1. Introduction to Molecules
Only a few elements (such as the noble gases He, Ne, Ar, Kr, Xe) exist in atomic form in nature.
The vast majority exist as atoms bonded to other atoms to form molecules.
These bonds may be with other atoms of the same element…
e.g., hydrogen, H2, H---H
nitrogen, N2, N---N
oxygen, O2, O---O
These are examples of two-atom or diatomic molecules – but you can get ozone O3 in which three O atoms are bonded together, for example. It is a triatomic molecule, which is a special case of a polyatomic molecule.

2. or with atoms of different elements…
e.g., water, H2O 2 H atoms and 1 O atom per molecule
ammonia, NH3 1 N atom and 3 H atoms per molecule
methane, CH4 1 C atom and 4 H atoms per molecule
ethanol, C2H5OH 2 C atoms, 1 O and 6 H atoms each
Note that the combinations of atoms are always ratios of whole
numbers (because we must have integer numbers of atoms).
Recall Dalton’s 5th postulate:
Atoms combine in whole number ratios to form compounds. (This follows logically if we accept that atoms are indivisible.)

3. Molecular Formulas
Chemists use different chemical and molecular formulas to convey a variety of different things:
An empirical formula gives simply the ratio of numbers of atoms of each element.
The empirical formula for glucose (a simple sugar) is CH2O indicating that carbon, hydrogen and oxygen atoms in glucose occur in the ratio 1:2:1
In a molecular formula the subscripts give the exact number of atoms of each element
In the case of glucose the molecular formula is C6H12O6
Later we will also see structural formulae.

4. Formula Units
Molecular formulas exist only for molecular substances – collections of identical, persistent well defined molecules.
Chemical bonds hold atoms together in molecules. Weaker, intermolecular forces then hold the molecules together.
But not all substances contain distinguishable molecules:
For example, sodium chloride (table salt) exists in a cubic lattice arrangement of sodium, Na+ and chloride, Cl- ions.
In such substances we cannot talk of molecular formulas but can identify formula units (in this case NaCl).

5. The empirical formula is thus Na2SO4
Determining/Using Empirical Formulas
Experimentally, any compound can be broken into its constituent elements to allow a determination of the relative amounts of each. This is known as Elemental Analysis.
Example:
Elemental analysis of 65.32g of a pure substance containing sodium, sulfur and oxygen only, shows the sample to contain 21.14g Na and 14.75g S. Determine the empirical formula.
21.14 g of sodium (M=22.99 g mol-1) is mol
14.75 g of sulfur (M=32.07 g mol-1) is mol
29.43 g of oxygen (M=16.00 g mol-1) is 1.84 mol
Ratio: 2 1 4
The empirical formula is thus Na2SO4

6. Determining Molecular Formulas
Several very different molecules can have the same empirical formula.
To determine the molecular formula we need the mass of the molecule (or the molar mass).
This is often done by mass spectrometry (see Oxtoby p23)
Example:
The empirical formula of squalene (shark liver oil) is C3H5. If the molar mass is g mol-1, what is its molecular formula?
The mass of 1 mole C3H5 unit is (3 x 12.01) +(5 x 1.01) = g / = 10 such units are required per mol squalene, and hence the molecular formula is C30H50.

7. Q: What is the empirical formula of a hydrocarbon that produces 2
Q: What is the empirical formula of a hydrocarbon that produces g CO2 and g H2O when combusted?
Solution: Because this is a hydrocarbon, the only elements it contains are carbon and hydrogen. The empirical formula will be the simplest mole ratio of these elements.
mol carbon = 2.703 g CO2 x [1 mol CO2/44.01 g] x [1 mol C/1 mol CO2] =  mol C
mol hydrogen  = 1.108 g H2O x [1 mol H2O/18.01 g] x [2 mol H/1 mol H2O] =  mol H
The empirical formula for this compound is CH2.

8. Q: What is the empirical formula of a substance containing carbon, hydrogen, and oxygen if g of substance produces g CO2 and g H2O upon combustion?
Solution: The mass of carbon dioxide and water can be used to find the moles of carbon and hydrogen. The mass of oxygen can be determined from the mass of the original substance. Because of conservation of mass
m(substance)=m(carbon)+m( hydrogen)+m( oxygen)
This requires that in addition to determining moles of carbon, hydrogen, and oxygen, the mass of carbon and hydrogen needs to be determined.

11. m(oxygen) = m(substance) – m(carbon) – m(hydrogen)  
g O x [1 mol/16.00 g] =  mol O
To obtain the simplest ratio, divide by the lowest value of moles. In this example, that is the moles of carbon

12. carbon =  /  = 1
hydrogen  = 0.0666/  = 2.00
oxygen =  /  = 0.998
The empirical formula is CH2O
Suppose that in addition we are told that the molar mass (molecular weight) of this compound is 120 g/mol. What is its molecular formula?

13. Apparent molar mass = 12.01 + 2(1.008) + 16.00 = 30.03 g/mol
Multiplication factor  = 120/30.03  = 3.996 = 4
Molecular formula = 4 x CH2O = C4H8O4

14. Normal Chemistry:
Conservation of mass
Conservation of number of elemental atoms
of a given kind (balanced chemical equation)
Conservation of energy
Nuclear Chemistry/Nuclear Physics:
Conservation of the sum of mass and energy
Conservation of mass number A