Humboldt State University ® Department of Chemistry

Richard A. Paselk
Chem 109

General Chemistry

Summer 2002
Lecture Notes:: 10 June

© R. Paselk 2002
  
     
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Stoichiometry

Stoichiometry: Moles, Atoms, & Formulae

Most samples of matter consist of combinations of atoms or ions to give compounds characterized by molecules or formulae. We are thus interested in molecular masses, formula masses etc.

Examples:

0.25 mole x (39.10 g/mole + 79.90 g/mole) = 29.75 g = 3.0 x 101g

First need formula = C6H12O6

so weight = 0.25 mole x (6 [12.01]g/mole + 12 x [1.008 g/mole] + 6 [16.00] g/mole) = 45.04 g = 45 g

so weight = 0.25 mole x (2 [12.01]g/mole + 6 x [1.008 g/mole] + 16.00 g/mole) = 11.52 g = 12 g

First need formula = C12H22O11

so MW = (12 [12.01]g/mole + 22 x [1.008 g/mole] + 11 [16.00 ] g/mole) = 342.30 g/mole

and moles = (23.5 g)/(342.30 g/mole) = 6.865 x 10-2mole = 6.87 x 10-2mole.

 

 

 

First need FW = (39.10 g/mole + 79.90 g/mole) = 119.00 g/mole

then find moles = (1.00 g)/(119.00 g/mole) = 8.403 x10-3mole

then multiply by Avogadro's Number: (8.403 x10-3mole) x (6.022 x 10-3 atoms/mole) = 5.061 x1021 KBr "units"

But - there are two atoms/formula, So 1.01 x1022 atoms

(Of course there are no atoms actually present, just ions. However, all the parts for the atoms are there, so we still ask the question in terms of atoms.)

 

 

 

Formula Stoichiometry

Determination of the Percentage Composition of Compounds: There are two common ways to describe the composition of compounds: by the ratios of elements by atom or mole in them, and by the ratio of elements by percentage. Let's look at the percentage elemental analysis of a compound.

First we need to determine the formula weight = 23.00 g/mol Na + 2 (16.00 g/mol O) = 55.00 g/mol NaO2

% Na = {(23.00 g/mol) / (55.00 g/mol)} x 100% = 41.81%

% O = {(32.00 g/mol) / (55.00 g/mol)} x 100% = 58.19%

Check: 41.81 % + 58.19% = 100.00% (Note that frequently the addition will not quite work out due to rounding errors.)

 

Determination of Empirical (Simplest) Formulae: Want to determine the ratios, in moles, of elements in an analysis.

First need to find the amount of oxygen: 100% - 61.80% - 8.63% = 29.57%

Next we need to find the number of moles. Easiest to assume 100 g total, and find moles of each:

Hg: (61.80g)/(200.6g/mole) = 3.081 x 10-1moles

N: (8.63g)/(14.01g/mol) = 6.16 x 10-1moles

O: (29.57g)/(16.00g/mol) = 1.848 moles

\ formula = Hg0.3081N0.616O1.848

But we want whole number ratios, so divide each coefficient by the smallest:

Hg0.3081/0.3081N0.616/0.3081O1.848/0.3081 to get:

Hg1N1.999O5.998, and rounding off

= HgN2O6

When should you round off? One of the problems in finding the simplest formula is determining how much error is legitimate in rounding off. Ultimately this is a decision determined by the error of the experimental data - how many significant figures do we have. For this course we generally have at least three sig figs, but I promise not to get too subtle, so as a rule of thumb for this class values such as x.2xx, x.33x, x.25x and x.5xx should be assumed to be not in error, and so must be multiplied to get the correct formula. (e.g. XY2.331 gives X3Y7)

Determination of Molecular Formulae: Notice that for molecular compounds that the empirical formula is not necessarily the molecular formula! That is the actual molecular formula could be a multiple of the simplest formula. Thus, to find molecular formulae we need two kinds of information, the empirical formula (from percentage composition) and the molecular weight (from physical characterization).

First we need to determine the formula weight = 3 (12.01) + 5 (1.008) = 41.07,

Then we divide the MW by the FW: 85/41 = 2.07.

Now we know the MW is not terribly precise, so we use it as a decision number, that is the the MW = FW, or 2 x FW, or 3 x FW etc. Obviously ours is 2 x, so:

Molecular formula = C6H10

 

Reaction Stoichiometry: Chemical Equations

Reaction Stoichiometry: Hydrogen stoichiometry reaction demo.

We now want to look at chemical equations. As implied in the name, there is an equality involved in the two sides of any chemical equation - each side must have the same numbers of the same kinds of atoms on each side (which also means, of course, that the total masses are identical on both sides). For example, consider the combustion of propane in excess oxygen to give carbon dioxide and water:

C3H8 + O2 Æ CO2 + H2O

Conservation of Mass tells us that we must have the same numbers of atoms on each side, so we need to Balance the equation. First look at the carbons (it is generally most effective to look at hydrogen and oxygen last). Propane has 3 carbons, so we need 3 carbon dioxides:

C3H8 + O2 Æ 3 CO2 + H2O

Now balance hydrogen. There are 8 H's in propane so need 8/2 = 4 waters:

C3H8 + O2 Æ 3 CO2 + 4 H2O

Finally, 3 carbon dioxides and 4 waters requires 3 (2) + 4 = 10 oxygen atoms/2 = 5 oxygen molecules:

C3H8 + 5 O2 Æ 3 CO2 + 4 H2O

Notice that this chemical equation gives both qualitative information (what things react to give what products) and quantitative information (how much stuff is produced if a particular amount of stuff reacts). This gives rise to various practical applications. (This is a good time to look at the Stoichiometry Problem set in the lab book).

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Last modified 10 June 2002