Types of Formulas

Download Report

Transcript Types of Formulas

Determining formulae
The percentage composition of a compound
leads directly to its empirical formula.
Recall: An empirical formula for a compound
is the formula of a substance written with the
smallest integer subscripts.
Eg. Consider hydrogen peroxide:
Molecular formula = H2O2
Empirical formula = HO
Timberlake LecturePLUS
1
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
2
Compounds with different molecular formulae
can have the same empirical formula, and
such substances will have the same
percentage composition.
Eg.
acetylene = C2H2
benzene = C6H6
both have the empirical formula = ?
Timberlake LecturePLUS
3
Empirical Formulas
Write your own one-sentence definition for
each of the following:
Empirical formula
Molecular formula
Timberlake LecturePLUS
4
• 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
5
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
6
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
7
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
8
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
9
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 formula10
Timberlake LecturePLUS
Empirical
Formula
Empirical
Mass
Timberlake LecturePLUS
Molecular
Formula
Molecular
Mass
11
Empirical formula from Composition
Consider the following flow-diagram:
Percent composition
Mass Composition
Number of moles of
each element
Divide by smallest number of
moles to find the molar ratios
Multiply by appropriate number to
get whole number
subscripts
Timberlake LecturePLUS
12
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
13
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
14
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
15
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
16
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
17
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
18
Why moles?
Why do you need the number of moles
of each element in the compound?
Timberlake LecturePLUS
19
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
20
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
21
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
22
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
23
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
24
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
25
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
26
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
27
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
28
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
29
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
30
STRUCTURAL FORMULA
The atoms in a molecule are connected or
chemically bonded in a precise way. A SF.
Shows how the atoms in a molecule are
arranged.
For ex: H2O H-O-H
C2H6
CH3
H-C- C- H
H H
Timberlake LecturePLUS
31
Empirical formula
•The simplest whole number ratio of atoms of
elements in a compound, described with the use of
subscripts.
•Ionic compounds are always shown as empirical
formulas.
Molecular Formula
The actual numbers of atoms in a molecule.
Structural Formula
Show the relative arrangements of atoms in a
molecule
Timberlake LecturePLUS
32
Timberlake LecturePLUS
33
HYDRATES
Solids which are found in combined
form with water in definite proportion
are called as HYDRATES. When
hydrates are heated, H2O evaporates,
and only solid is obtained in
amorphous .
(w/o a certain geometric structure,
generally in powdered form. H2O
molecules surround ionic substances
with certain amounts.
Timberlake LecturePLUS
34
WATER OF HYDRATION : Water molecules of a hydrate.
• Na2CO3.10H2O
Na2CO3(s) + 10H2O(g)
• DEHYDRATION: Evaporation of water
• Na2CO3.10H2O,
of hydration.
• CaSO4.2H2O,
• CuSO4.5H2O
Timberlake LecturePLUS
35
Timberlake LecturePLUS
36