\(a.m_C=\dfrac{8,8}{44}.12=2,4g\\ m_H=\dfrac{3,6}{18}.2=0,4g\\ m_C+m_H=2,4+0,4=2,8g< m_A\)
Trong A có C,H,O
\(b.m_O=4,4-2,8=1,6g\\ M_A=\dfrac{6,6}{1,68:22,4}=88g/mol\\ CTPT\left(A\right):C_xH_yO_z\\ \dfrac{12x}{2,4}=\dfrac{y}{0,4}=\dfrac{16z}{1,6}=\dfrac{88}{4,4}\\ \Rightarrow x=4;y=8;z=2\\ CTPT\left(A\right):C_4H_8O_2\)