We must calculate the masses of #"C, H"#, and #"N"# from the masses of carbon dioxide, water, and nitrogen dioxide.
#"Mass of C" = 6.019 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "1.6425 g C"#
#"Mass of H" = 4.313 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.482 52 g H"#
#"Mass of N" = 9.438 color(red)(cancel(color(black)("g NO"_2))) × "14.01 g C"/(46.01 color(red)(cancel(color(black)("g NO"_2)))) = "2.8739 g N"#
Now, we must convert these masses to moles and find their ratios.
From here on, I like to summarize the calculations in a table.
#ulbb("Element"color(white)(X) "Mass/g"color(white)(X) "Amt/mol"color(white)(m) "Ratio"color(white)(ml)×2color(white)(m)"Integers")#
#color(white)(mm)"C"color(white)(mmml)1.6425color(white)(mml)"0.136 76"color(white)(Xm)1color(white)(mmmll)2color(white)(Xmmmm)2#
#color(white)(mm)"H" color(white)(XXXl)0.48252 color(white)(mll)"0.478 69" color(white)(mm)3.5001 color(white)(ml)7.0002color(white)(Xml)7#
#color(white)(mm)"N" color(white)(XXXl)2.8739 color(white)(mml)"0.205 13" color(white)(mm)1.4999 color(white)(ml)2.9997color(white)(Xml)3#
The empirical formula is #"C"_2"H"_7"N"_3#.
