Answer :
Answer:
[tex]\rm CH_2[/tex].
Explanation:
Carbon and hydrogen are the only two elements in a hydrocarbon. When a hydrocarbon combusts completely in excess oxygen, the products would be [tex]\rm CO_2[/tex] and [tex]\rm H_2O[/tex]. The [tex]\rm C[/tex] and [tex]\rm H[/tex] would come from the hydrocarbon, while the [tex]\rm O[/tex] atoms would come from oxygen.
Look up the relative atomic mass of these three elements on a modern periodic table:
- [tex]\rm C[/tex]: [tex]12.011[/tex].
- [tex]\rm H[/tex]: [tex]1.008[/tex].
- [tex]\rm O[/tex]: [tex]\rm 15.999[/tex].
Calculate the molar mass of [tex]\rm CO_2[/tex] and [tex]\rm H_2O[/tex]:
[tex]M(\mathrm{CO_2}) = 12.011 + 2 \times 15.999 = 44.009\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\mathrm{H_2O}) = 2 \times 1.008 + 15.999 = 18.015\; \rm g \cdot mol^{-1}[/tex]
Calculate the number of moles of [tex]\rm CO_2[/tex] molecules in [tex]33.01\; \rm g[/tex] of [tex]\rm CO_2\![/tex]:
[tex]\displaystyle n(\mathrm{CO_2}) = \frac{m(\mathrm{CO_2})}{M(\mathrm{CO2})} = \frac{33.01\; \rm g}{44.009\; \rm g\cdot mol^{-1}} \approx 0.7501\; \rm mol[/tex].
Similarly, calculate the number of moles of [tex]\rm H_2O[/tex] molecules in [tex]13.51\; \rm g[/tex] of [tex]\rm H_2O\![/tex]:
[tex]\displaystyle n(\mathrm{H_2O}) = \frac{m(\mathrm{H_2O})}{M(\mathrm{H_2O})} = \frac{13.51\; \rm g}{18.015\; \rm g\cdot mol^{-1}} \approx 0.7499\; \rm mol[/tex].
Note that there is one carbon atom in every [tex]\rm CO_2[/tex] molecule. Approximately[tex]0.7501\; \rm mol[/tex] of [tex]\rm CO_2\![/tex] molecules would correspond to the same number of [tex]\rm C[/tex] atoms. That is: [tex]n(\mathrm{C}) \approx 0.7501\; \rm mol[/tex].
On the other hand, there are two hydrogen atoms in every [tex]\rm H_2O[/tex] molecule. approximately [tex]0.7499\; \rm mol[/tex] of [tex]\rm H_2O[/tex] molecules would correspond to twice as many [tex]\rm H\![/tex] atoms. That is: [tex]n(\mathrm{H}) \approx 2 \times 0.7499 \; \rm mol\approx 1.500\; \rm mol[/tex].
The ratio between the two is: [tex]n(\mathrm{C}): n(\mathrm{H}) \approx 1:2[/tex].
The empirical formula of a compound gives the smallest whole-number ratio between the elements. For this hydrocarbon, the empirical formula would be [tex]\rm CH_2[/tex].