Ta có: \(\%^{26}Mg=100\%-78,6\%-10,9\%=10,5\%\)
`=>` \(\overline{M_{Mg}}=78,6\%.24+10,9\%.25+10,5\%.26=24,319\left(g/mol\right)\)
Giả sử có 1 mol MgX2 \(\Rightarrow\left\{{}\begin{matrix}m_{^{26}Mg}=1.10,5\%.26=2,73\left(g\right)\\m_{MgX_2}=24,319+2M_X\left(g\right)\end{matrix}\right.\)
`=>` \(\dfrac{2,73}{24,319+2M_X}.100\%=1,4811\%\)
`=>` \(M_X=80\left(g/mol\right)\)
`=> X: Brom(Br)`