Ta có: \(n_{C_5H_9NO_4}+n_{C_5H_{11}NO_2}=\dfrac{9,125}{36,5}=0,25\left(mol\right)\left(1\right)\)
\(2n_{C_5H_9NO_4}+n_{C_5H_{11}NO_2}=\dfrac{7,7}{22}=0,35\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{C_5H_9NO_4}=0,1\left(mol\right)\\n_{C_5H_{11}NO_2}=0,15\left(mol\right)\end{matrix}\right.\)
⇒ m = 0,1.147 + 0,15.117 = 32,25 (g)