Calculate the percent nitrogen in ammonium hexacyano iron (II) which has the formula (NH4)4Fe(CN)6. [ N = 14.01, H = 1.008, Fe =55.85, C=12.01 ]

Answer 1

% N = grams of N / total grams (x100%)

(NH₄)₄Fe(CN)₆

Mass of N = 10 x 14.01 = 140.10

Mass of H = 16 x 1.008 = 16.128

Mass of Fe = 1 x 55.85 = 55.85

Mass of C = 6 x 12.01 = 72.06

Total mass (molar mass) = 284.138 g / mol

% N = 140.10 g / 284.138 g (x100%) = 49.31%