A student heated 1.228 grams of Mg(ClO3)2 until a stable weight was determined. the remaining residue weighed 0.584 grams. solve for the experimental percentage of oxygen.

llltkl | Student

Upon heating, `Mg(ClO_3)_2` decomposes to produce magnesium chloride and oxygen gas:

`Mg(ClO_3)_2 stackrel Delta rarr MgCl_2+3O_2(g)`

From the stoichiometry of its decomposition,

(24.305+(35.453+3*16)*2), i.e. 191.2 g `Mg(ClO_3)_2` produces 3*16*2, i.e. 96 g oxygen.

Therefore, 1.228 g `Mg(ClO_3)_2` produces 96*1.228/191.2 =0.61657 g oxygen.

Again, when (24.305+2*35.453), i.e. 95.2 g `MgCl_2` is obtained, amount of oxygen produced is 96 g.

So, when 0.584 g `MgCl_2` is obtained, amount of oxygen produced is 96*0.584/95.2 g=0.5889 g.

Two differing values of oxygen produced implies that the original sample of `Mg(ClO_3)_2 ` was impure, hence calculations on its basis returned an inflated value for oxygen. The most common source of impurity in these salts is moisture absorbed from air.

The actual amount of oxygen obtainable from the sample is thus, 0.5889 g.

Hence percentage of oxygen in the given sample of `Mg(ClO_3)_2` =0.5889*100/1.228=47.96.

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question