A hydrocarbon contains 10.5 g of carbon for each one gram of hydrogen. The mass of 1 litre of hydrocarbon vapours at `127^@C` and 1 atmospheric pressure is 2.8 g. Find out the molecular formula.

Options

A. `C_6H_(12)`
B. `C_4H_8`
C. `C_7H_8`
D. `C_4H_10`

Answer

C

Explanation

Carbon : hydrogen : : 10. 5.: 1
Calculation of empirical fomula:
`{:("Element" , "Percentage " , "At . mass " ,"Relative number of atoms","Simplest ratio of atoms"),("Carbon",10.5/11.5xx100=91.3,12,91.3/12=7.6,7.6/7.6=1xx7=7),("Hydrogen",1/11.5xx100=8.7,1,8.7/1=8.7,8.7/7.6=1.14xx7=):}`
Empirical formula `=C_7H_8 " "` Empirical formula mass `=(12xx7)+(1 xx8)=92`
Calculation of molecular mass:
Experimental conditions: `V_1=1` litre, `P_1 = 1` atm, `T_1 = 127 +273 = 400 K`
NTP conditions = `V_2=? ,P_(2) = 1atm , T_2 = 273 K`
Applying gas equation `V_2 = (P_1V_1)/(T_1) xxT_2/P_2 = (1xx1xx273)/(400xx1) = 0.6825` litre .
0.6825 litre of the gas weigh = 2.8 g `" "` 22.4 litre of the gas weigh `= (2.8)/(0.6825) xx22.4 = 91.89 ~~92`
`n = ("Molecular mass")/("Empirical formula mass ") = 92/92 :1" "` Molecular formula = Empirical formula `C_7H_8`