One mole of a hydrocarbon contains 48 g of carbon. If it vapour density is 28, the hydrocarbon is?
an alkane
an alkene
an alkyne
aromatic (C = 12, H = 1)
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Vapor density=28, molar mass=2×28=56
Mass of carbon=48, n.o of C atom= mass÷molar mass = 48÷12 =4.
Mass of hydrogen=56-48=8, n.o of hydrogen atoms=mass÷molarmass of hydrogen atom = 8÷1 = 8
Therefore, compound is C4H8 (butene).

Since one mole of the hydrocarbon contains 48 grams of carbon, the number of atoms in the hydrocarbon is 4812=4
.
Let the formula of the hydrocarbon be C4Hx
. The molar mass of this vapor is 12∗4+2∗x=48+2x
.
Now, we can write, vapor density = (molar mass of vapor)/2 =48+2x2
.
But it is given that the vapor density is equal to 28. Hence,
48+2x2=28⇒x=4
.
Hence, the formula of the hydrocarbon is C4H4 (alkene).


