Question

Calculate the percentage of iron in K₃Fe(CN)₆ (K = 39, Fe = 56, C = 12, N = 14).

Answer 1

Step 1: Write the given terms:
K = 39, Fe = 56, C = 12, N = 14
Step 2: Find the molar mass
Molar mass of K₃Fe(CN)₆ = 3 x (molar mass of K) + molar mass of Fe + 6 x (molar mass of C) + 6 x (molar mass of N)
= 3 x (39) + (56) + 6 x (12) + 6 x (14)
= 329 g
Step 3: Find the percentage:
One molecule of K₃Fe(CN)₆ contains 1 atom of Fe.

So the percentage composition of Fe is
= $\frac{massofironincomplex}{molarmassofcomplex}$ X 100
= $\frac{56x100}{329}$
= 17.02 %
Step 4: Result
Hence, 17.02% is the percentage of iron in K₃Fe(CN)₆