Question

uestion 23 of 34 > An herbicide contains only C, H, Cl, and N. The complete combustion of a 275.0 mg sample of the herbicide in excess oxygen produced 287.6 mL. of CO2 and 167.8 mL of H,0 vapor at STP. A separate analysis determined the 275.0 mg sample contained 75.82 mg Cl. Determine the percent composition of the herbicide. percent carbon: % percent hydrogen: % percent chlorine: % narcant nitmnen.

Answer 1

I at stp Moles of 1 litse E 1000MI: I mote occupies = 22.4 Litse -3 . 287.671 287.670) = 12.83X10 co₂ = 22. unitse. 22.4x Hos 1678 M) = M eme 167.871 -749 X10 22.4 Litse. 22:4810 ni Moley of 7:49 xichts Weight of Casson = morns of con X mo land mass of carbon = 12:83x10 MON 12.01g TAOT = .0.154.g. 198=1000mg s 0.154 x 1000mg = 154.1 mg. I weight of t = moles of H₂0 X2 x Atomic weight of Hi = 7.48x7oʻx2x1 % M01 = branio io long sgohl =

22.4เy Atomic weight Divid. - breast no symy =1283 12.01 12.83 26 -215 4.98 ? . Weignt of c 154.ing. H . 14.98mg . 75082 may N 30.1mg Total 275.00mg UŤ JE = 14.58 .flags 11 2:15 213 = 2.1 2.13 = 1 : 75.82 35.55 30-1ng =2.15.52 15 ly ong Emperical Formula a. Goty GIN Percent carbon - Massof carbon Total mass of torbico X100 = 15h role 275.00 0.560xloo = 56% dus was 56 Percent Hydsolen! % 14.98ng/X100 of 0.054uxloo 2nong percent Chloring: Auswas [sonyll. 75-82 Xloo . : ī 215 275 Answys 1 27.577%. I percent nitrogenin - 30.1mg xioo = 0.109 Xioo is ng Auswas [loog] % | Empesical Formula ? Auswas 7 Cotty CINI

Empesical briolecules Weight: 6xC+7XH + 1x21 +iXN! Cetty CIN 6 X12.01 X7X1 +1X35045 xil kly = 128.51 Mole cales weight = 2570167 To =2x Emporsical 128.518 Forinnlal Mol Mo Molecules Formula [ Gattina