Вычислить массовые доли элементов в:
CaCO3, AgNO3, ZnCL2

формула W = (n*Ar) / Mr  * 100%

в каждом действии я не стала писать умножение на 100%.  

H2S:

w(H) = 2 * 1 / 34 = 5,88%

w(S) = 1*32 / 34 = 94,1%

NH4Cl:

w(N) = 1 * 14 / 53,5 = 26,2%
w(H) = 4 * 1 / 53,5 = 7,5%

w(Cl) = 1 * 35,5 / 53,5 = 66,4%

Ca(H2PO4)2:

w(Ca) = 1*40 / 234 = 17%

w(H) = 4*1 / 234 = 1,7%

w(P) = 2 * 31 / 234 = 26,5%

w(O) = 8*16 / 234 = 54,7%

CaCl2:

w(Ca) = 1*40 / 111 = 36%

w(Cl) = 2*35,5 / 111 = 64%

Cu2CO3(OH)2

w(Cu) = 2*64 / 222 = 57,65%

w(C) = 1*12 / 222 = 5,4%

w(o) = 3+2 * 16 / 222 = 36%

w(H) = 2 * 1 / 222 = 0,9%

H2SO4:

w(H)= 2*1 / 98 = 2,04%

w(S) = 1*32 / 98 = 32,65%

w(O) = 4*16 / 98 = 65.3%

H2S:

M(H2S)=34

w(H) = 2•1 / 34 = 5,88%

w(S) = 1•32 / 34 = 94,1%

NH4Cl:

M(NH4Cl)=53,5

w(N) = 1•14 / 53,5 = 26,2%
w(H) = 4•1 / 53,5 = 7,5%

w(Cl) = 1•35,5 / 53,5 = 66,4%

Ca(H2PO4)2:

M(Ca(H2PO4)2)=234

w(Ca) = 1•40 / 234 = 17%

w(H) = 4•1 / 234 = 1,7%

w(P) = 2•31 / 234 = 26,5%

w(O) = 8•16 / 234 = 54,7%

CaCl2:

М(CaCl2)=111

w(Ca) = 1•40 / 111 = 36%

w(Cl) = 2•35,5 / 111 = 64%

Cu2CO3(OH)2:

М(Cu2CO3(OH)2)=222

w(Cu) = 2•64 / 222 = 57,65%

w(C) = 1•12 / 222 = 5,4%

w(o) = 3+2•16 / 222 = 36%

w(H) = 2•1 / 222 = 0,9%

H2SO4:

M(H2SO4)=98

w(H)= 2•1 / 98 = 2,04%

w(S) = 1•32 / 98 = 32,65%

w(O) = 4•16 / 98 = 65.3%

ах..да..еще нужно нужно умножать на 100%..