suppose you had let some magnesium oxide gas out of your test tube
while burning it in a crucible. How would the Mg:O ratio have been
affected? Would the Ratio have increased, decreased, or remained
unchanged? Explain using sample calculations?

next part is How would your calculated value for the empirical formula
of MAgnesium Oxide have been affected if all the MAgnesium in the
crucible had not burned? support your answer with some calculations

and finally could either situations mentioned in parts 1 and 2 have
affected your results? explain

The chemical equation for burning magnesium in air is "2 Mg + O2 -> 2
MgO."  Therefore, loss of magnesium oxide would result in the loss of
an equal amount of both magnesium and oxygen atoms.  As a result, the
magnesium to oxygen ratio would remain unchanged.

"Using Balanced Equations" by Sue Eggling, Clackamas Community
College(2002) http://dl.clackamas.cc.or.us/ch104-04/using.htm

"Since the number of mols of oxygen is directly proportional to the
mass of oxygen that reacted, the number of mols of oxygen calculated
would be lower which would result in a higher Mg:O ratio."  Instead of
just having a one-to-one ratio of magnesium and oxygen atoms in the
form of magnesium oxide, instead you have a higher proportion of
magnesium resulting from the unreacted magnesium and the magnesium in
the magnesium oxide.

"AP Chemistry II Laboratory Manual" by Patrick Gormley, Lapeer East
High School (November 15, 2004)
http://chem.lapeer.org/chem2docs/APChem2Manual.html

Search terms: calculated value empirical formula "magnesium oxide" magnesium

Sincerely,

Wonko

Well, no one can answer the third part but you, as you're the one who
did the experiment ;)

But, on the other hand, for part two consider what actually happens to
the ratio when you for instance have 1 atom of magnesium to one
"molecule" of magnesium oxide. A similar caveat may apply to the first
part too, but also consider melting and boiling points of the elements
and compounds you're dealing with.

re: wonko's answer.

Nicely put, although I'm not sure it's right! ;)

My take on the first bit is that *if* a quantity of MgO(g) escapes,
then the mass of material left is obviously less than it should be.
So, if you're not aware of the loss, you would assume that less oxygen
combined with the known amount of magnesium than ought to, so the
empirical formula would be something like 1 equivalent of Mg to 1-x of
O where x is small.

I can't find a boiling point for MgO in my textbooks at the minute
(have not done a vigorous search); but having carried out a similar,
if not identical, experiment in the past its boiling point is probably
in excess of a bunsen's flame so this is probably an unlikely source
of error.

Of course, I may have misinterpreted the question, too..

S