Question

A sample of 0.15g of the compound $\left[Pt\right(N{H}_3{)}_{x}B{r}_y{]}^{z+}.zB{r}^{-}$, is ignited and heated to decomposition, produced 0.0502g of Pt. A second 0.15 g sample was dissolved in water and titrated rapidly with 0.01 M $AgN{O}_3$ solution. 51.50 mL was required to precipitate all the ionic bromide. A third 0.15 g sample was heated for two hours on a steam bath in a solution to which 0.2 mole of $AgN{O}_3$ has been added. This precipitated all the bromide (not just the free ionic $B{r}^{-}$) as AgBr. The weight of the precipitate thus produced was 0.20 g. Find x, y and z :

(Pt = 195, Ag = 108, Br = 80, N = 14 and H = 1)

• A. $x$ = 3, $y$ = 3, $z$ = 4
• B. $x$ = 4, $y$ = 2, $z$ = 2
• C. $x$ = 2, $y$ = 4, $z$ = 4
• D. $x$ = 4, $y$ = 4, $z$ = 2

Answer 1

The correct option is B $x$ = 4, $y$ = 2, $z$ = 2

$\left[Pt\right(N{H}_3{)}_{x}B{r}_y{]}^{z+}.zB{r}^{-}\stackrel{△}{⟶}Pt$

Applying POAC for Pt atoms,
$1\times$ mole of the compound $=1\times$ mole of Pt
$\frac{0.15}{M}=\frac{0.0502}{195}$
$M=582.7$ ...(1)
Where, $M=195+17x+80y+80z$
$582.7=195+17x+80y+80z$
$387.7=17x+80y+80z$ ...(2)

For 0.15g of the second sample containing $zBr$ atoms per molecule (only the ionic bromide), one molecule of the compound shall combine with $z$ molecules of $AgN{O}_3$ to give $z$ molecules of $AgBr$.
$\left(Pt\right(N{H}_3{)}_{x}B{r}_y{)}^{z+}.zB{r}^{-}+zAgN{O}_3⟶zAgBr$

Applying mole ratio method to reactants,
$z\times$ mole of the compound $=1\times$ mole of $AgN{O}_3$
$z\times \frac{0.15}{M}=\frac{0.01\times 51.5}{1000}$
$z\times \frac{0.15}{582.7}=\frac{0.01\times 51.5}{1000}$
$z=2$ ...(3)

For the 0.15g of the third sample, all bromine atoms $(y+z)$ in the compound combine with $AgN{O}_3$ to give $(y+z)$ molecules of $AgBr$.
$\left[Pt\right(N{H}_3)xB{r}_y{]}^{z+}.zB{r}^{-}+(y+z)AgN{O}_3\to (y+z)AgBr$

Applying mole ratio method,
$(y+z)$ $\times$ mole of the compound $=1\times$ mole of AgBr
$(y+z)\times \frac{0.15}{M}=1\times \frac{0.20}{188}$
$(y+2)\times \frac{0.15}{582.7}=1\times \frac{0.20}{188}$
$y=2$ ...(4)
Combining all the equations,
$387.7=17x+80y+80z$
$387.7=17x+80\times 2+80\times 2$
$x=4$
Hence, $x=4,y=2$ and $z=2$.