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wariber [46]
3 years ago
9

Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h

ydrocarbon?
Chemistry
1 answer:
balu736 [363]3 years ago
8 0

CxHy     +  O2    -->    x CO2     +    y/2  H2O

 

Find the moles of CO2 :     18.9g  /  44 g/mol   =    .430 mol CO2   = .430 mol of C in compound

Find the moles of H2O:      5.79g / 18 g/mol     =     .322 mol H2O   = .166 mol of H in compound

 

Find the mass of C and H in the compound:

                             .430mol  x 12  =  5.16 g C

                              .166mol  x 1g   = .166g H   

 

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

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Radio active decay reactions follow first order rate kinetics.

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