Question
A 0.509 g sample pure platinum metal was reacted with hcl to form 0.889 g of a compound containing only platinum and chlorine?
Answers
bobpursley
So the product must be of the form PtCl<sub>n</sub>, and the question is what is n, the ratio of the chloride to platinum.
Moles of Pt: .509/atomicmassPt =.509/198=.00257
moles of Cl: (.889-.509)/atomicmassCl=.38/35.45=.0107
dividing each by the lowest, to get the ratio
Cl: .107/.00257=4.2
Pt:; .00257/.00257=1
Empirical formula: PtCl<sub>4</sub>
Moles of Pt: .509/atomicmassPt =.509/198=.00257
moles of Cl: (.889-.509)/atomicmassCl=.38/35.45=.0107
dividing each by the lowest, to get the ratio
Cl: .107/.00257=4.2
Pt:; .00257/.00257=1
Empirical formula: PtCl<sub>4</sub>