Solve Molecular Formula of Silicon-Fluorine Compound

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SUMMARY

The molecular formula of the newly discovered silicon-fluorine compound is Si2F6. Analysis reveals that the compound consists of 33.01 mass % silicon and 66.99 mass % fluorine. Using the ideal gas law (PV=nRT), the molecular weight was calculated to be 170.7 g/mol. The empirical formula was determined to be SiF3, leading to the conclusion that the molecular formula is derived by multiplying the empirical formula by 2.

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  • Basic chemistry concepts regarding mass percent composition
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Analysis of a newly discovered gaseous silicon-fluorine compound shows that it contains 33.01 mass % silicon. At 27oC 2.60g of the compound exerts a pressure of 1.50 atm in a 0.250-L vessel. What is the molecular formula of the compound?
I guess PV=nRT needs to be used. P,V,R, and T are given, i guess the only thing to sove for is n. That gives me the total moles of silicon fluorine. But i don't know where to go after that.

Thanks
 
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First of all find molecular wt of comp

now find empirical formula

n= mol. wt / empirical formula

and hence the formula for compound as (Si_xF_y)_n
 
Does this look right??

m=RT/PV = ((2.60)(.0821)(300))/((1.50)(.250)) = 170.7
Mol wt=170.7

33.01 %Si -100 = 66.99%F

33.01g Si and 66.99gF

33.01gSi(1mol/28.09gSi)=1.17mol
66.99gF(1mol/19.00gF)=3.52mol
Si1.17F3.52
emp formula = SiF3
Mol wt/emp wt = 170.7/85 = 2
mol formula = Si2F6

Is this correct.
 
Yes it is
 

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