a)
CTHH: CaHbClc
Có: \(a:b:c=\dfrac{23,8\%}{12}:\dfrac{5,9\%}{1}:\dfrac{70,3\%}{35,5}=1:3:1\)
=> CTHH: (CH3Cl)n
Mà PTKB = 2,805.18 = 50,5 (đvC)
=> n = 1
=> CTHH: CH3Cl
b)
Có: \(\left\{{}\begin{matrix}2p+n=93\\2p=1,657.n\end{matrix}\right.\)
=> p = 29
=> X là Cu