Friday, October 18, 2013

Empirical formula and molecular formula CBSE 9th Chemistry

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. C3H7NOempirical formula , C3H7NOmolecular formula 
Related Stydy material

Practice paper and Notes  
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DETERMINING EMPIRICAL AND MOLECULAR FORMULAE       Download File
Empirical and molecular formaul practice paper with solution                     Download File
Empirical_formula_worksheet                                                                   Download File

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