A student prepared $3.8 \mathrm{g}$ of $\mathrm{Co}_{x}\left(\mathrm{NH}_{3}\right)_{y} \mathrm{Cl}_{x}$. She then analyzed the compound by the procedure in this experiment.
A. In the gravimetric determination of chloride, she weighed out $0.3011 \mathrm{g}$ of the compound. The following data were obtained:
Mass of crucible plus AgCl
Mass of crucible
Mass of AgCl
Mass of $\mathrm{Cl}^{-}$ in $\mathrm{AgCl}$
Moles $\mathrm{Cl}^{-}$ in sample
Moles $\mathrm{Cl}^{-}$ in 100 -g sample
$18.7137 \mathrm{g}$
$18.2316 \mathrm{g}$
______________g MM $\mathrm{AgCl}=143.34 \mathrm{g}$
______________g $\mathrm{MM} \mathrm{Cl}^{-}=35.45 \mathrm{g}$
______________
______________
B. In the colorimetric determination of cobalt, she used a sample weighing $0.4802 \mathrm{g}$. The molarity of cobalt ion in the solution from the volumetric flask was $0.072 \mathrm{M}$.
Moles cobalt ion in 25 mL solution = moles cobalt in sample _____________________
Moles cobalt per 100 -g sample _____________________
In the volumetric determination of ammonia, the sample weighed $1.0014 \mathrm{g}$. In the titration, $0.1000 \mathrm{M}$ HCl was used. She found that $40.00 \mathrm{mL}$ of the $\mathrm{NH}_{3}$ solution required $30.13 \mathrm{mL}$ of the HCl to reach the end point.
No. moles HCl uscd ____________________ = no. moles $\mathrm{NH}_{3}$ in $40 \mathrm{mL} \mathrm{NH}_{3}$ solution
No. moles $\mathrm{NH}_{3}$ in $250 \mathrm{mL} \mathrm{NH}_{3}$ ___________=no. moles $\mathrm{NH}_{3}$ in sample
D. Calculation of the formula of $\operatorname{Co}_{x}\left(\mathrm{NH}_{3}\right)_{y} \mathrm{Cl}_{x}:$
In 100 g of sample,
Dividing by smallest number,
Formula of compound
no. moles Co ion =
no. moles $\mathrm{NH}_{3}=$
no. moles $\mathrm{Cl}^{-}=$
no. moles Co ion
no. moles $\mathrm{NH}_{3}$
no. moles $\mathrm{Cl}^{-}$
No. moles $\mathrm{NH}_{3}$ per 100 -g sample