# How to solve Problems based on Law of Definite Proportions / Constant Composition

Problem 1 : One mole of  V2Ocontains 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:

V2O 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 V2O=   ——————————–            x                   100
2 (50.9) + 5 (16.0)

%  oxygen by weight  in V2O5 = 44.0%

### Don’t Miss Problems based on Mole Concept, Atomic Mass, Molar Mass, Avogadro number

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

Same as using the Law of Conservation of Mass.

On the basis of your understanding of the above topic, answer the following questions –