рассчитайте массовые доли (в%) элементов в веществах

H2S

NH4CL

CO2

Ca (H2PO4) 2

CaCL2

Cu2CO3 (OH) 2

H2SO4

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% ...

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%