Cách 1: \(M_E=\dfrac{18}{0,3}=60\left(g/mol\right)\)
BTNT C: \(n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)
Ta có: \(m_{tăng}=m_{CO_2}+m_{H_2O}=20,4\left(g\right)\)
`=>` \(n_H=2.n_{H_2O}=2.\dfrac{20,4-0,3.44}{18}=0,8\left(mol\right)\)
`=>` \(n_{O\left(E\right)}=\dfrac{6-0,8-0,3.12}{16}=0,1\left(mol\right)\)
Ta có: \(n_C:n_H:n_O=0,3:0,8:0,1=3:8:1\)
`=>` CTPT của E có dạng \(\left(C_3H_8O\right)_n\)
`=>` \(n=\dfrac{60}{60}=1\)
`=> E` là `C_3H_8O`
Cách 2: \(M_E=\dfrac{18}{0,3}=60\left(g/mol\right)\)
`=>` \(n_E=\dfrac{6}{60}=0,1\left(mol\right)\)
BTNT C: \(n_C=n_{CO_2}=n_{CaCO_3}=\dfrac{30}{100}=0,3\left(mol\right)\)
Ta có: \(m_{tăng}=m_{CO_2}+m_{H_2O}=20,4\left(g\right)\)
`=>` \(n_H=2.n_{H_2O}=2.\dfrac{20,4-0,3.44}{18}=0,8\left(mol\right)\)
`=>` \(n_{O\left(E\right)}=\dfrac{6-0,8-0,3.12}{16}=0,1\left(mol\right)\)
Trong 1 mol E có: \(\left\{{}\begin{matrix}n_C=\dfrac{0,3}{0,1}=3\left(mol\right)\\n_H=\dfrac{0,8}{0,1}=8\left(mol\right)\\n_O=\dfrac{0,1}{0,1}=1\left(mol\right)\end{matrix}\right.\)
`=>` CTPT của E là `C_3H_8O`
