Assume 100 g of the compound

Then the masses of elements are:

• Carbon $= 14.5\text{ g}$
• Chlorine $= 64.46\text{ g}$
• Hydrogen $= 1.8\text{ g}$

Since total must be $100\%$, oxygen percentage is: $100 - (14.5 + 64.46 + 1.8) = 19.24\%$ So oxygen mass $= 19.24\text{ g}$.

Convert masses into moles

$n_C = \frac{14.5}{12} = 1.2083$ $n_H = \frac{1.8}{1} = 1.8$ $n_{Cl} = \frac{64.46}{35.5} \approx 1.8169$ $n_O = \frac{19.24}{16} = 1.2025$

Find simplest mole ratio

Divide all by the smallest value, which is about $1.2025$:

$\frac{n_C}{1.2025} \approx \frac{1.2083}{1.2025} \approx 1.00$ $\frac{n_H}{1.2025} \approx \frac{1.8}{1.2025} \approx 1.50$ $\frac{n_{Cl}}{1.2025} \approx \frac{1.8169}{1.2025} \approx 1.51 \approx 1.5$ $\frac{n_O}{1.2025} = 1.00$

So the ratio is approximately: $C : H : Cl : O = 1 : 1.5 : 1.5 : 1$

Convert into whole numbers

Multiply all by $2$: $2 : 3 : 3 : 2$

Therefore, the empirical formula is: $\boxed{C_2H_3Cl_3O_2}$

Calculate empirical formula mass

$\text{EFM} = 2(12) + 3(1) + 3(35.5) + 2(16)$ $= 24 + 3 + 106.5 + 32$ $= 165.5$

Match with the asked format

The question says: $\text{Empirical formula mass} = \underline{\hspace{2cm}} \times 10^{-1}$

Since: $165.5 = 1655 \times 10^{-1}$

Hence the required integer to fill in the blank is: $\boxed{1655}$

Comparison with stored answer

Stored correct answer = $1655$

Our derived answer matches it exactly.