gchem emperical formula

[113zami]

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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

 

[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