Question

A 1.018 g sample pure platinum metal was reacted with HCl to form 1.778 g of a compound containing only platinum and chlorine. Determine the empirical formula of this "Pt-Cl" Compound.

Answer 1

`PtCl_4`

Step-by-step explanation:

Hello!

In this case, since HCl and Pt react according to the following chemical equation:

`HCl+Pt\rightarrow PtCl_x+H_2`

Whereas PtClx is the compound containing Pt and Cl; thus, since 1.018 g out of 1.778 g correspond to Pt and therefore 0.760 g to chlorine, so we determine the empirical formula of this compound by firstly computing the moles of each element:

`n_(Pt)=1.018gPt*(1molPt)/(195.084gPt)=0.00522molPt\\\\\\n_(Cl)=0.760gCl*(1molCl)/(35.45gCl) =0.0214molCl`

Now, we divide the each moles by those of Pt as the fewest ones in order to compute their subscripts in the empirical formula:

`Pt=(0.00522)/(0.00522)=1 \\\\Cl=(0.0214)/(0.00522) =4`

Thus, the required formula is:

`PtCl_4`

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