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?

xMg + yN2 ==> MgxNy

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.

To determine the empirical formula of a compound, you need to determine the ratio of the elements present in the compound. Here's how you can approach it:

1. Start by finding the moles of each element in the compound.

- Mass of magnesium (Mg) = 2.00g
- Mass of nitrogen (N) = 2.77g - 2.00g = 0.77g

To convert these masses to moles, divide each by their respective molar masses:

- Molar mass of Mg = 24.31 g/mol
- Molar mass of N = 14.01 g/mol

Moles of Mg = 2.00g ÷ 24.31 g/mol = 0.0822 mol
Moles of N = 0.77g ÷ 14.01 g/mol = 0.055 mol

2. Next, find the simplest whole-number ratio of moles by dividing each by the smallest number of moles to get the empirical formula.

Divide both moles by 0.055 mol (the smallest):

Moles of Mg = 0.0822 mol ÷ 0.055 mol = 1.49 ≈ 1
Moles of N = 0.055 mol ÷ 0.055 mol = 1

The resulting ratio is 1:1.

3. Finally, write the empirical formula using the found ratio. In this case, the empirical formula would be MgN.

Therefore, the empirical formula for the compound is MgN, which indicates that 1 atom of magnesium is combined with 1 atom of nitrogen.