Em tham khảo nha :
\(\begin{array}{l}
b)\\
CTHH:CuC{l_2}\\
{M_{CuC{l_2}}} = {M_{Cu}} + 2{M_{Cl}} = 64 + 35,5 \times 2 = 135dvC\\
\% Cu = \dfrac{{{M_{Cu}}}}{{{M_{CuC{l_2}}}}} \times 100\% = \dfrac{{64}}{{135}} \times 100\% = 47,4\% \\
\% Cl = 100 - 47,4 = 52,6\% \\
c)\\
CTHH:CaC{O_3}\\
M = {M_{Ca}} + {M_C} + 3{M_O} = 40 + 12 + 16 \times 3 = 100dvC\\
\% Ca = \dfrac{{{M_{Ca}}}}{M} \times 100\% = \dfrac{{40}}{{100}} \times 100\% = 40\% \\
\% C = \dfrac{{{M_C}}}{M} \times 100\% = \dfrac{{12}}{{100}} \times 100\% = 12\% \\
\% O = 100 - 40 - 12 = 48\% \\
f)\\
CTHH:MgS{O_4}\\
M = {M_{Mg}} + {M_S} + 4{M_O} = 24 + 32 + 16 \times 4 = 120dvC\\
\% Mg = \dfrac{{{M_{Mg}}}}{M} \times 100\% = \dfrac{{24}}{{120}} \times 100\% = 20\% \\
\% S = \dfrac{{{M_S}}}{M} \times 100\% = \dfrac{{32}}{{120}} \times 100\% = 26,7\% \\
\% O = 100 - 20 - 26,7 = 53,3\% \\
h)\\
CTHH:AgN{O_3}\\
M = {M_{Ag}} + {M_N} + 3{M_O} = 108 + 14 + 3 \times 16 = 170dvC\\
\% Ag = \dfrac{{{M_{Ag}}}}{M} \times 100\% = \dfrac{{108}}{{170}} \times 100\% = 63,5\% \\
\% N = \dfrac{{{M_N}}}{M} \times 100\% = \dfrac{{14}}{{170}} \times 100\% = 8,2\% \\
\% O = 100 - 63,5 - 8,2 = 28,3\%
\end{array}\)
