Empirical Formulas
The empirical formula of a substance indicates the simplest whole number ratios of the
different kinds of atoms that make up the substance. The empirical formula of a
substance with the molecular formula N2O4, for example. is NO2.
1.
Fill in the table with the empirical formula for each of the following
hydrogen-carbon compounds.
Compound
methane
benzene
ethane
ethylene (ethane)
octane
acetylene (ethyne)
naphthalene
acetylene (ethyne)
Molecular Formula
CH4
C6H6
C2H6
C2H4
C8H18
C2H2
C10H8
C2H12
Which of the above compounds have the same empirical formulas.
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Empirical Formula
Chemistry 11
Percent Composition
1. Determine the empirical formula for each of the following compounds.
(a) The compound is 72.4% iron and 27.6% oxygen.
(b) The compound is 46.3% lithium and the remainder is oxygen.
(c) The compound is 50.5 % carbon, 5.3% hydrogen and the remainder is
nitrogen.
Chemistry 11
Self Quiz on Percent Composition
1.
What is the empirical formula of a compound that contains 3.66%
hydrogen, 37.94% phosphorus, and 58.40% oxygen?
2. What is the percent composition of each element in K2CrO4?
Empirical Formulas
The empirical formula of a substance indicates the simplest whole number ratios of the
different kinds of atoms that make up the substance. The empirical formula of a
substance with the molecular formula N2O4, for example, is NO2.
1. Fill in the table with the empirical formula for each of the following hydrogen-carbon
compounds.
Compound
methane
benzene
ethane
ethylene(ethane)
octane
acetylene (ethyne)
naphthalene
butane
Molecular Formula
CH4
C6H6
C2H6
C2H4
C8H18
C2H2
C10H8
C4H10
Empirical Formula
CH4
CH
CH3
CH2
C4H9
CH
C5H4
C2H5
Which of the above compounds have the same empirical formula?
benzene and acetylene
Chemistry 11 Coolschool
Answer key
Percent Composition
Note the sym bol “∝” is read as “… is proportional to…”. It is a m
relationship. (other relationships include “=” or “<” )
athematical
1. a) mol of Fe ∝ (% composition/molar mass) = (72.4/55.8)
= 1.30
Now repeat this calculation for all elements in the formula...
mol of O ∝ (27.6/16.00) = 1.73
Now make a ratio of the calculated values which represent num bers proportional
to the actual number of m oles, usually it is s implest if you put th e larger value
over the smaller value (but this is a tr ial and error approach so this is not
essential)…
moles of O = 1.73 = 1.33 … just perform this division
moles of Fe 1.30
1.00
Now multiply this ratio which has a denominator of 1.00, with whichever fraction, whose
value is 1, results in a whole number ratio. This is to say… try 1/1, or 2/2, or 3/3 and
so on, until one fraction results in a whole number over a w hole number. These
whole numbers represent the subscripts in the empirical formula of the compound.
1.33 x 3 = 4 …so the empirical formula is Fe3O4
1.00
3
3
b) The empirical formula is Li2O
c) mol of C ∝ (50.5)/(12.01) = 4.2
mol of H ∝ (5.3)/(1.00) = 5.3
mol of N ∝ (54.2)/(14.01) = 3.87
moles of C = 4.20 = 1.09 … so # of C’s = # of N’s
moles of N
3.87
1.00
moles of H = 5.3 = 1.26 x 4 = 5
moles of C
4.20
1.00 4
4
…so there are 5 H’s for every 4 C’s and checking the
above statement causes us to say there must then also be
4 N’s. Therefore, the empirical formula is C4H5N4
Chemistry 11
Self Quiz on Percent Composition
U03L07
1. What is the empirical formula of a compound that contains 3.66% hydrogen, 37.94%
phosphorus and 58.40% oxygen by mass?
moles of H ≃ 3.66 = 3.66
1.00
moles of O ≃ 58.40 = 3.65
16.00
moles of P ≃ 37.94 = 1.225
30.97
moles H = 3.66 ≃ 1
moles O 3.65
moles H = 3.66 ≃ 3
moles P 1.225
Therefore the Empirical Formula is H3PO3.
2. What is the percent composition by mass of each element in K2CrO4?
Assume a 1 mole sample is assessed.
Total
% K = 78.20g = 40.3%
194.2g
Mass of K = 2 mol • (39.10g/mol) = 78.20g
Mass of Cr = 1 mol • (52.00g/mol) = 52.00g
Mass of O = 4 mol • (16.00g/mol) = 64.00g
Mass
=
194.2g
% Cr = 52.00g = 27.8%
194.2g
% O = 64.00g = 33.0%
194.2g