class11 ,Explain law of constant composition or definite proportions with numericals and examples.
Answers
Answer:
The law of constant proportion was given by Joseph Louis proust. It says that in a chemical substance (compound) elements are always present in definite proportion by mass. For example: In a compound such as water the ratio of mass of hydrogen to the mass of oxygen is always 1
Problem 1 : One mole of V2O5 contains 5 moles of oxygen atoms, as it is clear from its formula. What is the % oxygen by weight in V2O5? (Atomic weights are; V – 50.9 g/mol, O – 16 g/mol)
Solution:
V2O5 contains 5 moles of oxygen atoms and 2 moles of vanadium atoms.
Atomic mass of V2O5 = 2 (50.9) + 5 (16.0)
5 (16.0)
% oxygen by wt. in V2O5 = ——————————– x 100
2 (50.9) + 5 (16.0)
% oxygen by weight in V2O5 = 44.0%
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Problem 2 : What is the experimental percent of oxygen in CO2 if 42.0 g of carbon reacted completely with 112.0 g of oxygen?
Solution: % O = (mass of O / mass of CO2) 100
% O = [112.0 g O / (42.0 g + 112.0 g) CO2] 100 = 72.7% O
Problem 3 : What is the theoretical percent of aluminum in aluminum oxide?
Solution: % Al = (Atomic mass of Al / Formula mass of Al2O3)× 100
% Al = (54 amu / 102 amu) 100 = 52.9%
Problem 4 : What is the percent composition of sodium chloride?
Solution : Atomic mass of sodium chloride = Atomic mass of sodium + Atomic mass of chlorine
% sodium by mass = ( 23 / 58.5 ) × 100 = 39.3%
% chlorine by mass = ( 35.5 / 58.5 ) × 100 = 60.7%
Sodium chloride contains 39.3% by mass sodium and 60.7% by mass chlorine.
Problem 5 : When 0.0976 g of magnesium was heated in air, 0.1618 g of magnesium oxide (MgO) was produced.
(a) What is the percent of Mg in MgO?
(b) Using only law of definite proportions, what mass of oxygen was needed to combine with the magnesium?
Solution :
(a) % Mg = (mass Mg / Mass MgO) 100
= (0.0976g / 0.1618 g) 100 = 60.3 %
(b) % O = 100% MgO – 60.3% Mg = 39.7% O
% O = (mass O / mass MgO) 100
39.7 % = (mass O / 0.1618 g) 100
mass O = 0.397 ( 0.1618 g) = 0.0642 g O