New answers tagged combustion
0
0.750//0.53732 = 1.396/1;Hydrocarbon, no heteroatoms, needs even number of H so ratio becomes 2.792/2. x4 = 11.18/8 X5= 13.96/10 X6= 16.72/12 best fit is 5times so becomes C14H10. Notice if one rounds off immediately the answer is nice and neat. Don't ever fall into that trap. Next step is MW determination; Rast method or freezing point in benzene or ...
3
Let Hydrocarbon be of form $\ce{C_xH_y}$
Our combustion reaction becomes $\ce{C_xH_y}+(2x+\frac{y}{2})\ce{O_2}\implies x\ce{CO_2}+\frac{y}{2}\ce{H_2O}$
$1$ mole of hydrocarbon produce $x*44g$ of $\ce{CO_2}$
If $\ce{CO_2}$ produced is $33.01$, then $x=\frac{3}{4}$
Similarly $1$ mole of hydrocarbon produces $\frac{y}{2}$ moles of $\ce{H_2O}$
If $\ce{H_2O}$ ...
Top 50 recent answers are included