Question #9c0f3

1 Answer
Oct 1, 2017


Here's what I got.


The first thing that you should notice here is that the problem provides you with the mass of magnesium and the mass of the oxide.

Since you know that

#"magnesium + oxygen gas " -> " magnesium oxide"#

you can say that the mass of oxygen gas that took place in the reaction must be equal to the difference between the mass of oxide and the mass of magnesium.

This means that you have

#overbrace("4.03 g")^(color(blue)("mass of oxide")) - overbrace("2.3 g")^(color(blue)("mass of magnesium")) = overbrace("1.7 g")^(color(blue)("mass of oxygen gas"))#

Since you subtracted two values to get this answer, the mass of oxygen gas must be rounded to one decimal place because that's how many decimal places you have for your least precise measurement, i.e. for #"2.3 g"#.

Now, to find the number of moles of oxygen present in the oxide, you need to use the molar mass of elemental oxygen, #"O"#, not of diatomic oxygen, #"O"_2#.

This is the case because the oxide does not contain any oxygen molecules, it only contains oxygen atoms!

#1.7 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0 color(red)(cancel(color(black)("g")))) = "0.10625 moles O"#

Do the same for magnesium--use its molar mass to find the number of moles of magnesium present in the oxide.

#2.3 color(red)(cancel(color(black)("g"))) * "1 mole Mg"/(24.305 color(red)(cancel(color(black)("g")))) = "0.09463 moles Mg"#

To find the mole ratio of the two elements in the oxide, simply divide the two values by the smallest one.

#"For Mg: " (0.09463 color(red)(cancel(color(black)("moles"))))/(0.09463color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.10625 color(red)(cancel(color(black)("moles"))))/(0.09463 color(red)(cancel(color(black)("moles")))) = 1.12#

Now, in order to find the empirical formula of the oxide, you should use the smallest whole number ratio that exists between the two elements in the oxide.

In this case, you can't really come up with a whole number ratio unless you multiply both values by #9#, in which case you'd end up with

#"Mg"_ ((1 * 9)) "O"_ ((1.12 * 9)) ~~ "Mg"_9"O"_10#

Ideally, the two elements should combine in a #1:1# mole ratio in the oxide, so you can say that a number of experimental/procedural errors have crept into your experiment.