Untitled - Kelly Walsh High School

Mass, Moles, and Equations 39
Quick Tip
As a good check, add all the percentages together. They should equal to 100 %
or be very, very close.
In the problem above, we determined the percentage data from the chemical
formula. We can determine the empirical formula if we know the percent compositions
of the various elements. The empirical formula tells us what elements
are present in the compound and the simplest whole-number ratio of elements.
The data may be in terms of percentage, or mass or even moles. However, the
procedure is still the same—convert each element to moles, divide each by the
smallest, and then use an appropriate multiplier if necessary. We can then
determine the empirical formula mass. If we know the actual molecular mass,
dividing the molecular formula mass by the empirical formula mass, gives an
integer (rounded if needed) that we can multiply each of the subscripts in the
empirical formula. This gives the molecular (actual) formula, which tells what
elements are in the compound and the actual number of each.
An analysis of a gas sample found 2.34 g of nitrogen and 5.34 g of oxygen present.
What was the empirical formula of this gas?
1 mol N
(2.34 g N)a b 0.167 mol N a0.167b
1.00
14.0 1 g 0.167
(5.34 g N)a
1 mol O
16.00 g
b 0.334 mol O a0.334b
2.00
0.167
The data indicates that the nitrogen and oxygen are present in a 1:2 ratio.
Therefore, the empirical formula is NO 2.
If we determined that the actual molecular mass of this gas was 92.00 g/mol, what
is the molecular formula?
The empirical formula mass is 46.00 g/mol. Dividing the actual molar mass by
the empirical molar mass gives:
92.00 g>mol
2
46.00 g>mol
Therefore, the molecular formula is twice the empirical formula or N 2O 4.