Understanding Formulas
Spring 2013
Dr. Yau
(loosely based on Chap. 1.5 & 1.6
in Jespersen, Brady & Hyslop, 6th edition)
1
Proof Of Atoms
• Early 1980’s, use
Scanning Tunneling
Microscope (STM)
• Surface can be
scanned for
topographical
information
• Image for all matter
shows spherical
STM of palladium
regions of matter
– Atoms
2
How Do We Visualize Atoms?
Atoms are too small for our eyes
to see, but we can use models
to help us understand the
concepts.
• Atoms represented by spheres
• Different atoms have different
colors
• Standard scheme given in Fig.
1.11 is represented on the
right.
3
Molecules
• Atoms combine to form more complex
substances
• Discrete particles
• Each composed of 2 or more atoms
Ex.
– Molecular oxygen, O2
– Carbon dioxide, CO2
– Ammonia, NH3
– Sucrose, C12H22O11
4
Chemical Formulas
• Specify composition of substance
• Chemical symbols
– Represent atoms of elements present
• Subscripts
– Given after chemical symbol
– Represents relative numbers of each type of atom
Ex.
Fe2O3 : iron & oxygen in 2:3 ratio
5
Chemical Formulas
Free Elements
– Element not combined with another in compounds
– Just use chemical symbol to represent
Ex. Iron
Sodium
Fe
Na
Neon
Ne
Aluminum Al
Diatomic Molecule
– Molecules composed of 2 atoms each
– Many elements found in nature
Ex. Oxygen
Hydrogen
O2
H2
Nitrogen
Chlorine
N2
Cl2
6
Depicting Molecules
• Want to show:
– Order in which atoms are attached to each other
– 3-dimensional shape of molecule
• Three ways of visualizing molecules:
1. Structural formula
2. Ball-and-Stick model
3. Space filling model
7
1. Structural Formulas
• Use to show how atoms are attached
– Atoms represented by chemical symbols
– Chemical bonds attaching atoms indicated by
lines
H
H
O
H2O
water
H
H
C
H
H
CH4
methane
8
3-D Representations of Molecules
Hydrogen
molecule,
H2
Oxygen
molecule,
O2
Nitrogen
molecule
N2
Chlorine
molecule,
Cl2
• Use touching spheres to indicate
molecules
• Different colors indicate different elements
• Relative size of spheres reflects differing
9
sizes of atoms
2. “Ball-and-Stick” Model
• Spheres = atoms
• Sticks = bonds
Methane,
CH4
Chloroform,
CHCl3
10
3. “Space-Filling” Model
• Shows relative sizes of atoms
• Shows how atoms take up space in molecule
Water
H2O
Methane
CH4
Chloroform, CHCl3
11
More Complicated Molecules
• Sometimes formulas contain parentheses
• How do we translate into a structure?
Ex. Urea, CO(NH2)2
– Expands to CON2H4
– Atoms in parentheses appear twice
Ball-and-stick
model
Space-filling model
12
Hydrates
• Crystals that contain water molecules
Ex. plaster: CaSO4∙2H2O calcium sulfate dihydrate
– Water is not tightly held
• Dehydration
– Removal of water by heating
– Remaining solid is anhydrous (without water)
Blue =
CuSO4 •5H2O
White =
CuSO4
13
Counting Atoms
1. Subscript following chemical symbol
indicates how many of that element are part
of the formula
– No subscript implies a subscript of 1.
2. Quantity in parentheses is repeated a
number of times equal to the subscript that
follows.
3. Raised dot in formula indicates that the
substance is a hydrate
– Number preceding H2O specifies how
many water molecules are present.
14
Counting Atoms
Ex. 1 (CH3)3COH
• Subscript 3 means 3 CH3 groups
So from(CH3)3, we get 3 × 1C = 3C
3 × 3H = 9H
#C = 3C + 1C = 4 C
#H = 9H + 1H = 10 H
#O = 1 O
Total # of atoms = 15 atoms
15
Counting Atoms
Ex. 2 CoCl2 · 6H2O
• The dot 6H2O means
you multiple both H2 & O by 6
• So there are:
#H
6 × 2 = 12 H
#O
6×1= 6O
#Co
1 × 1 = 1 Co
#Cl
2 × 1 = 2 Cl
Total # of atoms = 21 atoms
16
Your Turn!
Count the number of each type of atom in the
chemical formula given below
a.
b.
c.
d.
e.
Na2CO3
(NH4)2SO4
Mg3(PO4)2
CuSO4∙5H2O
(C2H5)2N2H2
a.
b.
c.
d.
e.
___Na,
___
2
1 C, ___
3 O
___N,
___H,
___S,
___O
2
8
1
4
___Mg,
___P,
___O
3
2
8
___Cu,
___S,
___O,
___H
9
10
1
1
___C,
___H,
___N
12
4
2
17
Dalton’s Atomic Theory
• We now have the tools to explain this theory
& its consequences
• All molecules of compound are alike & contain
atoms in same numerical ratio.
Ex. Water, H2O
Ratio of oxygen to hydrogen is 1 : 2
1 O atom : 2 H atoms in each molecule
O weighs 16 times as much as H
1 H = 1 mass unit
1 O = 16 mass units
18
Atoms in Fixed Ratios by Mass
For water in general:
 mass O = 8 mass H
 Regardless of amount of water present
19
Dalton’s Atomic Theory
Successes:
• Explains Law of Conservation of Mass
– Chemical reactions correspond to rearranging
atoms.
