Empirical formula :The
empirical formula is the simplest formula for a compound in which atoms of
different elements are present in simple ratio. It shows the relative number of
atoms of each element. For example CH2O is the empirical formula of
Glucose C6H12O6
Molecular formula : It
is the formula in which the actual number of atoms of
different element are present. For example, if the empirical formula of
benzene is CH where as molecular formula is C6H6 ,
etc.
An empirical formula
is often calculated from elemental composition data. The weight percentage of
each of the elements present in the compound is given by this elemental
composition.
Let's determine the
empirical formula for a compound with the following elemental
composition:40.00% C, 6.66% H, 53.34% O.
Element
|
percentage
|
Atomic mass
|
Relative number of
Atoms
|
Dividing by least
number
|
Simple ratio
|
C
|
40
|
12
|
40/12 = 3.33
|
3.33/3.33
|
1
|
H
|
6.66
|
1
|
6.66/1 =6.66
|
6.66/3.33
|
2
|
O
|
53.34
|
16
|
53.34/16 =3.33
|
3.33/3.33
|
1
|
Empirical
formula =C1H2O1 ; Empirical formula mass
= 12 + 2x1 +16 = 30 a.m.u
Given relative
molecular mass = 180
Divide the relative
molecular mass by the Empirical formula mass to find a multiple: 180/30 = 6
The molecular
formula is a multiple of 6 times the empirical formula: (C1H2O1)x6
= C6H12O6
EMPIRICAL AND
MOLECULAR FORMULAE WORKSHEET
1.What’s the
empirical formula of a molecule containing 65.5% carbon, 5.5% hydrogen, and
29.0% oxygen?
2. If the molar mass
of the compound in problem 1 is 110 grams/mole, what’s the molecular formula?
3 What’s the
empirical formula of a molecule containing 18.7% lithium, 16.3% carbon, and
65.0% oxygen?
4. If the molar mass
of the compound in problem 3 is 73.8 grams/mole, what’s the molecular formula?
5.The percentage
composition of acetic acid is found to be 39.9% C, 6.7% H, and 53.4% O.
Determine the empirical formula of acetic acid.
6. The molar mass
for question #9 was determined by experiment to be 60.0 g/mol. What is the
molecular formula?
7. A 50.51 g sample
of a compound made from phosphorus and chlorine is decomposed. Analysis of the
products showed that 11.39 g of phosphorus atoms were produced. What is the
empirical formula of the compound?
8. When 2.5000 g of
an oxide of mercury, (HgxOy) is decomposed into the
elements by heating, 2.405 g of mercury are produced. Calculate the empirical
formula.
9.The compound
benzamide has the following percent composition. What is the empirical formula?
C = 69.40 % H= 5.825
% O = 13.21 % N= 11.57 %
10.A component of
protein called serine has an approximate molar mass of 100 g/mole. If the
percent composition is as follows, what is the empirical and molecular formula
of serine?
C = 34.95 % H= 6.844
% O = 46.56 % N= 13.59 %
Answer.
1. C3H3O
mass = 55 g/mole
2. C6H6O2
3 Li2CO3
4 Li2CO3
5. CH2O
6. C2H4O2
7. PCl3
8. Hg2O
9. C7H7NO
10. C3H7NO3 empirical formula ,
C3H7NO3 molecular formula
Related Stydy material
Related Stydy material
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