Answer
$NCl_{3}$
Work Step by Step
In the 6.61 mg sample of chloride, 0.77 mg nitrogen and (6.61-0.77)mg chlorine are present.
Moles of N=$\frac{0.77\times10^{-3}g}{14.0067g/mol}=5.5\times10^{-5}\,mol$
Moles of Cl=$\frac{5.84\times10^{-3}g}{35.453g/mol}=1.65\times10^{-4}\,mol$
Since $5.5\times10^{-5}\,mol$ is the smallest value, division by it gives a ratio of 1:3 for N:Cl.
The empirical formula is obtained by mentioning the numbers in the ratio after writing the respective symbols.
Thus, the empirical formula is $NCl_{3}$.