Để đốt cháy hoàn toàn 4,45 g chất hữu cơ X phải dùng hết 4,20 lít khí oxi (đktc). Sản phẩm cháy gồm có 3,15 g H2O và 3,92 lít hỗn hợp khí gồm CO2 và N2 (đktc)

\(\begin{array}{l} {n_{C{O_2}}} = 0,1875{\rm{ }}mol;{\rm{ }}{{\rm{n}}_{{H_2}O}} = 0,175{\rm{ }}mol;{n_{hh{\rm{ khi}}}} = 0,175{\rm{ }}mol\\ \left\{ \begin{array}{l} {n_{C{O_2}}}(a{\rm{ }}mol)\\ {n_{{N_2}}}(b{\rm{ }}mol) \end{array} \right.\\ a + b = 0,175{\rm{ }}(1)\\ {m_{C{O_2}}} + {m_{{N_2}}} + {m_{{H_2}O}} = {m_X} + {m_{{O_2}}}\\ \Rightarrow {m_{C{O_2}}} + {m_{{N_2}}} = {m_X} + {m_{{O_2}}} - {m_{{H_2}O}} \end{array}\)

\(\begin{array}{l} \begin{array}{*{20}{l}} { \Leftrightarrow {\rm{ }}44a{\rm{ }} + {\rm{ }}28b{\rm{ }} = {\rm{ }}4,45{\rm{ }} + 32.{\rm{ }}0,1875{\rm{ }} - {\rm{ }}3,15}\\ { \Leftrightarrow {\rm{ }}44a{\rm{ }} + {\rm{ }}28b{\rm{ }} = {\rm{ }}7,3\;{\rm{ }}\left( 2 \right)}\\ {\left( 1 \right),{\rm{ }}\left( 2 \right){\rm{ }} \Rightarrow {\rm{ }}a{\rm{ }} = {\rm{ }}0,15{\rm{ }}mol;{\rm{ }}b{\rm{ }} = {\rm{ }}0,025{\rm{ }}mol} \end{array}\\ \Rightarrow {n_C} = 0,15mol;{n_{{N_2}}} = 2.0,025 = 0,05mol\\ {n_H} = 2{n_{{H_2}O}} = 2.0,175 = 0,35mol \end{array}\)

\(\begin{array}{l} {m_{O(X)}} = {m_X} - ({m_C} + {m_H} + {m_N})\\ \Rightarrow {m_{O(X)}} = 1,6g \to {n_{O(X)}} = 0,1{\rm{ }}mol\\ \begin{array}{*{20}{l}} {C{\rm{ }}:{\rm{ }}H{\rm{ }}:{\rm{ }}O{\rm{ }}:{\rm{ }}N{\rm{ }} = {\rm{ }}{n_C}:{\rm{ }}{n_H}:{\rm{ }}{n_O}:{\rm{ }}{n_{N\;}} = {\rm{ }}0,15{\rm{ }}:{\rm{ }}0,35{\rm{ }}:{\rm{ }}0,1{\rm{ }}:{\rm{ }}0,05\; = {\rm{ }}3{\rm{ }}:{\rm{ }}7{\rm{ }}:{\rm{ }}2{\rm{ }}:{\rm{ }}1}\\ { \Rightarrow CTPT:{C_3}{H_7}{O_2}N} \end{array} \end{array}\)