Question

A chemist dissolve 1.406g pure platinum (Pt) in an excess of a mixture of hydrochloric and nitric acid and then;- after a series of subsequent step involving several other chemicals, isolates a compound of molecular formula ${\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$. Determine the maximum possible yields of the compound.

Short answer

The maximum yield of ${\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$is 2.582g.

Step-by-step solution

Calculating molar mass of  Pt2C10H18N2S2O6                                       

Given mass of pure platinum (Pt) = 1.406g

We know that molar masses of Pt, C, O, H, N, S are 195.08g,12.01g,16.00g ,1.0079g ,14.01g and 32.07g respectively.

${\text{Molar mass of Pt}}_{\text{2}}{\text{C}}_{\text{10}}{\text{H}}_{\text{18}}{\text{N}}_{\text{2}}{\text{S}}_{\text{2}}{\text{O}}_{\text{6}}$

$=\left(2\times 195.08+10\times 12.01+18\times 1.0079+2\times 14.01+2\times 32.07+6\times 16.00\right)$g

$$\begin{gathered} =\left(390.16+120.01+18.1422+28.02+64.14+96\right)g \\ =716.56g \end{gathered}$$

Calculating the maximum possible yield of Pt2C10H18N2S2O6

1.406g of pure platinum in excess of hydrochloric acid and nitric acid yields a compound, ${\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$

The entire 1.406g of Pt is used to form${\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$_.Then only the yield will be maximum.

The maximum mass of ${\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$ formed from 1.406gm Pt is:

$=1.406g.Pt\times \frac{1mol.Pt}{196.08g.Pt}\times \frac{1mol.{\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}}{2mol.Pt}\times \frac{716.56gg.{\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}}{1mol.{\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}}$

$=2.528g.{\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$

Therefore, the maximum yield of ${\mathbf{Pt}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{10}}{\mathbf{H}}_{\mathbf{18}}{\mathbf{N}}_{\mathbf{2}}{\mathbf{S}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{6}}$is 2.582g.