams of each atom
you have:
Step 2: Convert grams to moles using the molar
mass of each
62.1g C
13.8g H
24.1 g N
Determining the EF
Now that you have the mol ratio, you need to
make them Whole-Numbers.
Step 3: Divide each mol by the smallest mol
value from step #2
5.17 mol C
13.7 mol H
1.72 mol N
Determining the EF
This results in 3 mol C, 8 mol H and 1 mol N
therefore the EF = C3H8N
Learning Check EF-1
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
Timberlake LecturePLUS
3) C3H6O3
12
Solution EF-1
A. What is the empirical formula for C4H8?
2) CH2
B. What is the empirical formula for C8H14?
1) C4H7
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
Timberlake LecturePLUS
3) C3H6O3
13
Learning Check EF-2
If the molecular formula has 4 atoms of
N, what is the molecular formula if SN is
the empirical formula? Explain.
1) SN
2) SN4
3) S4N4
Timberlake LecturePLUS
14
Solution EF-2
If the molecular formula has 4 atoms of
N, what is the molecular formula if SN is
the empirical formula? Explain.
3) S4N4
If the actual formula has 4 atoms of N,
and S is related 1:1, then there must also
be 4 atoms of S.
Timberlake LecturePLUS
15
Empirical and Molecular Formulas
molar mass
=
simplest mass
a whole number = n
n = 1 molar mass = empirical mass
molecular formula = empirical formula
n = 2 molar mass = 2 x empirical mass
molecular formula =
2 x empirical formula
molecular formula
= or > empirical formula16
Timberlake LecturePLUS
EF to MF
To go from the EF to the MF, you need two additional
pieces of information:
1 – calculate the mass from your EF
2 – You must be given the mass of the MF
(X) EFmass = MFmass
X = MFmass / EFmass
EF to MF
Example: You found the EF to be HO, and the MFmass =
34.02g
1.
2.
3.
4.
Calculate the EFmass = 17.01g
Calculate X = 34.02g / 17.01g
X=2
EF = HO
MF = H2O2
Learning Check EF-3
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What
is the molecular formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
Timberlake LecturePLUS
19
Solution EF-3
A compound has a formula mass of 176.0
and an empirical formula of C3H4O3. What is
the molecular formula?
2) C6H8O6
C3H4O3 = 88.0 g/EF
176.0 g
=
2.00
88.0
Timberlake LecturePLUS
20
Learning Check EF-4
If there are 192.0 g of O in the
molecular formula, what is the true
formula if the EF is C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12
Timberlake LecturePLUS
21
Solution EF-4
If there are 192.0 g of O in the
molecular formula, what is the true
formula if the EF is C7H6O4?
3) C21H18O12
192 g O
= 3 x O4 or 3 x C7H6O4
64.0 g O in EF
Timberlake LecturePLUS
22
Finding the Molecular 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. State mass percents as grams in a
100.00-g sample of the compound.
Cl 71.65 g
C 24.27 g
H 4.07 g
Timberlake LecturePLUS
23
2. Calculate the number of moles of each
element.
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
=
Timberlake LecturePLUS
4.04 mol H
24
Why moles?
Why do you need the number of moles
of each element in the compound?
Timberlake LecturePLUS
25
3. Find the smallest whole number ratio by
dividing each mole value by the smallest
mole values:
Cl: 2.02
=
1 Cl
2.02
C:
H:
2.02
2.02
=
1C
4.04
=
2H
2.02
4. Write the simplest or empirical formula
CH2Cl
Timberlake LecturePLUS
26
5. EM (empirical mass)
= 1(C) + 2(H) + 1(Cl) = 49.5
6. n = molar mass/empirical mass
Molar mass
EM
=
99.0 g/mol
49.5 g/EM
= n=2
7.Molecular formula
(CH2Cl)2
= C2H4Cl2
Timberlake LecturePLUS
27
Learning Check EF-5
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.
Timberlake LecturePLUS
28
Solution EF-5
60.0 g C x
4.5 g H
___________= ______ mol C
x ___________ = _______mol H
35.5 g O x ___________ = _______mol O
Timberlake LecturePLUS
29
Solution EF-5
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
Timberlake LecturePLUS
30
Divide by the smallest # of moles.
5.00 mol C =
________________
______ mol O
4.5 mol H
=
______ mol O
________________
2.22 mol O =
________________
______ mol O
Are are the results whole numbers?_____
Timberlake LecturePLUS
31
Divide by the smallest # of moles.
5.00 mol C =
___2.25__
2.22 mol O
4.5 mol H
2.22 mol O
=
___2.00__
2.22 mol O =
___1.00__
2.22 mol O
Are are the results whole numbers?_____
Timberlake LecturePLUS
32
Finding Subscripts
A fraction between 0.1 and 0.9 must not
be rounded. Multiply all results by an
integer to give whole numbers for
subscripts.
(1/2)
(1/3)
(1/4)
(3/4)
0.5
0.333
0.25
0.75
x2
x3
x4
x4
=
=
=
=
Timberlake LecturePLUS
1
1
1
3
33
Multiply everything x 4
C: 2.25 mol C
x 4 = 9 mol C
H: 2.0 mol H
x 4 = 8 mol H
O: 1.00 mol O
x 4 = 4 mol O
Use the whole numbers of mols as the
subscripts in the simplest formula
C9H8O4
Timberlake LecturePLUS
34
Learning Check EF-6
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?
Timberlake LecturePLUS
35
Solution EF 6
0.853 mol S /0.853 = 1 S
0.857 mol N /0.853 = 1 N
1.71 mol Cl /0.853
= 2 Cl
Empirical formula = SNCl2 = 117.1 g/EF
Mol. Mass/ Empirical mass
351/117.1 = 3
Molecular formula = S3N3Cl6
Timberlake LecturePLUS
36
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