Question

14. If a solution is made by dissolving 14.5 g of Fe(NO3)3 in 455 ml of water, what is the mass percent of iron and nitrate in this solution? Assume 1.00 g/ml density of the water. 4 pts

Answer 1

Ans 14 :

Density = mass / volume

1.00 g/mL = m / 455 mL

m = 455 g

mass of solution = mass of water + mass of Fe(NO3)3

= 455 g + 14.5 g

= 469.5 g

Now one mole of Fe(NO3)3 weighing 241.86 grams has 55.845 g of Fe iron and 186.0147 g of nitrate

so 14.5 g of Fe(NO3)3 will have : ( 14.5 g x 55.845 g ) / 241.86 g = 3.348 g iron

and : (14.5 g x 186.0147 g ) / 241.86 g = 11.152 g of nitrate

so mass percent of iron = ( mass iron / mass solution ) x 100

= (3.348 g / 469.5 g ) x 100

= 0.713 %

mass percent of nitrate = ( mass nitrate / mass solution ) x 100

= (11.152 g / 469.5 g ) x 100

= 2.38 %