Chemical Formulas

1
Chemical Formulas
Empirical Formula - Shows the relative number of
atoms of each element in the compound.
It is the simplest formula, and is
derived from masses of the elements. Molecular
Formula - Shows the actual number of atoms
of each element in the molecule of the
compound. Structural Formula - Shows the actual
number of atoms, and the bonds between
them that is, the arrangement of
atoms in the molecule.
2
Empirical and Molecular Formulas
Empirical Formula - The simplest formula for a
compound that agrees
with the elemental analysis! The
smallest set of whole numbers of
atoms. Molecular Formula - The formula of the
compound as it exists,
it may be a multiple of the Empirical
formula.
3
Calculating the Moles and Number of Formula
Units in a Given Mass of Cpd.
Problem Sodium Phosphate is a component of some
detergents. How many moles and
formula units are in a 38.6 g sample? Plan We
need to determine the formula, and the molecular
mass from the atomic masses of each
element multiplied by the coefficients. Solution
The formula is Na3PO4. Calculating the molar
mass M 3x Sodium 1 x
Phosphorous 4 x Oxygen
3 x 22.99 g/mol 1 x 30.97 g/mol 4 x 16.00
g/mol 68.97 g/mol 30.97
g/mol 64.00 g/mol 163.94 g/mol
Converting mass to moles
Moles Na3PO4 38.6 g Na3PO4 x (1 mol Na3PO4)

163.94 g Na3PO4 0.235 mol
Na3PO4
Formula units 0.235 mol Na3PO4 x 6.022 x 1023
formula units
1 mol Na3PO4
1.42 x 1023 formula units
4
Steps to Determine Empirical Formulas
Mass (g) of Element
M (g/mol )
Moles of Element
use no. of moles as subscripts
Preliminary Formula
change to integer subscripts
Empirical Formula
5
Some Examples of Compounds with the Same
Elemental Ratios
Empirical Formula
Molecular Formula
CH2(unsaturated Hydrocarbons) C2H4 ,
C3H6 , C4H8 OH or HO
H2O2 S

S8 P
P4
Cl
Cl2 CH2O
(carbohydrates)
C6H12O6
6
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
Moles of Cr
6.420 g Cr x
Moles of O 7.902 g O x
1 mol Na 22.99 g Na
7
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
Moles of Cr
6.420 g Cr x
Moles of O 7.902 g O x
1 mol Na 22.99 g Na
1 mol Cr 52.00 g Cr
1 mol O 16.00 g O
8
Determining Empirical Formulas from
Masses of Elements - I
Problem The elemental analysis of a sample
compound gave the following results 5.677g Na,
6.420 g Cr, and 7.902 g O. What is the empirical
formula and name of the compound? Plan First we
have to convert mass of the elements to moles of
the elements using the molar masses. Then we
construct a preliminary formula and name of the
compound. Solution Finding the moles of the
elements Moles of Na 5.678 g Na x
0.2469 mol Na
Moles of Cr 6.420 g Cr x
0.12347 mol Cr Moles of O 7.902 g
O x 0.4939 mol O
1 mol Na 22.99 g Na
1 mol Cr 52.00 g Cr
1 mol O 16.00 g O
9
Determining Empirical Formulas from
Masses of Elements - II
Constructing the preliminary formula
Na0.2469 Cr0.1235 O0.4939
Converting to integer subscripts (dividing all by
smallest subscript)
Na1.99 Cr1.00 O4.02
Rounding off to whole numbers
Na2CrO4 Sodium Chromate
10
Determining the Molecular Formula from
Elemental Composition and Molar Mass - I
Problem The sugar burned for energy in cells of
the body is Glucose (M 180.16 g/mol), elemental
analysis shows that it contains 40.00 mass C,
6.719 mass H, and 53.27 mass O. (a)
Determine the empirical formula of glucose.
(b) Determine the molecular formula. Plan We are
only given mass , and no weight of the compound
so we will assume 100g of the compound,
and becomes grams, and we can do as
done previously with masses of the
elements. Solution Mass Carbon
40.00 x 100g/100 40.00 g C Mass
Hydrogen 6.719 x 100g/100 6.719g H
Mass Oxygen 53.27 x 100g/100 53.27 g
O
99.989 g Cpd
11
Determining the Molecular Formula from
Elemental Composition and Molar Mass - II
Converting from Grams of Elements to moles
Moles of C Mass of C x
3.3306 moles C Moles of H Mass of H x
6.6657 moles H Moles
of O Mass of O x 3.3294
moles O Constructing the preliminary formula
C 3.33 H 6.67 O 3.33 Converting to integer
subscripts, divide all subscripts by the
smallest C 3.33/3.33 H 6.667 / 3.33 O3.33 /
3.33 CH2O
1 mole C 12.01 g C
1 mol H 1.008 g H
1 mol O 16.00 g O
12
Two Compounds with Molecular Formula C2H6O
Property Ethanol
Dimethyl Ether
M (g/mol) 46.07
46.07 Color
Colorless
Colorless Melting point - 117oC
- 138.5oC Boiling point
78.5oC -
25oC Density (at 20oC) 0.789 g/mL
0.00195 g/mL Use
Intoxicant in In
refrigeration
alcoholic beverages
H H
H H H C C O
H H C O C H
H
H H
H
Table 3.4
13
Some Compounds with Empirical Formula CH2O
(Composition by Mass 40.0 C, 6.71 H, 53.3O)
Molecular M Formula
(g/mol) Name Use or Function
CH2O 30.03 Formaldehyde
Disinfectant Biological

