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 :

sebassandin

Answer:

[tex]PtCl_4[/tex]

Explanation:

Hello!

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

[tex]HCl+Pt\rightarrow PtCl_x+H_2[/tex]

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:

[tex]n_{Pt}=1.018gPt*\frac{1molPt}{195.084gPt}=0.00522molPt\\\\\\n_{Cl}=0.760gCl*\frac{1molCl}{35.45gCl} =0.0214molCl[/tex]

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

[tex]Pt=\frac{0.00522}{0.00522}=1 \\\\Cl=\frac{0.0214}{0.00522} =4[/tex]

Thus, the required formula is:

[tex]PtCl_4[/tex]

Best regards!

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