Q23.136CP - Chemistry: Molecular Nature Of Matter And Change

Question

Werner prepared two compounds by heating a solution of  ${\mathrm{PtCl}}_2$ with triethyl phosphine, $\mathrm{P}{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_3$ which is an excellent ligand for $\mathrm{Pt}$. The two compounds gave the same analysis: $\mathrm{Pt}$,   $38.8\%$, $\mathrm{Cl},14.1\%$, $\mathrm{C},28.7\%$, $\mathrm{P},12.4\%$and $\mathrm{H},6.02\%$ . Write formulas, structures, and systematic names for the two isomers.

 

Step-by-Step Solution

Answer

 Formula : ${\mathrm{PtCl}}_2\mathrm{P}{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_3$

names : cis/trans -dicholorobis(tri-ethylphosphine)platinum(II)

Step 1: Given information

Assume that the sample is $100.0$ grams. Then, the mass percent given will easily be converted to grams.

Elements	Mass (g)
Pt	38.8
Cl	14.1
C	28.7
P	12.4
H	6.02

Step 2: Molecular formula

An expression which states the quantity and kind of atoms present in a very molecule of a substance.

Step 3: Convert the mass to moles to get the amount in moles

Convert the mass to moles to get the amount in moles of each atom present.

$\begin{array}{rcl}\mathrm{molPt}& =& 38.8\mathrm{g}\times \frac{1\mathrm{mol}}{195.08\mathrm{g}}\\ & =& 0.19889\mathrm{molCl}\\ & =& 14.1\mathrm{g}\times \frac{1\mathrm{mol}}{35.45\mathrm{g}}\\ & =& 0.3977433\mathrm{molC}\end{array}$

Solve further

$\begin{array}{rcl}\mathrm{molPt}& =& 28.7\mathrm{g}\times \frac{1\mathrm{mol}}{12.01\mathrm{g}}\\ & =& 2.389675\mathrm{molP}\\ & =& 12.4\mathrm{g}\times \frac{1\mathrm{mol}}{30.974\mathrm{g}}\\ & =& 0.40425\mathrm{molH}\end{array}$

Solve further

$\begin{array}{rcl}\mathrm{molPt}& =& 6.02\mathrm{g}\times \frac{1\mathrm{mol}}{1.008\mathrm{g}}\\ & =& 5.9722\end{array}$

Step 4: Number of atoms present in the molecule

Divide each molar amount by the smallest amount in moles  Pt) to get the number of atoms present in the molecule.

Elements	Amount mol	Number of atoms
Pt	0.19889	1
Cl	0.3977433	2
C	2.389675	12
P	0.40425	2
H	5.9722	30

Then, you can get the molecular formula.

Molecular Formula: ${\mathrm{PtCl}}_2{\mathrm{P}}_2{\mathrm{C}}_{12}{\mathrm{H}}_{30}$

Considering that one of the ligands is $\mathrm{P}{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_3$  , the formula is:

${\mathrm{PtCl}}_2{\left[\mathrm{P}{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_3\right]}_2$