Posted by Anonymous on Tuesday, November 20, 2007 at 9:30pm.
Heating a 2.00g sample of magnesium metal with an excess of pure nitrogen gas produce 2.77g of a compound containing only magnesium and nitrogen. What is the empirical formula for this compound?

chemistry  DrBob222, Tuesday, November 20, 2007 at 9:41pm
xMg + yN2 ==> Mg_{x}N_{y}
Convert 2.00 g Mg to mols. mols = g/atomic mass.
g nitrogen = 2.77  2.00 = 0.77 g.
Convert 0.77 g nitrogen to mols.
Now determine the ratio of Mg to N.
The easy way to do this is to divide the smaller number of moles by itself which will give exactly 1.000. Then, to keep things equal, divide the other number of mols by that same small number. SMALL difference between the numbers and whole numbers may be rounded to the nearest whole number BUT numbers between 0.25 and 0.75 CAN'T be rounded. In those cases, multiply both numbers by a factor to obtain a whole number and those whole numbers will be the values that go in for x and y in the empirical formula.