Question

Analysis of a 2.016 g sample of pure magnesium oxide gave 1.216 g magnesium and 0.800 g of oxygen. A second sample weighing 4.479 g was found to contain 2.701 g magnesium and 1.776 g oxygen. Show that these data are in agreement with the law of definite composition.

Answer 1

Law of definite composition

The elements in a compound always be in fixed ratio by mass.

For MgO

% O = molar mass of O x 100 / molar mass of MgO

= 16*100/40.3044 = 39.697 %

% Mg = 100 - 39.697 = 60.303%

First sample

The balanced reaction

MgO = Mg + (1/2)O2

40.3044 g MgO gives = 24.305 g Mg

2.016 g MgO gives = 24.305 x 2.016 / 40.3044

= 1.216 g Mg

40.3044 g MgO gives = 16 g O2

2.016 g MgO gives = 16 x 2.016 / 40.3044

= 0.800 g O2

For the first sample, the given amount of MgO, Mg and O2 are same as by stoichiometric.

For the given sample

% O = mass of O x 100 / mass of MgO

= 0.800 x 100 / 2.016 = 39.69%

% Mg =mass of Mg x 100 / mass of MgO

= 1.216 x 100 / 2.016 = 60.3%

% O and % Mg is same as stoichiometrically for the given mass

For the second sample

% O = mass of O x 100 / mass of MgO

= 1.776 x 100 / 4.479 = 39.6%

% Mg =mass of Mg x 100 / mass of MgO

=2.701 x 100 / 4.479 = 60.3%

% O and % Mg is same as stoichiometrically for the given mass

Both samples follow the law of definite composition.