Рассчитайте массовые доли  ( в%)элементов в веществах H2S NH4CL CO2 Ca(H2PO4)2 CaCL2 Cu2CO3(OH)2 H2SO4

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%..