Hydrocarbon-contain-80-by-mass-of-carbon-and-the-rest-is-for-hydrogen-calculate-the-empirical-formula-of-the-compound-given-H-1-C-12-

Question Number 178011 by Spillover last updated on 12/Oct/22

$$\mathrm{Hydrocarbon}\:\mathrm{contain}\:\mathrm{80\%}\:\mathrm{by} \\ $$$$\mathrm{mass}\:\mathrm{of}\:\mathrm{carbon}\:\mathrm{and}\:\mathrm{the}\:\mathrm{rest} \\ $$$$\mathrm{is}\:\mathrm{for}\:\mathrm{hydrogen}.\mathrm{calculate}\:\mathrm{the} \\ $$$$\mathrm{empirical}\:\mathrm{formula}\:\mathrm{of}\:\mathrm{the}\:\mathrm{compound} \\ $$$$\left[\mathrm{given}\:\mathrm{H}=\mathrm{1}\:\:\mathrm{C}=\mathrm{12}\right] \\ $$

Answered by a.lgnaoui last updated on 12/Oct/22

the composant C_x H_y masse carbine(M_c = 12x →y(fir hudrigene) masse titaleM=12x+y Mc=(4/5)M=(4/5)(12x+y) wuth C_n H_(2n) (alcane)[x=2y] m(csebine=12→4fir hydrogene(exclu−:m(carbone=3×masse hydrigene) so the composant is: Alcene C_n H_(2n+2) (with n>1) n=2 C_2 H_6 (masse carbone =24→6for hydrogene) masse carbone 24=4×masse hydrogene) masse totale=30 → (((24)/(30))=(4/5) )=80% M(satisfait) so the composant is C_2 H_6

$${the}\:{composant}\:{C}_{{x}} {H}_{{y}} \\ $$$${masse}\:{carbine}\left(\mathrm{M}_{{c}} =\:\mathrm{12}{x}\:\:\:\:\rightarrow{y}\left({fir}\:{hudrigene}\right)\right. \\ $$$${masse}\:{titale}\mathrm{M}=\mathrm{12}{x}+{y} \\ $$$$\mathrm{M}{c}=\frac{\mathrm{4}}{\mathrm{5}}\mathrm{M}=\frac{\mathrm{4}}{\mathrm{5}}\left(\mathrm{12}{x}+{y}\right)\:\: \\ $$$${wuth}\:\:\:{C}_{{n}} {H}_{\mathrm{2}{n}} \left({alcane}\right)\left[{x}=\mathrm{2}{y}\right] \\ $$$${m}\left({csebine}=\mathrm{12}\rightarrow\mathrm{4}{fir}\:{hydrogene}\left({exclu}−:{m}\left({carbone}=\mathrm{3}×{masse}\:{hydrigene}\right)\right.\right. \\ $$$$ \\ $$$${so}\:\:{the}\:{composant}\:{is}:\:\boldsymbol{{Alcene}}\:\boldsymbol{{C}}_{\boldsymbol{{n}}} \boldsymbol{{H}}_{\mathrm{2}\boldsymbol{{n}}+\mathrm{2}} \left({with}\:{n}>\mathrm{1}\right) \\ $$$${n}=\mathrm{2}\:\:\:\:\:\:{C}_{\mathrm{2}} {H}_{\mathrm{6}} \left({masse}\:{carbone}\:=\mathrm{24}\rightarrow\mathrm{6}{for}\:{hydrogene}\right) \\ $$$$\left.{masse}\:{carbone}\:\mathrm{24}=\mathrm{4}×{masse}\:{hydrogene}\right) \\ $$$${masse}\:{totale}=\mathrm{30}\:\:\:\rightarrow\:\left(\frac{\mathrm{24}}{\mathrm{30}}=\frac{\mathrm{4}}{\mathrm{5}}\:\:\right)=\mathrm{80\%}\:{M}\left({satisfait}\right) \\ $$$${so}\:{the}\:{composant}\:{is}\:{C}_{\mathrm{2}} {H}_{\mathrm{6}} \\ $$$$ \\ $$

Commented by Spillover last updated on 12/Oct/22