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3.Definite proportions.

posted Mar 8, 2016, 8:05 PM by Upali Salpadoru   [ updated Mar 23, 2016, 4:14 PM ]

3.Percentage composition.

3. This no. represents the Chapter no. of the book "Step by step Chemical Calculations"  This lesson is a revised translation of that. 

According to Proust’s Law, (published in 1795) which states that any pure compound should consist of the same elements combined in a definite proportion irrespective of the process of getting it.

   We can calculate the ratio of the constituents in a compound  using the formula. The formula is generally obtained by combining chemicals.

 Here is an example of Magnesium burning in air.

When the silver coloured metal strip burns it gives out a brilliant light which can even damage the eyes if you directly look at the flame.  

gr 2.jpg

When magnesium burns, a white coloured solid, magnesium oxide is formed. The formula is shown here.

The equation for the reaction:

Magnesium + Oxygen = Magnesium oxide

Mg               +   O2     =    MgO  +  MgO

Here are the readings of an experiment performed.

Mass of crucible                             =  10.5 g
Mass of crucible + Magnesium    =   13.5 g
Mass after burning                        =    15.5 g


Fig.1 Magnesium burning inside a crucible.


 

Calculation: for Mass Ratio

Finding the percentages:

Mass of Magnesium                 =   3.0 g

Mass of Oxygen combined     =   2.0 g

          Mass ratio  of  Mg: O    =  3 :2

 

Total number of parts             = 3 + 2  =  5

Proportion of Magnesium       =  3/5

Percentage of Mg                     =  3 ÷ 5 x 100 =  60%

 

Proportion of  Oxygen        =  2/5

Percentage of Oxygen         = 2 ÷ 5 x 100 =40%

 

Finding the percentage composition of Ammonium sulphate.

Molecular mass of (NH4)2SO4…………………………………=  132

i)  Mass of two atoms of N…………………………=    28

     Percentage mass of Nitrogen…………………= 28 ÷ 132 x 100   = 21.21%

ii) Mass of 8 atoms of H

     Percentage mass of Hydrogen…… ………… =  8 ÷ 132 x 100   =  6.06 %

iii) Mass of one atom of S………………………..=  32

      Peercentage mass of Sulphur ………………..=  32 ÷  132 x 100 =  24,24%

iv) Mass of 4 atoms of O…………………………=  64

      Percentage mass of Oxygen………………….=   64÷ 132 x 100 = 48.48 %

 

Verification

The total of percentages must add up to 100.

Nitrogen……… =  21.21

Hydrogen……...=    6.06

Sulphur ……….=  24.24

Oxygen………..=  48.48

                               99.99    gets rounded up to 100.

 practice.jpg

You are sure to enjoy the filling of these blank spaces:
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Here the masses are in amu.

Question

Working

Answer

1.0  Molecular mass of Lead oxide  PbO2

207+ 16.x 2 = 239 amu.

 

 

1.1 Percentage of Pb.

Pbx100       =   207 x100 /  239  

PbO2

…………%

1.2 Percentage of O2

 

…………%

2.0 Molecular mass of CaCO3

 

 

2.1 Percentage of Ca,

 

 

2.2 Percentage of C.

 

 

2.3 Percentage of O2.

 

 

3.0 Molecular mass of CO2

 

 

3.1 Percentage of C

 

 

4.0 Molecular mass of KMnO4

 

 

4.1 Percentage of O2

 

 

4.2 mass of Oxygen in 100g of KMnO4

 

 

5.0 Molecular mass of Potassium ferri cyanide K3Fe(CH)6

 

5.1 Percentage of Potassium.

3K X100   = 

 K3Fe(CN)6      

35.8

5.2 Percentage of (CN) group.

6(CN) X 100              

 K3Fe(CN)6

47.6

Answers

Q.No.

1.1

1.2

2.1

2.2

2.3

3.1

4.1

4.2

5.1

5.2

Answer

86.61%

13.38%

40 %

12%

48 *

27.27 %

40.51 %

40.51 g

35.8

47.6

 
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