HELP! As quick possible. Question : A hydrocarbon compound is burnt completely in air to form 7.2cm^3 of CO2 and 7.2 of water vapour. What is the molecular formula of the hydrocarbon compound? [ 24 dm^3 at room temp]
Show working please
7.2 (unit?) of water vapour???
7.2 cm^3
no. of mole of CO2 formed = (7.2/1000) / 24.0 = 0.0004 no. of mole of H2O formed = (7.2/1000)/24.0 = 0.0004 CxHy + (x+y/4)O2 -> xCO2 +(y/2) H2O x = 0.0004 y/2 = 0.004 => y = 0.0008 => CH2?
I think we can only get this but not the exact one, as there is insufficient information
But the answer is C3H6
Is that all you've got for the question? No missing information like the molecular mass of the hydrocarbon?
Nope. thats it. :(
Callisto is right something is meesing with these gata all you can get is CxH2x
My teacher says try solve it using the ratio. But i dont understand how.
it has to be an alkene (CnH2n) coz the ratio of the products are balanced...the ratio of products divided by 2 gives the number of moles of hydrocarbon used...but u need the mass of the hydrocarbon to get the molecular formula?
eg : if u take the complete combustion of C2H4 \[C_{2}H_{4} + O_{2} \rightarrow 2CO_{2} + 2H_{2}O\] number of moles of products equal = means the hydro carbon belongs to the "alkene" group Then u can find the no of moles of hydrocarbon used by divide the number of moles of Product divided by 2 = In your case 0.0003= no of moles of product formed, then u get the no of moles of hydrocarbon used by dividing t with 2 = u get 0.00015 Now the point is u need to know the Mass of hydrocarbon to get the molecular formula \[n= \frac{mass}{molecular formula}\]
I got it now. Sorry because its an objective question. So, i guess we can pick according to the ratio. a. C2H5 b. C3H6 c. C3H8 d. C6H6 thanks everybody. sorry because i didnt include the answer :/ sorry
its B:)
yeap
only B is the compound that C:H is 1:2, as I've calculated before
thts why it is B :P
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