# In the gravimetric analysis of magnesium, when combusted to magnesium oxide, why should care be taken to collect all of the magnesium oxide is retained by the crucible?

Jun 25, 2017

When magnesium burns in air.........

#### Explanation:

.....the following reaction occurs.

$M g \left(s\right) + \frac{1}{2} {O}_{2} \left(g\right) \rightarrow M g O \left(s\right) + \Delta$

The magnesium oxide is a finely divided white solid, and CAN be lost during the exothermic combustion reaction. Nevertheless, given complete combustion (which is a given in these circumstances), the ratio of magnesium atoms to oxygen atoms will be 1:1 whatever mass of the combustion product you collect. And thus $\text{option D}$ is the winner.

Ratio of Mg:O would be too high if significant smoke escapes.

#### Explanation:

When magnesium burns the reaction is as follows:

$2 M g \left(s\right) + {O}_{2} \to 2 M g O \left(s\right) + h e a t$

The video below shows a demonstration of this reaction. Note the smoke that is given off during the reaction.

The smoke is some of the white MgO formed in the reaction.

Now here is video of the experiment you are describing.

Starting this reaction with 24.3g of Mg should result in formation of 40.3g of MgO. This is due to the addition of 16g of oxygen.

24.3g = 1 mole Mg
16g = 1 mole O

If a significant amount of smoke is allowed to escape the final measured mass of MgO will be lower than 40.3g. Let's say you
actually collect 32.3g of MgO.

You would do calculations assuming no lost Mg. So you would have these numbers:

Mg = 24.3g
O = 8g

24.3g = 1 mole Mg
8g = 0.5 mole O

This would result in a calculation of a ratio of 2:1 and you would determine the formula of magnesium oxide to be $M {g}_{2} O$.

So I would say that A is the answer based on the wording of this question.

Hope this helps!