Dimensional analysis question

Question

How many grams of Fe³⁺ are present in 3.49 grams of iron(iii) cyanide?

Answer 1

The formula for iron(III) cyanide is Fe(CN)₃, meaning it contains 1 Fe, 3 C, and 3 N atoms per molecule.  So just take the molar mass of Fe and divide it by the molar mass of the molecule to find its percentage, then use that to solve your question.

 

Fe = 55.85 g/mol

C = 12.01 g/mol

N = 14.01 g/mol

 

55.85 + 3(12.01) + 3(14.01) = 133.91 g/mol

55.85/133.91 x 100 = 41.707%

 

3.49 x .41707 = 1.456

 

So 3.49 g of iron(III) cyanide contains 1.46 g of Fe³⁺.