Câu 1 : 

\(CT:P_xO_y\)

Ta có : 

\(\dfrac{m_P}{m_O}=\dfrac{31}{24}\Rightarrow\dfrac{31x}{16y}=\dfrac{31}{24}\)

\(\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\)

\(CT:P_2O_3\)

Câu 2 : 

\(CT:Na_xO_y\)

\(\%Na=\dfrac{23x}{62}\cdot100\%=74.2\%\)

\(\Rightarrow x=2\)

\(M_A=23\cdot2+16\cdot y=62\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow y=1\)

\(CTHH:Na_2O\)

\(m_{Na}=\dfrac{74,19.62}{100}=46\left(g\right)=>n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)

\(m_O=\dfrac{25,81.62}{100}=16\left(g\right)=>n_O=\dfrac{16}{16}=1\left(mol\right)\)

=> CTHH: Na₂O

Gọi CTHH của hợp chất là: \(\left(Na_xO_y\right)_n\)

Ta có: \(x:y=\dfrac{74,19\%}{23}:\dfrac{25,81\%}{16}=3,23:1,61=2:1\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Vậy CTHH của hợp chất là: \(\left(Na_2O\right)_n\)

Theo đề, ta có: \(M_{hợp.chất}=\left(23.2+16\right).n=62\left(\dfrac{g}{mol}\right)\)

\(\Leftrightarrow n=1\)

Vậy CTHH của hợp chất là: Na₂O

\(m_{Na}=\dfrac{62.74,19\%}{100\%}=46\left(g\right)\)

\(n_{Na}=\dfrac{46}{23}=2\left(mol\right)\)

\(m_O=62-46=16\left(g\right)\\ n_O=\dfrac{16}{16}=1\left(mol\right)\\ =>CTHH:Na_2O\)

Câu 1 : 

\(CT:P_xO_y\)

\(\%P=\dfrac{31x}{142}\cdot100\%=43.66\%\)

\(\Rightarrow x=2\)

\(M_B=31\cdot2+16\cdot y=142\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow y=5\)

\(CTPT:P_2O_5\)

Câu 2 : 

\(a.\)

\(M_A=22\cdot2=44\left(\dfrac{g}{mol}\right)\)

\(CT:C_xO_y\)

\(\%C=\dfrac{12x}{44}\cdot100\%=27.3\%\)

\(\Rightarrow x=1\)

\(M_A=12+32\cdot y=44\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow y=2\)

\(CT:CO_2\)

\(m_{Fe}=\%Fe.M_X=28\%.400=112\left(g\right)\\ m_S=\%S.M_X=24\%.400=96\left(g\right)\\ m_O=m_X-m_{Fe}-m_S=400-112-96=192\left(g\right)\)

\(\Rightarrow n_{Fe}=\dfrac{m}{M}=\dfrac{112}{56}=2\left(mol\right)\)

\(m_S=\dfrac{m}{M}=\dfrac{96}{32}=3\left(mol\right)\)

\(m_O=\dfrac{m}{M}=\dfrac{192}{16}=12\left(mol\right)\)

\(CTHH:Fe_2\left(SO_4\right)_3\)

\(m_{Fe}=\dfrac{28.400}{100}=112g\\ m_S=\dfrac{24.400}{100}=96g\\ m_O=400-112-96=192g\\ \Rightarrow n_{Fe}=\dfrac{112}{56}=2mol\\ n_S=\dfrac{96}{32}=3mol\\ n_O=\dfrac{192}{16}=12\\ CTHH:Fe_2S_3O_{12}\)

Bài 2:

\(Đặt.CTTQ.của.A:H_xS_yO_z\left(x,y,z:nguyên,dương\right)\\ Ta.có:\left\{{}\begin{matrix}x=\dfrac{98.2,04\%}{1}=2\\y=\dfrac{98.32,65\%}{32}=1\\z=\dfrac{98.\left(100\%-2,04\%-32,65\%\right)}{16}=4\end{matrix}\right.\\ \Rightarrow x=2;y=1;z=4\\ \Rightarrow CTHH:H_2SO_4\)

Bài 1: Sửa đề 59,2% Al thành 52,9% Al

\(Đặt.CTTQ:Al_xO_y\left(x,y:nguyên,dương\right)\\ x=\dfrac{52,9\%.102}{27}\approx2\\ \Rightarrow y\approx\dfrac{\left(100\%-52,9\%\right).102}{16}\approx3\\ \Rightarrow CTHH:Al_2O_3\)

phần trăm còn lại của oxi là : 100%-28,57%-14,2%=57,23%

\(m_{Mg}=\dfrac{85\cdot28,57}{100}\approx24\left(g\right)\)

\(m_C=\dfrac{85\cdot14,2}{100}\approx12\left(g\right)\)

\(m_O=\dfrac{85\cdot57,23}{100}\approx48\left(g\right)\)

=> \(n_{Mg}=\dfrac{24}{24}=1\left(mol\right);n_C=\dfrac{12}{12}=1\left(mol\right);n_O=\dfrac{48}{16}=3\left(mol\right)\)

=> trong 1 phân tử có 1 nguyên tử Mg, 1 nguyên tử O, 3 nguyên tử Mg

=> CTHH:MgCO3