General Chemistry
M. R. Naimi-Jamal
Faculty of Chemistry
Iran University of Science & Technology
‫فصل سوم‪:‬‬
‫استوکیومتری ‪I‬‬
‫ترکیبات شیمیائی‬
Contents
3-1
3-2
3-3
Molecular and Ionic Compounds
Molecular Mass
Composition
Molecules, Compounds, and Formulas
• Molecule – smallest identifiable unit in which a
pure substance can be divided and still retain the
composition and chemical properties of the
substance
Formulas
• Molecular Formula – formula that expresses the
number of atoms of each type within one molecule
of a compound, e.g. H2O2, P4.
• Condensed Formula – variation of the molecular
formula that shows groups of atoms showing how
atoms are grouped e.g. CH3CH2OH.
– C2H6O could be ethanol or dimethylether .
Condensed formula tells us the difference.
Formulas
• Structural Formula – shows how the atoms
in a compound are connected
– Lines between atoms represent chemical bonds
• A chemical bond is the interaction between two or
more atoms that holds them together
Models
• Since the particulate level is to small to be
seen, we use models to represent particles
– Models are a representation of something else
• Two common models
– Ball-and-stick model
• Atoms are balls
• Electrons are sticks that connect atoms
– Space-filling model
• Shows outer boundries of the particle in 3D space
Models (continued)
Ball and Stick Model
Space-filling model
How to look at molecules
• Different ways to look
at compounds
– Name
– Molecular formula
– Condensed formula
– Structural formula
H
H
H
C
C
H
H
O
H
– Molecular model
• Example
– Ethanol
– C2H6O
– CH3CH2OH
Standard color scheme
Some molecules
H2O2
CH3CH(OH)CH3
CH3CH2Cl
HCO2H
P4O10
Inorganic molecules
S8
P4
Ions and Ionic Compounds
• An atom that either gains or loses electron(s) is
an ion.
• There is no change in the number of protons or
neutrons in the nucleus of the atom.
• Cation – has a positive charge from loss of
electron(s).
• Anion – has a negative charge from gain of
electron(s).
Cations
• If an atom loses an electron (the electron is
transferred to an atom of another element in
a reaction), the atom has one more proton
than electrons.
– When this happens, a cation is formed
– Cation is the positively charged ion
– Li
1 e- + Li+
Anions
• If an atom gains an electron (the electron is
transferred to the atom from another
element in a reaction), the atom has one
more electron than protons.
– When this happens, an anion is formed
– Anion is the negatively charged ion
– O + 2eO2-
How to determine if atom will gain or lose
electrons.
• Metals generally lose electrons in the course
of their reactions to form cations
• Nonmetals frequently gain one or more
electrons to form anions in the course of
their chemical reactions.
Sodium chloride
Extended array of Na+ and Cl- ions
Simplest formula unit is
NaCl
Monoatomic Ions
• Monoatomic ions – single atoms that have lost or
gained electrons
• Metals form ions having a positive charge of their
group number
• Transition metals form cations, but they can form
several different cations
• Nonmetals form ions having a negative charge = 8
minus the group number
• Noble gases lose or gain electrons very rarely
Polyatomic Ions
• Polyatomic ions are made up of two or
more atoms
– Example: CO32- , so it has a carbon and 3
Oxygens
Atoms and the Mole
• Mole (mol) – the amount of substance that
contains as many elementary entities (atoms,
molecules, or other particles) as there are atoms in
exactly 12 g of the carbon-12 isotope
– So, one mole has the same amount of particles no
matter what substance we are talking about
• Avogadro’s number – 6.022 x 1023 atoms/mol
Molar Mass
• Molar mass – mass in grams of one mole of
atoms
– Sodium (Na) – 22.99 g/mol = 6.022 x 1023 Na
atoms
– Lead (Pb) – 207.2 g/mol = 6.022 x 1023 Pb
atoms
Molecular mass
H OH
Glucose
H O
HO
HO
H
H
H
OH
OH
Molecular formula C6H12O6
Empirical formula CH2O
Molecular Mass: Use the naturally occurring mixture of isotopes,
6 x 12.01 + 12 x 1.01 + 6 x 16.00 = 180.18
Exact Mass:
Use the most abundant isotopes,
6 x 12.000000 + 12 x 1.007825 + 6 x 15.994915
= 180.06339
Example:
Determine (a) the number of NH4+ ions in a 264 g
sample of (NH4)2SO4 and (b) the volume of 1,2,3propanetriol (glycerol, d = 1.261 g/mL) that contains
1.00 mol O atoms.
Example: An Estimation Example
Which of the following is a reasonable value for the
number of atoms in 1.00 g of helium?
