\(CaCO_3\\ \%m_{Ca}=\dfrac{40}{40+12+3.16}.100=40\%\\ \%m_C=\dfrac{12}{40+12+16.3}.100=12\%\\ \Rightarrow\%m_O=100\%-\left(40\%+12\%\right)=48\%\\ H_2SO_4\\ \%m_H=\dfrac{2.1}{2.1+32+4.16}.100\approx2,041\%\\ \%m_S=\dfrac{32}{2.1+32+4.16}.100\approx32,653\%\\ \%m_O=\dfrac{4.16}{2.1+32+4.16}.100\approx65,306\%\\ Fe_2O_3\\ \%m_{Fe}=\dfrac{56.2}{56.2+16.3}.100=70\%\\ \Rightarrow\%m_O=100\%-70\%=30\%\)

CaCO₃
\(\%M_{\dfrac{Ca}{CaCO_3}}=\dfrac{40}{100}.100\%=40\%\)
\(\%M_{\dfrac{C}{CaCO_3}}=\dfrac{12}{100}.100\%=12\%\)
\(\%M_{\dfrac{O}{CaCO_3}}=100\%-\left(40\%+12\%\right)=48\%\)
H₂SO₄
\(\%M_{\dfrac{H_2}{H_2SO_4}}=\dfrac{2}{98}.100\%=2,04\%\)
\(\%M_{\dfrac{S}{H_2SO_4}}=\dfrac{32}{98}.100\%=32,65\%\)
\(\%M_{\dfrac{O}{H_2SO_4}}=100\%-\left(2,04\%+32,65\%\right)=65,31\%\)
Fe₂O3
\(\%M_{\dfrac{Fe}{Fe_2O_3}}=\dfrac{112}{160}.100\%=70\%\)
\(\%M_{\dfrac{O}{Fe_2O_3}}=100\%-70\%=30\%\)
 

\(M_{NaOH}=40(g/mol)\\ \%_{Na}=\dfrac{23}{40}.100\%=57,5\%\\ \%_O=\dfrac{16}{40}.100\%=40\%\\ \%_H=\dfrac{1}{40}.100\%=2,5\%\\ M_{H_2CO_3}=2+12+16.3=62(g/mol)\\ \%_H=\dfrac{2}{62}.100\%=3,23\%\\ \%_C=\dfrac{12}{62}.100\%=19,35\%\\ \%_O=100\%-3,23\%-19,35\%=77,42\%\)

\(M_{CaCO_3}=40+12+16.3=100(g/mol)\\ \%_{Ca}=\dfrac{40}{100}.100\%=40\%\\ \%_C=\dfrac{12}{100}.100\%=12\%\\ \%_O=\dfrac{48}{100}.100\%=48\%\\ M_{KNO_3}=39+14+16.3=101(g/mol)\\ \%_K=\dfrac{39}{101}.100\%=38,61\%\\ \%_N=\dfrac{14}{101}.100\%=13,86\%\\ \%_O=100\%-38,61\%-13,86\%=47,53\%\)

\(a.CTHH:K_2CO_3:\\ \%K=\dfrac{78}{138}=56,52\%\\ \%C=\dfrac{12}{138}=8,69\%\\ \%O=100\%-56,52\%-8,69\%=34,79\%\)

\(b.CTHH:H_2SO_4:\\ \%H=\dfrac{2}{98}=2,04\%\\ \%S=\dfrac{32}{98}=32,65\%\\\%O=100\%-2,04\%-32,65\%=65,31\% \)

Trong $CO_2$ : $\%O = \dfrac{16.2}{44}.100\% = 72,73\%$

Trong $Al_2O_3$ : $\%O = \dfrac{16.3}{102}.100\% = 47,06\%$

Suy ra:  $\%O : CO_2 > Al_2O_3$

\(n_{CO_2}=\dfrac{5.28}{44}=0.12\left(mol\right)\Rightarrow n_C=0.12\left(mol\right)\Rightarrow m_C=0.12\cdot12=1.44\left(g\right)\)

\(n_{H_2O}=\dfrac{0.9}{18}=0.05\left(mol\right)\Rightarrow n_H=0.05\cdot2=0.1\left(mol\right)\)

\(n_{N_2}=\dfrac{0.224}{22.4}=0.01\left(mol\right)\Rightarrow n_N=0.02\left(mol\right)\Rightarrow m_N=0.02\cdot14=0.28\left(g\right)\)

\(m_O=m_A-m_C-m_H-m_N=2.46-1.44-0.1-0.28=0.64\left(g\right)\)

\(\%C=\dfrac{1.44}{5.28}\cdot100\%=27.27\%\)

\(\%H=\dfrac{0.1}{5.28}\cdot100\%=1.89\%\)

\(\%N=\dfrac{0.28}{5.28}\cdot100\%=5.3\%\)

