An organic compound has $$42.1\%$$ carbon, $$6.4\%$$ hydrogen and remainder is oxygen. If its molecular weight is 342, then its molecular formula is :

Organic compound has 42.1% carbon, 6.4% hydrogen, and the remainder is oxygen, with a molecular weight of 342.

First, we calculate the percentage of oxygen: % O = 100 - 42.1 - 6.4 = 51.5%.

Next, we determine the moles of each element in one mole of the compound:

C: $$\frac{42.1}{100} \times 342 / 12 = \frac{143.98}{12} = 12.0$$ → 12 atoms

H: $$\frac{6.4}{100} \times 342 / 1 = 21.89$$ → 22 atoms

O: $$\frac{51.5}{100} \times 342 / 16 = \frac{176.13}{16} = 11.0$$ → 11 atoms

Therefore, the molecular formula is $$C_{12}H_{22}O_{11}$$ (sucrose). We can verify this by checking the molar mass: $$12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342$$, which matches the given molecular weight.

The correct answer is Option C: $$C_{12}H_{22}O_{11}$$.