Answer to Question #119857 in General Chemistry for Ellena

Question

In the analysis of an iron ore by precipitation of the iron as Fe(OH)3, what weight of the sample should be taken so that each centigram of the ignited precipitation present 0.100% Fe.

Answer 1

Molecular weight of of Fe(OH)3 = 107 g/ mole

In 107 g of Fe(OH)3, Fe present = 56g

So in 1 g of Fe(OH) 3 Fe present = 56/ 107

= 0.52

In 1 centigram Fe present = 0.52/ 100

= 0.0052

That is 0.52%

So 0.52% got from 1 g

0.1% should present in 0.1/ 0.52

= 0.019 g