Question

1 oxit của kim loại m . Tìm CTHH M trong 2 trường hợp sau :

a) mM : mO = 9 : 8

b) %mM : %mO = 7 : 3

Answer 1

a) CTHH: M_xO_y

\(\dfrac{m_M}{m_O}=\dfrac{9}{8}\)

=> \(\dfrac{x.M_M}{16y}=\dfrac{9}{8}=>M_X=\dfrac{18y}{x}=9.\dfrac{2y}{x}\left(g/mol\right)\)

Xét \(\dfrac{2y}{x}=3=>M_M=27\left(Al\right)=>\dfrac{x}{y}=\dfrac{2}{3}=>CTHH:Al_2O_3\)

b) CTHH: M_xO_y

\(\dfrac{\%m_M}{\%m_O}=\dfrac{m_M}{m_O}=\dfrac{7}{3}\)

=> \(\dfrac{x.M_M}{16y}=\dfrac{7}{3}\)

=> \(M_M=\dfrac{56}{3}.\dfrac{2y}{x}\left(g/mol\right)\)

Xét \(\dfrac{2y}{x}=3=>M_M=56\left(Fe\right)=>\dfrac{x}{y}=\dfrac{2}{3}=>CTHH:Fe_2O_3\)