Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 9: 9 February

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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.

• What is the percentage composition of sodium superoxide (NaO₂)?

Determination of Empirical (Simplest) Formulae:

Want to determine the ratios, in moles, of elements in an analysis.

• A compound of mercury, oxygen, and nitrogen gave an analysis of 61.80% mercury and 8.63% nitrogen by weight. What is the empirical formula for this compound? (Note that oxygen is commonly not given in these analyses because it is used to react with other substances in the analysis process, and so is difficult to measure itself.)

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

therefore: formula = Hg_0.3081N_0.616O_1.848

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

Hg_{0.3081/0.3081}N_0.616/0.3081O_1.848/0.3081 to get:

Hg₁N_1.999O_5.998, 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 due to error, and so must be multiplied to get the correct formula. (e.g. XY_2.331 gives X₃Y₇)

Determination of Molecular Formulae:

Notice that for molecular compounds 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).

• Example: A molecule is found to have an empirical formula of C₃H₅ by elemental analysis and a MW of 85 by freezing point depression. What is its molecular formula?

Reaction Stoichiometry: Chemical Equations

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 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 the atom with the least number of atoms, with oxygen last). Propane has 3 carbons, so we need 3 carbon dioxides:

C₃H₈ + O₂ 3 CO₂ + H₂O

C₃H₈ + O₂ 3 CO₂ + 4 H₂O

C₃H₈ + 5 O₂ 3 CO₂ + 4 H₂O

Notice that a 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.

• Example: How many grams of carbon dioxide are produced during the complete combustion of 475.5 g of natural gas (methane, CH₄)

First need to find moles of CH₄:

MW = 12.01 + 4 (1.008) = 16.04

Moles of CH₄ = 475.5 g / 16.04 g/mol = 29.64 mole.

Next need balanced equation:

CH₄ + 2 O₂ CO₂ + 2 H₂O

Note from the equation we have a 1:1 ratio of moles methane to mole carbon dioxide

So we have 29.64 moles CO₂

& MW of CO₂ = 12.01 + 2 (16.00) = 44.01

and (29.64 mol)(44.01 g/mol) = 1.304 kg

Let's look at a slightly more complicated reaction.

• Example: How many moles of oxygen will be needed for the complete combustion of 275.5 g of butane (C₄H₁₀).

2 C₄H₁₀ + 13 O₂ 8 CO₂ + 10 H₂O

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© R A Paselk

Last modified 9 February 2015