Question #d1163

1 Answer
Nov 3, 2016

Answer:

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

Explanation:

The theoretical percentage of #"Mg"# in #"MgO"# is 60.31 %.

If the #"Mg"# reacted with #"N"_2#, the other major gas in the air, it would have formed #"Mg"_3"N"_2#.

The percentage of #"Mg"# in #"Mg"_3"N"_2# is 72.24 %.

Thus, any formation of #"Mg"_3"N"_2# would increase the apparent percentage of #"Mg"# in the product.

In your experiment,

#"Mass of Mg" = "2.993 g - 2.868 g" = "0.125 g"#

#"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 %#

This is equal to the theoretical value within the experimental uncertainty.

Your sample contains no #"Mg"_3"N"_2#.