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# A deug marijuana owes its activity to tetrahydrocannabinol, which contains 70%as many carbon atoms as hydrogen atoms and 15 times as many oxygen atoms. The no. Of moles in a gram of tetrahydracannabinol is 0.00318.  Determine its molecular formula

Samyak Jain
333 Points
one year ago
This question looks difficult but is pretty simple.
Solution :
Let number of hydrogen atoms in tetrahydrocannabinol be x and number of oxygen atoms be y.
According to given conditions, number of carbon atoms = 70% of x = 70% of 15y.
We get x = 15y i.e. number of hydrogen atoms = 15 times number of oxygen atoms.
And number carbon atoms = (7/10) x   =   (7/10).15y

Make a table :
Atomic ratio   Simplest atomic ratio
C      21 / 2                    21
H      15                         30
O       1                          2

Empirical formula of compound  =  C21H30O2
Its molecular formula will be n(C21H30O2).
Molar mass of the compound = n(21 x 12 + 30 x 1 + 2 x 16) = 314 n         …(1)
Now, 1 g contains 0.00318 mole of compound i.e. 1 mole of compound will have mass of (1 / 0.00318) g $\dpi{100} \approx$ 314.5 g
i.e. Molar mass = 314.5 g                      ...(2)

From (1) & (2), 314 n $\dpi{100} \approx$ 314.5   $\dpi{100} \Rightarrow$  n = 1.
$\dpi{100} \therefore$  Molecular formula of tetrahydrocannabinol is  C21H30O2 .