Question

A compound which contains one atom of $X$ and two atoms of $Y$ for each three atoms of $Z$ is made by mixing $5.00\, g$ of $X , 1.15 \times 10^{23}$ atoms of $Y$ and $0.03$ mole of $Z$ atoms. Given that only $4.40 \,g$ of compound results. Calculate the atomic weight of $Y$ if the atomic weight of $X$ and $Z$ are $60$ and $80\, a. m. u$. respectively.

Answer 1

Emprical formula $= XY _{2} Z _{3}$
According to the $Q$.

It is clear that here $Z$ is limiting reagent
hence $X +2 Y + \underset{0.03}{3 Z} \rightarrow \underset{.01\,mole}{XY _{2} Z _{3}}$
According to the condition,
$0.01$ mole $=4.40 \,gm$
$1\,mole = 440\,gm$
given that atomic wt. of $X\, \&\, Z$ are $60\, \&\, 80\, amu$ respectively
hence
$ 60+2 y+3 \times 80=440$
or $ 2 y+60+240=440 $
or $ 2 v=440-300$
or $y=\frac{140}{2}=70$
Atomic wt. of $y =70$ amu.