(a) 4.1 × 10–23
(c) 1.5 × 1023
(b) 4.0
(d) 1.5 × 1024
Percent Composition
• Law of constant composition – any sample of a
pure compound always consists of the same
elements combined in the same proportion by
mass
• Three ways to express this
– Number of atoms of each type per molecule
– Mass of each element per mole of compound
– Mass percent - mass of each element in the compound
relative to the total mass
• Percent Composition – mass percent of each
element
Mass Percent Composition
from Chemical Formulas
The mass percent composition of a compound refers
to the proportion of the constituent elements,
expressed as the number of grams of each element
per 100 grams of the compound. In other words …
g element X
% element X = –––––––––––––– OR …
100 g compound
g element
% element = ––––––––––– × 100
g compound
Percentage Composition of Butane
Chemical Composition
Halothane
C2HBrClF3
Mole ratio
nC/nhalothane
Mass ratio
mC/mhalothane
M(C2HBrClF3) = 2MC + MH + MBr + MCl + 3MF
= (2 x 12.01) + 1.01 + 79.90 + 35.45 + (3 x 19.00)
= 197.38 g/mol
Example
Calculating the Mass Percent Composition of a Compound
Calculate the molecular mass
M(C2HBrClF3) = 197.38 g/mol
For one mole of compound, formulate the mass
ratio and convert to percent:
Example
(2 12.01) g
%C 
100%  12.17%
197.38g
1.01g
%H 
100%  0.51%
197.38g
79.90g
% Br 
100%  40.48%
197.38g
35.45g
%Cl 
100%  17.96%
197.38g
(3 19.00) g
%F 
100%  28.88%
197.38g
Example:
Calculate, to four significant figures, the mass percent
of each element in ammonium nitrate.
Example:
How many grams of nitrogen are present in 46.34 g
ammonium nitrate?
Example:
An Estimation Example
Without doing detailed calculations, determine which of
these compounds contains the greatest mass of sulfur
per gram of compound: barium sulfate, lithium sulfate,
sodium sulfate, or lead sulfate.
Formulas From Percent Composition
• Just the REVERSE of what we did
weight %
to mass
%A
%B
gA
gB
mass to
moles
Mole
ratio
Ratio gives
formula
X mol A
Y mol B
AxBy
x mol A
y mol B
Formulas From Percent Composition
• This tells us the ratio not the total number of
atoms
• In a molecular formula, we must know the
total number of atoms
• So, NH2 is the empirical formula of hydrazine
– Hydrazine could be NH2, N2H4, N3H6 ….
– If given, molar mass of hydrazine is 32.0 g/mol,
we know the correct molecular formula is N2H4
Empirical formula
5 Step approach:
1.
2.
3.
4.
5.
Choose an arbitrary sample size (100g).
Convert masses to amounts in moles.
Write a formula.
Convert formula to small whole numbers.
Multiply all subscripts by a small whole number
to make the subscripts integral.
Relating Molecular Formulas
to Empirical Formulas
• A molecular formula is a simple integer multiple
of the empirical formula.
• That is, an empirical formula of CH2 means that
the molecular formula is CH2, or C2H4, or C3H6,
or C4H8, etc.
• So: we find the molecular formula by:
molecular formula mass
= integer (nearly)
empirical formula mass
We then multiply each subscript in the empirical formula by the
integer.
Example 3.11
The empirical formula of hydroquinone, a
chemical used in photography, is C3H3O, and
its molecular mass is 110 u. What is its
molecular formula?
Example:
Determining the Empirical and Molecular Formulas of a
Compound from Its Mass Percent Composition.
Dibutyl succinate is an insect repellent used against household
ants and roaches. Its composition is 62.58% C, 9.63% H and
27.79% O. Its experimentally determined molecular mass is
230 u. What are the empirical and molecular formulas of
dibutyl succinate?
Step 1: Determine the mass of each element in a 100g sample.
C 62.58 g
H 9.63 g O 27.79 g
Example:
Step 2: Convert masses to amounts in moles.
1 m olC
 5.210 m olC
12.011g C
1 m olH
nH  9.63 g H 
 9.55 m olH
1.008 g H
nC  62.58 g C 
nO  27.79 g O 
1 m olO
 1.737 m olO
15.999 g O
Step 3: Write a tentative formula.
C5.21H9.55O1.74
Step 4: Convert to small whole numbers.
C2.99H5.49O
Example:
Step 5: Convert to a small whole number ratio.
Multiply x 2 to get C5.98H10.98O2
The empirical formula is C6H11O2
Step 6: Determine the molecular formula.
Empirical formula mass is 115 u.
Molecular formula mass is 230 u.
The molecular formula is C12H22O4
Example:
Phenol, a general disinfectant, has the composition
76.57% C, 6.43% H, and 17.00% O by mass.
Determine its empirical formula.
Example:
Diethylene glycol, used in antifreeze, as a softening
agent for textile fibers and some leathers, and as a
moistening agent for glues and paper, has the
composition 45.27% C, 9.50% H, and 45.23% O by
mass. Determine its empirical formula.
Elemental Analysis …
• … is one method of determining empirical
formulas in the laboratory.
• This method is used primarily for simple organic
compounds (that contain carbon, hydrogen,
oxygen).
– The organic compound is burned in oxygen.
– The products of combustion (usually CO2 and H2O)
are weighed.
– The amount of each element is determined from the
mass of products.
Elemental Analysis (cont’d)
… H2O, which is absorbed
by MgClO4, and …
The sample is
burned in a
stream of oxygen
gas, producing …
… CO2, which is
absorbed by NaOH.
Elemental Analysis (cont’d)
If our sample
were CH3OH,
every two
molecules of
CH3OH …
… would give two
molecules of CO2 …
… and four
molecules of H2O.
Example:
Burning a 0.1000-g sample of a carbon–
hydrogen–oxygen compound in oxygen yields
0.1953 g CO2 and 0.1000 g H2O. A separate
experiment shows that the molecular mass of
the compound is 90 u. Determine (a) the mass
percent composition, (b) the empirical formula,
and (c) the molecular formula of the compound.
Chapter 3 Questions
1, 4, 19, 30, 35, 40,
41, 46, 54, 57, 60,
69