Відповідь:

w = n*Ar/ Mr * 100%

1. Mr (CaO) = 40+ 16 = 56 г/моль

w(Ca) = 1* 40/56 * 100% = 71.4 %

w( O) = 100 - 71.4 = 28.7 %

2. Mr( FeS2) = 56+ 2* 32 = 120 г/моль

w( Fe) = 1*56/120 * 100% = 46.6%

W(S2) = 100- 46.6 = 53.4 %

3. Mr (Cr(OH)3)  = 52+ 3*16+ 3 = 52+48+ 3 =103 г/моль

w( Cr) = 1*52/103 * 100% = 50.48%

w(O) =3*16 / 103 * 100%= 46.6 %

w ( H) = 3*1/103 * 100% = 2.9%