Question

What is the total sample size (in grams) for a sample of magnesium nitrite which contains 0.92 g of oxygen?

My answer was wrong (32.34g)

Answer 1

ok lets see if i got it right, i just woke up so maybe im not right:
Mg(NO2)2 ---> 4 * Ar(O) + 2 * Ar(N) + 1 * Ar(Mg)
So
X(O)= 4*Ar(O)/M(Mg(NO2)2)=64/116,3=0,55
m(Mg(NO2)2= 0,92 g/0,55=1,67 g