Question

Analysis of a compound formed between magnesium and nitrogen showed it contained 14.4g of magnesium and 5.6.g of nitrogen.

What is the empirical formula of the compound?

A Mg2N3

B Mg3N2

CMg4N6

D Mg6N4

And why is it that way?

Answer 1

B. Mg3N2

Step-by-step explanation:

There are 20.0 grams total material (14.4g Mg and 5.6 g N). We can convert grams into moles of each element by dividing by that element's molar mass (g/mole) to yield the number of moles of each element. To find the empirical formula, we need to convert the moles of each to a whole number by using a fracor that provides a whole number for each element. 5 works well, and we get 0.6x5 = 3 Mg atoms and 0.4x5 = 2 N atoms.

Mass % g/mole grams moles Factor of 5

Mg 72.00% 24.305 14.4 0.6 2.962 or 3

N 28.00% 14.007 5.6 0.4 1.999 or 2

100.0% 20 1.0

There are 3 Mg and 2 N atoms per molecule. The empirical formula is Mg3N2.