Giải thích các bước giải:

$\begin{array}{l} M_{CaCO_3}=40+12+16\times 3=100\ (đvC)\\ \%Ca=\dfrac{40}{100}\times 100\%=40\%\\ \%C=\dfrac{12}{100}\times 100\%=12\%\\ \%O=\dfrac{16\times 3}{100}\times 100\%=48\%\\ \\ M_{CuSO_4}=64+32+16\times 4=160\ (đvC)\\ \%Cu=\dfrac{64}{160}\times 100\%=40\%\\ \%S=\dfrac{32}{160}\times 100\%=20\%\\ \%O=\dfrac{16\times 4}{160}\times 100\%=40\%\\ \\ M_{NH_4NO_3}=14\times 2+1\times 4+16\times 3=80\ (đvC)\\ \%N=\dfrac{14\times 2}{80}\times 100\%=35\%\\ \%H=\dfrac{4}{80}\times 100\%=5\%\\ \%O=\dfrac{16\times 3}{80}\times 100\%=60\%\\ \\ M_{SO_3}=32+16\times 3=80\ (đvC)\\ \%S=\dfrac{32}{80}\times 100\%=40\%\\ \%O=\dfrac{16\times 3}{80}\times 100\%=60\%\\ \\ M_{P_2O_5}=31\times 2+16\times 5=142\ (đvC)\\ \%P=\dfrac{31\times 2}{142}\times 100\%=43,66\%\\ \%O=\dfrac{16\times 5}{142}\times 100\%=56,34\%\\ \\ M_{SO(NH_2)_2}=32+16+14\times 2+1\times 4=80\ (đvC)\\ \%S=\dfrac{32}{80}\times 100\%=40\%\\ \%O=\dfrac{16}{80}\times 100\%=20\%\\ \%N=\dfrac{14\times 2}{80}\times 100\%=35\%\\ \%H=\dfrac{1\times 4}{80}\times 100\%=5\%\\ \\ M_{(NH_4)_2SO_4}=14\times 2+1\times 8+32+16\times 4=132\ (đvC)\\ \%N=\dfrac{14\times 2}{132}\times 100\%=21,21\%\\ \%H=\dfrac{1\times 8}{132}\times 100\%=6,06\%\\ \%S=\dfrac{32}{132}\times 100\%=24,24\%\\ \%O=\dfrac{16\times 4}{132}\times 100\%=48,49\%\\ \\ M_{C_{12}H_{22}O_{11}}=12\times 12+1\times 22+16\times 11=342\ (đvC)\\ \%C=\dfrac{12\times 12}{342}\times 100\%=42,2\%\\ \%H=\dfrac{22}{342}\times 100\%=6,43\%\\ \%O=\dfrac{16\times 11}{342}\times 100\%=51,46\%\\ \\ M_{Al_2(SO_4)_3}=27\times 2+32\times 3+16\times 12=342\ (đvC)\\ ⇒\%Al=\dfrac{27\times 2}{342}\times 100\%=15,8\%\\ \%S=\dfrac{32\times 3}{342}=28,1\%\\ \%O=\dfrac{16\times 12}{342}\times 100\%=56,1\%\end{array}$

Giải thích các bước giải:

(Lưu ý: Chỗ \(H_{22}O_{11}\) bạn sửa thành \(C_{12}H_{22}O_{11}\) nha)

\(\begin{array}{l} M_{CaCO_3}=40+12+16\times 3=100\ (đvC)\\ \%Ca=\dfrac{40}{100}\times 100\%=40\%\\ \%C=\dfrac{12}{100}\times 100\%=12\%\\ \%O=\dfrac{16\times 3}{100}\times 100\%=48\%\\ \\ M_{CuSO_4}=64+32+16\times 4=160\ (đvC)\\ \%Cu=\dfrac{64}{160}\times 100\%=40\%\\ \%S=\dfrac{32}{160}\times 100\%=20\%\\ \%O=\dfrac{16\times 4}{160}\times 100\%=40\%\\ \\ M_{NH_4NO_3}=14\times 2+1\times 4+16\times 3=80\ (đvC)\\ \%N=\dfrac{14\times 2}{80}\times 100\%=35\%\\ \%H=\dfrac{4}{80}\times 100\%=5\%\\ \%O=\dfrac{16\times 3}{80}\times 100\%=60\%\\ \\ M_{SO_3}=32+16\times 3=80\ (đvC)\\ \%S=\dfrac{32}{80}\times 100\%=40\%\\ \%O=\dfrac{16\times 3}{80}\times 100\%=60\%\\ \\ M_{P_2O_5}=31\times 2+16\times 5=142\ (đvC)\\ \%P=\dfrac{31\times 2}{142}\times 100\%=43,66\%\\ \%O=\dfrac{16\times 5}{142}\times 100\%=56,34\%\\ \\ M_{SO(NH_2)_2}=32+16+14\times 2+1\times 4=80\ (đvC)\\ \%S=\dfrac{32}{80}\times 100\%=40\%\\ \%O=\dfrac{16}{80}\times 100\%=20\%\\ \%N=\dfrac{14\times 2}{80}\times 100\%=35\%\\ \%H=\dfrac{1\times 4}{80}\times 100\%=5\%\\ \\ M_{(NH_4)_2SO_4}=14\times 2+1\times 8+32+16\times 4=132\ (đvC)\\ \%N=\dfrac{14\times 2}{132}\times 100\%=21,21\%\\ \%H=\dfrac{1\times 8}{132}\times 100\%=6,06\%\\ \%S=\dfrac{32}{132}\times 100\%=24,24\%\\ \%O=\dfrac{16\times 4}{132}\times 100\%=48,49\%\\ \\ M_{C_{12}H_{22}O_{11}}=12\times 12+1\times 22+16\times 11=342\ (đvC)\\ \%C=\dfrac{12\times 12}{342}\times 100\%=42,2\%\\ \%H=\dfrac{22}{342}\times 100\%=6,43\%\\ \%O=\dfrac{16\times 11}{342}\times 100\%=51,46\%\\ \\ M_{Al_2(SO_4)_3}=27\times 2+32\times 3+16\times 12=342\ (đvC)\\ ⇒\%Al=\dfrac{27\times 2}{342}\times 100\%=15,8\%\\ \%S=\dfrac{32\times 3}{342}=28,1\%\\ \%O=\dfrac{16\times 12}{342}\times 100\%=56,1\%\end{array}\)

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