What is the percent of iron in iron (II) hydroxide? (round your answer to 3 sig figs)

Answer 1

Fe = 55.85 g/mol

O = 15.999 g/mol

H = 1.008 g/mol

Iron(II) hydoxide = Fe(OH)₂

molar mass Fe(OH)2 = 55.85 + (2*15.999) + (2*1.008) = 55.88 + 31.998 + 2.016 = 89.86 g/mol

Percent Fe = 55.85 / 89.86 (x100%) = 62.2% (3 sig. figs.)

Answer 2