gchem emperical formula

[113zami]

combustion of a 50 gram sample of an unknown hydrocarbon yields 132 grams of CO2 and 126 grams of H2O. what is the emperical formula of the original sample hydrocarbon?

please explain your answer, thanks for the help

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[jdent]

Well Id start wit balancing the equation by converting the grams to moles. My balanced products is 3 CO2 and 7 H2O. Thats step 1. The next step is to balance it to the first part of the equation. I got the hydrocarbon to have a empirical formula of C3H14. However it looks wrong because it is not the correct formula for a hydrocarbon. Someone lease comment to help. Thanks.

 

[mddang]

50g hydrocarbon + x grams oxygen ->> 132gCO2 + 126gH2O


We have 158 g of O2 reacting 50g of hydrocarbon, yielding 132gCO2 + 126gH2O.

We will use percent mass composition orto determine how much carbon and hydrogen we started with.

Molar mass of CO2 is 12+16+16 = 32g/mol
Percent mass of C in CO2 = 12/32
132g*12/(32+12) = 36g of carbon.


Molar mass of H2O = 18g/mol
Percent mass of H in H2) = 2/18
126g*2/18 = 14g of hydrogen.

Divide each of these amounts by their respective molar masses..

C: 36g/(3g/mol) = 3mol
H: 14g/(1g/mol) = 14mol

C3H14


Same thing as jdent... Bad problem maybe?

 

[Mstoothlady2012]

132 g CO2 x 1mol/44g = 3 mol CO2

126 g H2O x 1 mol/18g = 7 mol H2O

C3H7

2C3H7 + 6O2 ---> 6CO2 + 7H2O

 

[jdent]

What was your last step to make it C3H7? Please explain it out!

 

[Mstoothlady2012]

What was your last step to make it C3H7? Please explain it out!

3 mol of CO2 has 3 mol of C & 6 mol of O

7 mol of H2O has 14 mol of H & 7 mol of O <---- aahh I see your point!!!

I guess it is C3H14...hmmmm

 

[mddang]

132 g CO2 x 1mol/44g = 3 mol CO2

126 g H2O x 1 mol/18g = 7 mol H2O

C3H7

2C3H7 + 6O2 ---> 6CO2 + 7H2O

An odd number of H isn't possible for a hydrocarbon.

 

[jdent]

Now that we agree like I originally posted that it is C3H14. What is this molecule? Something is still fishy?!?

 

[Mstoothlady2012]

So the answer is.....................

 

[DentalDeac]

Now that we agree like I originally posted that it is C3H14. What is this molecule? Something is still fishy?!?

Yeah, C3H14 is not possible...how would this work?

 

[dleodyddlek]

C3H8 + 6.5(O2) + 3(H2) -> 3(CO2) + 7(H2O)








lol j/k


seriously, what the hell can C3H14 give? it's impossible