How to find the empirical formula of a compound from given data. Also calculating the molecular formula. **More good stuff available at: www.wsautter.com and http://www.youtube.com/results?search_query=wnsautter&aq=f

1. Copyright Sautter 2015 1

2. EMPIRICAL FORMULAE
• An empirical formula is the simplest formula for a
compound. For example, H2O2 can be reduced to a
simpler formula, HO. Similarly, N2O4 can be reduced to
NO2. Both HO and NO2 are empirical formulae.
• Some formulae cannot be reduced and are the empirical
formula. An example is H2O which cannot be reduced to
lower terms.
• Empirical formulae for compounds can be determined
from experimental information such as percent
composition data.
2

3. The molecular formula
is always a whole
number multiple of the
empirical formula 3

4. PERCENT COMPOSITION
• Percent composition refers to the mass percentage of each element
contained in a compound.
• Percent composition can be determined experimentally or from the
formula of a substance. For example, the percentage of carbon and
hydrogen contained in methane can be found from its formula CH4.
• One mole of CH4 contains 1 mole of carbon and 4 moles of
hydrogen. One mole of carbon has a mass of 12.0 grams and 4
moles of hydrogen have a mass of 4.0 grams (4 x 1.0 grams per
mole). The mass of one mole of CH4 is 16.0 grams (12.0 + 4.0 =
16.0 g)
• % C = (12.0 / 16.0) x 100 % = 75.0 % carbon
• % H = (4.0/ 16.0) x 100 % = 25.0 % hydrogen
• The total of the percents for each element must = 100 %
• 75.0 % + 25.0 % = 100 % 4

5. PERCENT COMPOSITION
• Problem: A compound is found to contain 0.1417 g of
nitrogen and 0.4045 g of oxygen. The sample has a mass
of 0.5462 g. What is the percent composition of the
compound?
• Solution: % of X = (grams of X / total mass) x 100 %
• % N = (0.1417 / 0.5462) x 100 % = 25.6 %
• % O = (0.4045 / 0.5462) x 100 % = 74.1 %
• Check : 25.6 % + 74.1 % = 99.7 % ~ 100 %
• It is often required to find the empirical formula of a
compound from data such as that given in this
problem. Remember, empirical formula is the lowest
whole number ratio of atoms in a compound.
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6. Empirical Formula Calculations
• Problem: A compound is found to contain 0.1417 g of nitrogen and
0.4045 g of oxygen. The sample has a mass of 0.5462 g. What is the
empirical formula of the compound?
• Solution: All formulae are based on moles. First find the moles of
each element present.
• Moles N = 0.1417 g / 14.0 grams per mole = 0.0101
• Moles O = 0.4045 g / 16.0 grams per mole = 0.0253
• Now divide the smallest mole value into the other mole values
• Moles O / moles N = 0.0253moles / 0.0101 = 2.50
• Moles N / moles N is of course 1.00
• The formula at this point is then NO2.5
• Empirical formula requires the lowest “whole number ratio”
• Multiplying the formula through by 2 to obtain whole numbers we get
N2O5 (dinitrogen pentaoxide) 6

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