Empirical and molecular formulae
A.
Introduction
Formula in Chemistry is very important to simplify the presentation of substances. There are
many types of formula, for example, chemical formula, molecular formula, structural
formula etc.
Empirical formula of a compound shows the simplest whole-number ratio of atoms or ions
present. It does NOT represent the actual number of atoms bonded together.
For example, FeS, NaCl, CH2
Molecular formula of a compound shows the actual number of each kind of atoms present
in one molecule (for covalent compounds only). How about ionic compounds?
For example, C2H4,C4H6, C2H4O2.
The major different between empirical and molecular formula is that the former may not exit
as actual substance.
For example, the empirical formula of C2H4 (ethene) is CH2 but it does not exist.
B.
Determining chemical formulae by experiments
To determine the empirical formula of black copper(II) oxide
The following data are obtained:
mass of test-tube
mass of test-tube + black copper(II) oxide
mass of test-tube + copper
mass of black copper(II) oxide
mass of copper
mass of oxygen present in the oxide
= 21.320g
= 22.950g
= 22.622g
= 1.6308g
= 1.302g
= (1.630-1.302) = 0.328g
Copper (Cu)
Oxygen (O)
masses in g
number of moles (divided
by atomic mass)
simplest ratio (divided by
the smallest value)
Thus the empirical formula of black copper(II) oxide is __________.
2
ii.
To determine the empirical formula of magnesium oxide
mass of crucible + lid
mass of crucible + lid + magnesium
mass of crucible + lid + magnesium oxide
mass of magnesium
mass of magnesium oxide
mass of oxygen present in the oxide
Magnesium (Mg)
masses in g
= 20.635g
= 21.555g
= 22.165g
= 0.92g
= 1.53g
= (1.53-0.92) = 0.61g
Oxygen (O)
number of moles
simplest ratio
Thus the empirical formula of magnesium oxide is __________.
C.
Examples of finding empirical formulae
Example 1: Hydrocarbon M contains 85.7% carbon. Calculate the empirical formula of M.
Answer:
percentage by mass of carbon = 85.7
percentage by mass of hydrogen = 100 - 85.7 = 14.3
Assume we have 100g of M,
Carbon (C)
Hydrogen (H)
masses in g
number of moles
simplest ratio
The empirical formula of M is ________.
Example 2:
Aspirin contains 60% carbon, 4.5% of hydrogen and 35.5% oxygen. What is
the empirical formula?
Answer:
Assume we have 100g of aspirin, then there is 60g of carbon, 4.5g of
hydrogen, 35.5g of oxygen.
Magnesium (Mg) Hydrogen (H)
Oxygen (O)
masses in g
number of moles
simplest ratio
simplest whole number ratio
Thus the empirical formula for aspirin is __________.
3
Classwork :
D.
1.
The oxide of metal M (relative atomic mass = 52) contains 31.6%
by mass of oxygen. What is the empirical formula?
2.
2.8g of X is completely reacted with excess Y to form 17g of
compound. What is the empirical formula of the compound?
(Relative atomic masses of X = 28, Y = 35.5).
Examples of finding molecular formulae
Example 1: 1.5g hydrocarbon M contains 85.7% carbon is found to occupy 600cm3 at
s.t.p. Calculate the empirical and molecular formula of M. (molar volume of
M at s.t.p. = 22.4dm3)
Answer:
percentage by mass of carbon = 85.7
percentage by mass of hydrogen = 100 - 85.7 = 14.3
Assume we have l00g of M,
Carbon (C)
Hydrogen (H)
masses in g
number of moles
simplest ratio
The empirical formula of M is _______.
no. of moles of M =
(1dm3 = 1000cm3)
relative molecular mass of M =
let the molecular formula of M be
relative molecular thus =
therefore,
Thus the molecular formula of M is __________.
Classwork :
1.
2.
E.
A compound L contains 52.18% carbon, 13.04% hydrogen and 34.78%
oxygen by mass. 9.2g of L are found to occupy 4,480cm3 at s.t.p. What is
the molecular formula of L?
Compound Q contains 69.8% carbon, 11.7% It, and some oxygen. 560cm3
of its vapor weighs 2.15g at s.t.p.
a.
Calculate the relative molecular mass of Q.
b.
Find the molecular formula of Q.
Molecular formulae of simple organic molecules
Example 1 :
26cm3 of a gaseous hydrocarbon required 91cm3 of oxygen for complete
combustion and gave 52cm3 of carbon dioxide. Find the formula of the
hydrocarbon.
Answer:
let the molecular formula be CxHy
26cm3 of CxHy require 91 cm3 of oxygen to give 52cm3 of carbon dioxide
1 cm3 of CxHy, requires 3.5cm3 of oxygen to give 2cm3 of carbon dioxide
consider the equation:
CxHy(g) +
O2(g) Æ CO2(g) +
H2O(l)
vol. ratio
vol. ratio
therefore
molecular formula is ________________.
from actual data
from equation
4
Classwork :
1.
17cm3 of a compound of carbon, hydrogen and oxygen required 68cm3 of
oxygen for complete combustion. The experiment was maintained at a
temperature of 120°C and the volume of carbon dioxide and steam formed
were 51 cm3 each. What is the formula of the original compound?
Consider the equation:
CXHYOZ +
F.
O2
Æ
CO2
+
H2O
Determine molecular formulae from combustion data
Example 1:
Upon complete combustion of 0.46g of a compound containing carbon,
hydrogen and oxygen, 0.88g of carbon dioxide and 0.54g of water were
formed. Calculate the empirical formula of the original compound.
0.88g of CO2 contain _______ fraction of carbon = _________ g
0.54g of H2O contain _______ fraction of hydrogen = __________ g
mass of oxygen = ______________________________
carbon (C)
Hydrogen (H)
Oxygen (O)
masses in g
number of moles
simplest ratio
The empirical formula is ________________.
Classwork :
1.
2.
A compound contains carbon, oxygen and hydrogen only. 4.5g of it when
completely burnt in oxygen, gives 4.4g of carbon dioxide and 0.9g of water.
Its molecular mass is found to be 90. Find the molecular formula.
A gaseous hydrocarbon with vapor density equals 21 was burnt completely in
air. The gaseous products were first passed through on U-tube (cooling under
water) then to another test tube containing potassium hydroxide solution. The
results of the experiment were recorded as below:
weight of U-tube before experiment
= 36.40g
weight of U-tube after experiment
= 36.94g
weight of test-tube & content before experiment
= 78.90g
weight of test-tube & content after experiment
= 80.22g
a.
What was the function of the U-tube?
b.
What was the function of potassium hydroxide solution?
c.
Find the empirical and molecular formulae of the hydrocarbon.
(Hint : 2 x vapor density of a gas = relative molecular mass of the gas)