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

Answer:

$PtCl_4$

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:

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

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

$Pt=\frac{0.00522}{0.00522}=1 \\\\Cl=\frac{0.0214}{0.00522} =4$

Thus, the required formula is:

$PtCl_4$

Best regards!