\(\%O=65.54\%\)

a) \(M_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(DvC\right)\)

\(\%Ca=\dfrac{40.1}{74}.100\%=54\%\)

\(\%O=\dfrac{16.2}{74}.100\%=43\%\)

\(\%H=100\%-54\%-43\%=3\%\)

a) \(\left\{{}\begin{matrix}\%m_{Ca}=\dfrac{40.1}{74}.100\%=54,054\%\\\%m_O=\dfrac{16.2}{74}.100\%=43,243\%\\\%m_H=\dfrac{2.1}{74}.100\%=2,703\%\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\%m_{Ba}=\dfrac{137.1}{208}.100\%=65,865\%\\\%Cl=\dfrac{35,5.2}{208}.100\%=34,135\%\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}\%m_K=\dfrac{39.1}{56}.100\%=69,643\%\\\%m_O=\dfrac{16.1}{56}.100\%=28,571\%\\\%m_H=\dfrac{1.1}{56}.100\%=1,786\%\end{matrix}\right.\)

d) \(\left\{{}\begin{matrix}\%m_{Al}=\dfrac{27.2}{102}.100\%=52,94\%\\\%m_O=\dfrac{16.3}{102}.100\%=47,06\%\end{matrix}\right.\)

e) \(\left\{{}\begin{matrix}\%m_{Na}=\dfrac{23.2}{106}.100\%=43,396\%\\\%m_C=\dfrac{12}{106}.100\%=11,321\%\\\%m_O=\dfrac{16.3}{106}.100\%=45,283\%\end{matrix}\right.\)

g) \(\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{56.1}{72}.100\%=77,78\%\\\%m_O=\dfrac{16.1}{72}.100\%=22,22\%\end{matrix}\right.\)

h) \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{65.1}{161}.100\%=40,373\%\\\%m_S=\dfrac{32.1}{161}.100\%=19,876\%\\\%m_O=\dfrac{16.4}{161}.100\%=39,751\%\end{matrix}\right.\)

i) \(\left\{{}\begin{matrix}\%m_{Hg}=\dfrac{201.1}{217}.100\%=92,627\%\\\%m_O=\dfrac{16}{217}.100\%=7,373\%\end{matrix}\right.\)

k) \(\%m_{Na}=\dfrac{23.1}{85}.100\%=27,06\%;\%m_N=\dfrac{14.1}{85}.100\%=16,47\%\%;\%m_O=\dfrac{16.3}{85}.100\%=56,47\%\)

Gọi CTPT của hợp chất hữu cơ là CxHyOz (x, y, z nguyên dương; z ≥ 0)

Ta có:

BT nguyên tố ⇒ nC = nCO2 = 0,03 mol ⇒ mC = 12. 0,03 = 0,36g

BT nguyên tố ⇒ nH = 2.nH2O = 2. 0,04 = 0,08 mol ⇒ mH = 0,08. 1 = 0,08 g

mO = 0,6 - 0,36 - 0,08 = 0,16(g)

⇒ Hợp chất A có chứa C, H, O

Phần trăm khối lượng các nguyên tố trong phân tử A là:

Câu 1:

a) Al₂O₃:

Phần trăm Al trong Al₂O₃:   \(\%Al=\dfrac{27.2}{27.2+16.3}.100=52,94\%\)

Phần trăm O trong Al₂O₃:   \(\%O=100-52,94=47,06\%\)

b) C₆H₁₂O:

Phần trăm C trong  C₆H₁₂O:  \(\%C=\dfrac{12.6}{12.6+12+16}.100=72\%\)

Phần trăm H trong  C₆H₁₂O:  \(\%H=\dfrac{1.12}{12.6+12+16}.100=12\%\)

Phần trăm O trong  C₆H₁₂O :  \(\%O=100-72-12=16\%\)

Câu 2: 

\(m_H=\dfrac{5,88.34}{100}\approx2\left(g\right)\)

\(m_S=\dfrac{94,12.34}{100}=32\left(g\right)\)

\(n_H=\dfrac{m}{M}=\dfrac{2}{1}=2\left(mol\right)\)

\(n_S=\dfrac{m}{M}=\dfrac{32}{32}=1\left(mol\right)\)

⇒ CTHH của hợp chất:  H₂S

\(PTK_{CaCO_3}=NTK_{Ca}+NTK_C+3.NTK_O=40+12+3.16=100\left(đ.v.C\right)\\ \%m_{Ca}=\dfrac{NTK_{Ca}}{PTK_{CaCO_3}}.100\%=\dfrac{40}{100}.100=40\%\\ \%m_C=\dfrac{NTK_C}{PTK_{CaCO_3}}.100\%=\dfrac{12}{100}.100=12\%\\ \%m_O=100\%-\left(\%m_{Ca}+\%m_C\right)=100\%-\left(40\%+12\%\right)=48\%\)