What is the empirical formula of a compound that contains 10.52 grams of Ni 4.38 grams of C and 5.1 grams of N

Answer 1

aren't these fun!

find the amounts of moles in the compond
10.52g / 58.69 (Ni's molar mass) = 0.18 mol Ni
4.38g / 12.01 = 0.36 mol C
5.1g / 14.01 = 0.36 mol N

from here, divide each by the value with the smallest amount of moles (Ni-- 0.18 moles)
0.18 mol Ni / 0.18 = 1 mol Ni
0.36 mol C / 0.18 = 2mol C
0.36 mol N / 0.18 = 2 mol N

so there's 2 C, 2 N, and 1 Ni. the formula could be (and order doesn't matter)...
NiC2N2

Answer 2

Ni = 10.52/58.69 = 0.18 mol
C = 4.38/12 = 0.36 mol
N - 5.1/14 = 0.36 mol

Formula = NiC2N2 or Ni(CN)2