The **empirical formula** is the simplest whole number molar ratio of the elements in the compound.

We must figure out the masses of #"N"# and #"Cl"#, convert to moles, and then find their ratio.

Your compound contains #"0.77 mg N"# and forms #"6.61 mg of chloride"#.

#"Mass of chloride = mass of N + mass of Cl"#

#"Mass of Cl" = "mass of chloride - mass of N" = "6.61 mg - 0.77 mg" = "5.84 mg of Cl"#

#"Moles of N" = 0.77 × 10^"-3" color(red)(cancel(color(black)("g N"))) × "1 mol N"/(14.01 color(red)(cancel(color(black)("g N")))) = 5.5 × 10^"-5"color(white)(l) "mol N"#

#"Moles of Cl" = 5.84 × 10^"-5" color(red)(cancel(color(black)("g Cl"))) × "1 mol Cl"/(35.45 color(red)(cancel(color(black)("g Cl")))) = 1.65 × 10^"-4"color(white)(l)"mol Cl"#

From this point on, I like to summarize the calculations in a table.

#"Element"color(white)(Ag) "Mass/mg"color(white)(Xm) "Moles"color(white)(Xm) "Ratio"color(white)(m)color(white)(l)"Integers"#

#stackrel(————————————————————)(color(white)(m)"N" color(white)(XXXmm)0.77 color(white)(Xml)5.5 color(white)(l)× 10^"-5"
color(white)(Xll)1color(white)(mmmmll)1)#

#color(white)(m)"Cl" color(white)(XXXXll)5.84 color(white)(mml)1.65 × 10^"-4" color(white)(Xl)3.00 color(white)(mmml)3#

There is 1 mol of #"N"# for 3 mol of #"Cl"#.

The empirical formula is #"NCl"_3#.

Here is a video that illustrates how to determine an empirical formula.