1. Understanding Formulas
Spring 2013
Dr. Yau
(loosely based on Chap. 1.5 & 1.6
in Jespersen, Brady & Hyslop, 6th edition)

2. Proof Of Atoms Early 1980’s, use Scanning Tunneling Microscope (STM)
Surface can be scanned for topographical information
Image for all matter shows spherical regions of matter
Atoms
Fig 1.10
STM of palladium 2

3. 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 is represented on the right.
Fig 1.11 in 3 pieces so fits on slide 3

4. 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

5. 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

6. Chemical Formulas Free Elements Ex. Iron Fe Neon Ne Diatomic Molecule
Element not combined with another in compounds
Just use chemical symbol to represent
Ex. Iron Fe Neon Ne
Sodium Na Aluminum Al
Diatomic Molecule
Molecules composed of 2 atoms each
Many elements found in nature
Ex. Oxygen O2 Nitrogen N2
Hydrogen H2 Chlorine Cl2

7. Depicting Molecules Want to show: Three ways of visualizing molecules:
Order in which atoms are attached to each other
3-dimensional shape of molecule
Three ways of visualizing molecules:
Structural formula
Ball-and-Stick model
Space filling model

8. 1. Structural Formulas Use to show how atoms are attached
Atoms represented by chemical symbols
Chemical bonds attaching atoms indicated by lines
H2O
water
CH4
methane

9. 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 sizes of atoms
Fig 1.12 9

10. 2. “Ball-and-Stick” Model
Spheres = atoms
Sticks = bonds
Chloroform, CHCl3
Methane, CH4

11. 3. “Space-Filling” Model
Shows relative sizes of atoms
Shows how atoms take up space in molecule
Methane CH4
Water H2O
Chloroform, CHCl3

12. 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

13. Hydrates Crystals that contain water molecules Dehydration
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

14. Counting Atoms
Subscript following chemical symbol indicates how many of that element are part of the formula
No subscript implies a subscript of 1.
Quantity in parentheses is repeated a number of times equal to the subscript that follows.
Raised dot in formula indicates that the substance is a hydrate
Number preceding H2O specifies how many water molecules are present.

15. 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

16. 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 = 6 O
#Co 1 × 1 = 1 Co
#Cl 2 × 1 = 2 Cl
Total # of atoms = 21 atoms

17. Your Turn!
Count the number of each type of atom in the chemical formula given below
___Na, ___ C, ___ O
___N, ___H, ___S, ___O
___Mg, ___P, ___O
___Cu, ___S, ___O, ___H
___C, ___H, ___N 2 1 3
Na2CO3
(NH4)2SO4
Mg3(PO4)2
CuSO4∙5H2O
(C2H5)2N2H2 2 8 1 4 3 2 8 1 1 9 10
Dialog: How do I count the atoms of elements in a chemical formula? 4 12 2 17

18. 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

19. Atoms in Fixed Ratios by Mass
For water in general:
mass O = 8 mass H
Regardless of amount of water present

20. 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

21. 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
Dialog: What is the law of multiple proportions? 21

22. Using Law Of Multiple Proportions
sulfur sulfur
dioxide trioxide
Mass S g g
Mass O g g
Use this data to prove law of multiple proportions
Each molecule has one sulfur atom, and therefore the same mass of sulfur. The oxygen ratio is 3 to 2 by both mass and atoms 22

23. Law of Multiple Proportions
Compound
Sample Size
Mass of Sulfur
Mass of Oxygen
Sulfur dioxide
64.06 g
32.06 g
32.06 g
Sulfur trioxide
80.06 g
32.06 g
48.00 g
Ratio of

24. 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

25. 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

26. Visualizing Mixtures Look at mixtures at atomic/molecular level
Different color spheres stand for 2 substances
Homogeneous mixture/solution – uniform mixing
Heterogeneous mixture – 2 phases a. b.
Fig. 1.21 26

27. 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
Fig. 1.23 27

28. 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”

29. 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
CH O  CO H2O
4 H + 4O + C = H + 4O + C

30. Balanced Chemical Equation
Ex C4H O2  8CO H2O
1 C & 2 O per molecule
2 H & 1 O per molecule
4 C & 10 H per molecule
2 O per molecule
Subscripts
Define identity of substances
Must not change when equation is balanced

31. Balanced Chemical Equation
Ex. 2C4H O2  8CO H2O
13 molecules of O2
8 molecules of CO2
10 molecules of C4H10
2 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

32. 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. 2C4H O2  8CO H2O
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

33. Learning Check Fe(OH)3 + 2 HNO3  Fe(NO3)3 + 2 H2O Not Balanced
Only Fe has same number of atoms on either side of arrow.
Reactants
Products Fe 1
3 + (2×3) = 9
(3×3) + 2 = 11 O
3 + 2 = 5
(2×2) = 4 H 2 3 N

34. 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

35. 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 Ba
(3 × 1) = 3