• Explains Law of Definite Proportions
– Given compound always has atoms of same
elements in same ratios.
• Predicted Law of Multiple Proportions
– Not yet discovered
– Some elements combine to give 2 or
more compounds
Ex. SO2 & SO3
20
Law Of Multiple Proportions
• When 2 elements form more than one
compound, different masses of one element
that combine with same mass of other
element are always in ratio of small whole
numbers.
– Atoms react as complete (whole) particles.
– Chemical formulas
• Indicate whole numbers of atoms
• Not fractions
21
Using Law Of Multiple Proportions
sulfur
dioxide
trioxide
Mass S
32.06 g
g
Mass O
32.00 g
g
sulfur
32.06
48.00
• Use this data to prove law
of multiple proportions
22
Law of Multiple Proportions
Compound Sample Mass of
Size
Sulfur
Sulfur
64.06
g
32.06
g
dioxide
Mass of
Oxygen
32.06 g
Sulfur
trioxide
80.06 g
Ratio of
O in SO 3 48.00g 3


O in SO 2 36.00g 2
32.06 g
48.00 g
23
Molecules Small and Large
• So far we’ve only discussed small molecules
• Some are very large, especially those found
in nature
• Same principles apply to all
Ex. DNA - short segment
24
How Do We Know Formulas?
• Hardly “out of the blue”
• Don’t know formula when compound 1st
isolated
• Formulas & structures backed by extensive
experimentation
• Use results of experiments to determine
– Formula
– Chemical reactivity
• Molecular Shape
– Can speculate once formula is known
– Determine from more experiments
25
Visualizing Mixtures
• Look at mixtures at atomic/molecular level
• Different color spheres stand for 2 substances
a. Homogeneous mixture/solution – uniform mixing
b. Heterogeneous mixture – 2 phases
a.
b.
26
Chemical Reactions
• When 1 or more substances
react to form 1 or more new
substances
Ex. Reaction of methane, CH4,
with oxygen, O2, to form
carbon dioxide, CO2, &
water, H2O.
Reactants = CH4 & O2
Products = CO2 & H2O
• How to depict?
– Words too long
– Pictures too awkward
27
Chemical Equations
• Use chemical symbols & formulas to
represent reactants & products.
– Reactants on left hand side
– Products on right hand side
– Arrow () means “reacts to yield”
Ex. CH4 + 2O2  CO2 + 2H2O
– Coefficients
• Numbers in front of formulas
• Indicate how many of each type of
molecule reacted or formed
– Equation reads “methane & oxygen
react to yield carbon dioxide & water”
28
Conservation of Mass in Reactions
• Mass can neither be created nor destroyed
• This means that there are the same number of each
type of atom in reactants & in products of reaction
– If # of atoms same, then mass also same
CH4 + 2O2

CO2 + 2H2O
4 H + 4O + C
=
4 H + 4O + C
29
Balanced Chemical Equation
Ex.
2C4H10 + 13O2  8CO2 + 10H2O
4 C & 10 H
per
molecule
2 O per
molecule
1C&2O
per
molecule
2H&1O
per
molecule
Subscripts
 Define identity of substances
 Must not change when equation is
balanced
30
Ex.
Balanced Chemical Equation
2C4H10 + 13O2  8CO2 + 10H2O
2 molecules
of C4H10
13 molecules
8 molecules
of O2
of CO2
10 molecules
of C4H10
Coefficients
 Number in front of formulas
 Indicate number of molecules of each type
 Adjusted so # of each type of atom is
same on both sides of arrow
 Can change
31
Balanced Chemical Equations
• How do you determine if an equation is balanced?
– Count atoms
– Same number of each type on both sides of equation?
• If yes, then balanced
• If no, then unbalanced
Ex. 2C4H10 + 13O2  8CO2 + 10H2O
Reactants
Products
2×4 = 8 C
8×1 = 8 C
2×10 = 20 H
10×2 = 20 H
13×2 = 26 O
(8×2)+(10×1)= 26 O
32
Learning Check
Fe(OH)3 + 2 HNO3  Fe(NO3)3 + 2 H2O
Reactants
Fe
1
Products
1
O
3 + (2×3) = 9
(3×3) + 2 = 11
H
3+2=5
(2×2) = 4
N
2
3
• Not Balanced
• Only Fe has same number of atoms
on either side of arrow.
33
Learning Check:
How many atoms of each element appear on
each side of the arrow in the following
equation?
4NH3 + 3O2 → 2N2 + 6H2O
Reactants
Products
N
(4 × 1) = 4
(2 × 2) = 4
O
(3 × 2) = 6
(6 × 1) = 6
H
(4 × 3) = 12
(6 × 2) = 12
34
Learning Check:
Count the number of atoms of each element on
both sides of the arrow to determine whether
the following equation is balanced.
2(NH4)3PO4 + 3Ba(C2H3O2)2 → Ba3(PO4)2 +
6NH4C2H3O2
Reactants
Products
N
(2 × 3) = 6
(6 × 1) = 6
H (2×3×4)+(3×3×2) = 42
(6×4) + (6×3) = 42
O (2×4) + (3×2×2) = 20
(2×4) + (6×2) = 20
P
(2 × 1) = 2
(2 × 1) = 2
Ba
(3 × 1) = 3
(3 × 1) = 3
35