(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.)
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.
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
therefore: 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
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. XY2.331 gives X3Y7)
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).
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 oxygen last). Propane has 3 carbons, so we need 3 carbon dioxides:
C3H8 + O2 3 CO2 + H2O
Now balance hydrogen. There are 8 H's in propane so need 8/2 = 4 waters:
C3H8 + O2 3 CO2 + 4 H2O
Finally, 3 carbon dioxides and 4 waters requires 3 (2) + 4 = 10 oxygen atoms/2 = 5 oxygen molecules:
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© R A Paselk
Last modified 8 February 2013