Question

A. If the Mg₃N₂ is not decomposed, will the reported magnesium to oxygen ratio be high or low? Explain. Hint: The Mg:N mass ratio is 1:0.38 and the Mg:O mass ratio is 1:0.66.

B. An analysis shows that 2.97 g of iron metal combines with oxygen to form 4.25 g of an     oxide of iron. What is the empirical formula of the compound?

Answer 1

A) It will be low, because the total mass will be lower, because Mg:N won't weigth as much as the Mg:O so have to have 2 molecules of Mg:N so can have the weigth of one molecule of Mg:O so now you need the doble of atoms of Mg to mantein the same ratio with the O.

B) X Fe + Y/2 O2 ========= Fe_xO_y

Atomic weigth of Iron is 55,85g/mol

1mol Fe ------------ 55,85g

X --------------------- 2,9g

X= 0,052mol

Xmol Fe ----------------- 1mol  Fe_xO_y

0,052mol Fe ------------ Z

Z its the mol of  Fe_xO_y that is produce in the reaction and mol= grams/molecular weigth, so i can substitute in the equation.

0,052mol Fe = Z = g/MW

MW= 4,25g/0,052mol = 81,73 g/mol

Now we know that the Iron weigth is 55,85g and 2 atoms of iron will be 111,7g/mol and that more than the MW of the compound so for the iron in the empirical formula we only have 1 atom.

The rest of the weigth has to come from the oxygen, that is 16g/mol.

81,73g/mol - 55,85g/mol =25,88 g/mol

25,88g/mol / 16g/mol = 1,6 atoms.

FeO_1,6

Because we can't have half an atom we multiply by 2 both values so the empirical formula will be Fe₂O₃