preservative C2H4O2 60.05
Acetic acid Acetate polymers vinegar

( 5 solution) C3H6O3
90.08 Lactic acid Causes milk
to sour forms
in muscle
during exercise C4H8O4 120.10
Erythrose Forms during sugar

metabolism C5H10O5
150.13 Ribose Component of
many nucleic
acids and
vitamin B2 C6H12O6 180.16
Glucose Major nutrient for energy

in cells
14
Determining the Molecular Formula
from Elemental Composition and Molar Mass - III
(b) Determining the Molecular Formula The
formula weight of the empirical formula is
1 x C 2 x H 1 x O 1 x 12.01 2 x 1.008 1
x 16.00 30.03
M of Glucose empirical formula mass
Whole-number multiple

6.00 6
180.16 30.03
Therefore the Molecular Formula is
C 1 x 6 H 2 x 6 O 1 x 6 C6H12O6
15
Adrenaline Is a Very Important Compound in the
Body - I

• Analysis gives
• C 56.8
• H 6.50
• O 28.4
• N 8.28
• Calculate the Empirical Formula

16
Adrenaline - II

• Assume 100g!
• C 56.8 g C/(12.01 g C/ mol C) 4.73 mol C
• H 6.50 g H/( 1.008 g H / mol H) 6.45 mol H
• O 28.4 g O/(16.00 g O/ mol O) 1.78 mol O
• N 8.28 g N/(14.01 g N/ mol N) 0.591 mol N
• Divide by 0.591
• C 8.00 mol C 8.0 mol C or
• H 10.9 mol H 11.0 mol H
• O 3.01 mol O 3.0 mol O C8H11O3N
• N 1.00 mol N 1.0 mol N

17
Combustion Train for the Determination of the
Chemical Composition of Organic Compounds.
m 2
m 2
CnHm (n ) O2 n CO(g) H2O(g)
Fig. 3.4
18
(No Transcript)
19
Ascorbic Acid ( Vitamin C ) - I Contains C , H ,
and O

• Upon combustion in excess oxygen, a 6.49 mg
  sample yielded 9.74 mg CO2 and 2.64 mg H2O
• Calculate its Empirical formula!
• C 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2)
• 2.65 x 10-3 g C
• H 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O)
• 2.92 x 10-4 g H
• Mass Oxygen 6.49 mg - 2.65 mg - 0.30 mg
• 3.54 mg O

20
Vitamin C Combustion - II

• C 2.65 x 10-3 g C / ( 12.01 g C / mol C )
• 2.21 x 10-4 mol C
• H 0.295 x 10-3 g H / ( 1.008 g H / mol H )
• 2.92 x 10-4 mol H
• O 3.54 x 10-3 g O / ( 16.00 g O / mol O )
• 2.21 x 10-4 mol O
• Divide each by 2.21 x 10-4
• C 1.00 Multiply each by 3 3.00 3.0
• H 1.32
  3.96 4.0
• O 1.00
  3.00 3.0

C3H4O3
21

Vitamin C
22
Determining a Chemical Formula from
Combustion Analysis - I
Problem Erythrose (M 120 g/mol) is an
important chemical compound as
a starting material in chemical synthesis, and
contains Carbon Hydrogen, and
Oxygen. Combustion analysis of
a 700.0 mg sample yielded 1.027 g CO2 and
0.4194 g H2O. Plan We find the masses
of Hydrogen and Carbon using the mass
fractions of H in H2O, and C in CO2. The mass of
Carbon and Hydrogen are subtracted from
the sample mass to get the mass of
Oxygen. We then calculate moles, and construct
the empirical formula, and from the
given molar mass we can calculate the
molecular formula.
23
Determining a Chemical Formula from Combustion
Analysis - II
Calculating the mass fractions of the elements
Mass fraction of C in CO2

0.2729 g C
/ 1 g CO2 Mass fraction of H in H2O


0.1119 g H / 1 g H2O Calculating masses of C
and H Mass of Element mass of compound x
mass fraction of element
mol C x M of C mass of 1 mol CO2
1 mol C x 12.01 g C/ 1 mol C 44.01 g
CO2
mol H x M of H mass of 1 mol H2O
2 mol H x 1.008 g H / 1 mol H 18.02
g H2O
24
Determining a Chemical Formula from
Combustion Analysis - III
0.2729 g C 1 g CO2
Mass (g) of C 1.027 g CO2 x
0.2803 g C Mass (g) of H 0.4194 g H2O x
0.04693 g H Calculating
the mass of O Mass (g) of O Sample mass -(
mass of C mass of H )
0.700 g - 0.2803 g C - 0.04693 g H 0.37277 g
O Calculating moles of each element C
0.2803 g C / 12.01 g C/ mol C 0.02334 mol C
H 0.04693 g H / 1.008 g H / mol H 0.04656 mol
H O 0.37277 g O / 16.00 g O / mol O
0.02330 mol O C0.02334H0.04656O0.02330 CH2O
formula weight 30 g / formula 120 g /mol / 30 g
/ formula 4 formula units / cpd C4H8O4
0.1119 g H 1 g H2O
25
Molecular Formula
Molecules
Atoms
Avogadros Number
6.022 x 1023
Moles
Moles