Question d1163

Nov 3, 2016

If some of the magnesium had reacted with nitrogen, The apparent percent composition would have increased

Explanation:

The theoretical percentage of $\text{Mg}$ in $\text{MgO}$ is 60.31 %.

If the $\text{Mg}$ reacted with ${\text{N}}_{2}$, the other major gas in the air, it would have formed ${\text{Mg"_3"N}}_{2}$.

The percentage of $\text{Mg}$ in ${\text{Mg"_3"N}}_{2}$ is 72.24 %.

Thus, any formation of ${\text{Mg"_3"N}}_{2}$ would increase the apparent percentage of $\text{Mg}$ in the product.

$\text{Mass of Mg" = "2.993 g - 2.868 g" = "0.125 g}$
$\text{Mass of MgO" = "3.076 g - 2.868 g" = "0.208 g}$
"% Mg" = "Mass of Mg"/"Mass of MgO" × 100 % = (0.125 color(red)(cancel(color(black)("g"))))/(0.208 color(red)(cancel(color(black)("g"))))× 100 % = 60.1 %#
Your sample contains no ${\text{Mg"_3"N}}_{2}$.