Quantitative analysis of an organic compound (X) shows following \% omposition. C : 14.5% Cl: 64.46\% H: 1.8 \%(Empirical formula mass of the compound (X) is__________$$\times 10^{-1}$$ (Given molar mass in $$gmil^{-1}$$ of C : 12, H : 1, O : 16, Cl : 35.5 )

Find the empirical formula mass of compound X with C: 14.5 %, Cl: 64.46 %, H: 1.8 %.

The total given percentages of C, Cl, and H add to 14.5 + 64.46 + 1.8 = 80.76 %, leaving 100 - 80.76 = 19.24 % for oxygen.

This corresponds to moles of each element per 100 g of compound: C: $$\frac{14.5}{12} = 1.208$$, H: $$\frac{1.8}{1} = 1.8$$, O: $$\frac{19.24}{16} = 1.2025$$, Cl: $$\frac{64.46}{35.5} = 1.816$$.

Dividing by the smallest mole value (1.2025 for O) gives the ratios C: $$1.208/1.2025 \approx 1.005 \approx 1$$, H: $$1.8/1.2025 \approx 1.497 \approx 1.5$$, O: $$1.2025/1.2025 = 1$$, Cl: $$1.816/1.2025 \approx 1.510 \approx 1.5$$.

Multiplying these ratios by 2 yields C = 2, H = 3, O = 2, Cl = 3, so the empirical formula is $$C_2H_3O_2Cl_3$$.

The empirical formula mass is calculated as $$2(12) + 3(1) + 2(16) + 3(35.5) = 24 + 3 + 32 + 106.5 = 165.5$$, which in units of $$\times 10^{-1}$$ becomes $$165.5 = 1655 \times 10^{-1}$$.

The answer is 1655.