Chemistry combustion analysis gas problem

AI Thread Summary
The discussion revolves around determining the molecular and empirical formulas of a hydrocarbon from combustion data. A 0.0150 mol sample produced 1.680L of CO2 and 0.810g of H2O, leading to calculations of moles of CO2 and H2O. The calculated molar ratio suggests a molecular formula of C5H6, but there is confusion about whether it could also be C4H5 or other variations. Participants confirm that C5H6 is correct based on the calculations provided, emphasizing the importance of accurate mole ratio determination. The thread highlights the complexities of deriving molecular formulas from combustion analysis.
Nellen2222
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Homework Statement



Complete combustion of a 0.0150 mol sample of a hydrocarbon, CxHy, gives 1.680L of CO2 at STP and 0.810g of H2O
a) what is the molecular formula
b) what is empirical formula

Homework Equations



pv=nRt

The Attempt at a Solution



a) mol of co2: n=pv/Rt=1 atm * 1.680L/0.0821*273= 0.074955495mol

- Mol of h2o: 0.810g/18.02g/mol * 2 = 0.08990011 mol H2

mol ratio: 0.08990011/0.074955495= 1.199379845... * 5 = 6

molecular formula: c5h6.

Is this wrong? is it possible to multiply by four and round the number to 5 instead of 6? so that it would be c4h5? Or is the above one correct?also: how do i get the molecular formula? i can't get a lowest whole number ratio here.
Thanks!
 
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Looks good to me provided you did the math correctly. Could also be C10H12 or C15H18...
 
Last edited:
Nellen2222 said:
0.0150 mol sample of a hydrocarbon

chemisttree said:
Could also be C10H12 or C15H18...

C5H6 it is, OP started with the number of moles.
 
Argh! That should have been Nellen's reply.
 
Oops.
 
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