- Hợp chất SO₃

%S = M_S : M_SO3 .100% = 32 : 80 .100% = 40%

%O = 100% - 40% = 60%

- Hợp chất Fe₂O₃

%Fe = 2M_Fe : M_Fe2O3 . 100% = 2.56 : 160. 100% = 70%

%O = 100% - 70% = 30%

- Hợp chất CO₂

%C = M_C : M_CO2 . 100% = 12 : 44 . 100% = 27,3 %

%O = 100% - 27,3% = 72,7%

\(\begin{array}{l} *SO_3:\\ \%S=\dfrac{32}{32+16\times 3}\times 100\%=40\%\\ \%O=\dfrac{16\times 3}{32+16\times 3}\times 100\%=60\%\\ *Fe_2O_3:\\ \%Fe=\dfrac{56\times 2}{56\times 2+16\times 3}\times 100\%=70\%\\ \%O=\dfrac{16\times 3}{56\times 2+16\times 3}\times 100\%=30\%\\ *CO_2:\\ \%C=\dfrac{12}{12+16\times 2}\times 100\%=27,27\%\\ \%O=\dfrac{16\times 2}{12+16\times 2}\times 100\%=72,73\%\end{array}\)

`@` `\text {MgO}`

\(\text{PTK = 24 + 16 = 40 < amu>}\)

\(\%\text{O}=\dfrac{16\cdot100}{40}=40\%\)

Vậy, khối lượng `%` của `\text {O}` trong `\text {MgO}` là `40%`

`@` `\text {Fe}_2 \text {O}_3`

\(\text{PTK = }56\cdot2+16\cdot3=160\text{ }< \text{amu}\text{ }>\)

\(\%\text{Fe}=\dfrac{56\cdot2\cdot100}{160}=70\%\)

Vậy, khối lượng `%` của `\text {Fe}` trong `\text {Fe}_2 \text {O}_3` là `70%`

`a,` \(K.L.P.T_{Fe_2O_3}=56.2+16.3=160< amu>.\) 

\(\%Fe=\dfrac{56.2.100}{160}=70\%\)

\(\%O=100\%-70\%=30\%\)

`b,`\(K.L.P.T_{CaCO_3}=40+12+16.3=100< amu>.\)

\(\%Ca=\dfrac{40.100}{100}=40\%\)

\(\%C=\dfrac{12.100}{100}=12\%\) 

\(\%O=100\%-40\%-12\%=48\%\)

`c,` \(K.L.P.T_{HCl}=1+35,5=36,5< amu>.\)

\(\%H=\dfrac{1.100}{36,5}\approx2,74\%\)

\(\%Cl=100\%-2,74\%=97,26\%\)

a: \(\%Fe=\dfrac{56\cdot2}{56\cdot2+16\cdot3}=70\%\)

=>%O=30%

b: \(\%Ca=\dfrac{40}{40+12+16\cdot3}=40\%\)

\(\%C=\dfrac{12}{100}=12\%\)

%O=100%-12%-40%=48%

c: %H=1/36,5=2,74%

=>%Cl=97,26%

\(NaNO_3\\ \%m_{Na}=\dfrac{23}{23+14+3.16}.100\approx27,059\%\\ \%m_N=\dfrac{14}{23+14+3.16}.100\approx16,471\%\\ \%m_O=\dfrac{3.16}{23+14+3.16}.100\approx56,471\%\)

Em tương tự làm cho các chất còn lại!

\(\%Fe=\dfrac{m_{Fe}}{M_{Fe_2O_3}}=\dfrac{112}{160}=70\%\\ \%O=100\%-\%Fe=100\%-70\%=30\%\)

a)

$PTK = 5M_{O_2} = 5.32 = 160$

b)

CTHH của hợp chất : $X_2O_3$

Ta có : 

$2X + 16.3 = 160 \Rightarrow X = 56$

Vậy X là nguyên tố sắt, KHHH : Fe

c)

$\%Fe  = \dfrac{56.2}{160} .100\% = 70\%$
$\%O = 100\% -70\% = 30\%$

\(M_{Fe_2O_3}=56.2+16.3=160\left(DvC\right)\)

\(\%Fe=\dfrac{56.2}{160}.100\%=70\%\\ \%O=100\%-70\%=30\%\)

\(M_{SO_2}=32+16.2=64\left(DvC\right)\)

\(\%S=\dfrac{32}{64}.100\%=50\%\\ \%O=100\%-50\%=50\%\)

\(M_{KHCO_3}=39+1+12+16.3=100\left(DvC\right)\)

\(\%K=\dfrac{39}{100}.100\%=39\%\\ \%H=\dfrac{1}{100}.100\%=1\%\\ \%C=\dfrac{12}{100}.100\%=12\%\\ \%O=100\%-39\%-1\%-12\%=48\%\)

Khối lượng của hợp chất Fe₂O₃ là:

M_{\(Fe_2O_3\)}= 56 . 2 + 16 . 3 = 160

%m_Fe = \(\dfrac{112.100\%}{160}=70\%\)

%m_O = 100% - 70% = 30%

__________________________________________________________________

%m_S = \(\dfrac{32.100\%}{64}=50\%\)

%m_O = 100 % - 50% = 50%

__________________________________________________________________

%m_K = \(\dfrac{39.100\%}{100}=39\%\)

%m_H = \(\dfrac{1.100\%}{100}=1\%\)

%m_C = \(\dfrac{12.100\%}{100}=12\%\)

%m_O = 100% - 39% - 1% - 12% = 48%

\(PTK_{NaNO_3}=23+14+3.16=85\left(đvC\right)\)

\(\%m_{Na}=\dfrac{23}{85}.100=27,05\%\)

\(\%m_N=\dfrac{14}{85}.100=16,47\%\)

\(\%m_O=\dfrac{3.16}{85}=56,47\%\)

\(PTK_{K_2CO_3}=2.39+12+3.16=138\left(đvC\right)\)

\(\%m_K=\dfrac{78}{138}.100=56,52\%\)

\(\%m_C=\dfrac{12}{138}.100=8,69\%\)

\(\%m_O=\dfrac{3.16}{138}.100=34,78\%\)

các ý còn lại làm tương tự