Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 7: 2 February

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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

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 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 1023 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 = 22.99 g/mol Na + 2 (16.00 g/mol O) = 54.99 g/mol NaO2

% Na = {(23.00 g/mol) / (54.99 g/mol)} x 100% = 41.81%

% O = {(32.00 g/mol) / (55.99 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, as we will see next time.

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© R A Paselk

Last modified 2 February 2011