COMPOSITION OF ORGANIC COMPOUNDS 41 



C. CALCULATION OF RESULTS. 



With the exception of oxygen all the elements present in an 

 organic compound are thus estimated. The amount of oxygen is 

 found by difference. 



From the figures obtained the percentage composition is calculated, 

 i.e. the amount given by 100 grams of substance, thus: 



0*2009 gm. substance gave o 2987 gm. CO 3 and 0*1092 gm. H 2 O. 

 0-1887 g m - M I 5'2 c.c. moist N at 16-5 and 767 mm. 



Now, 0-2987 gm. CO 2 = 0-2987 x H gm. C = 0-2987 x -1- gm. C = 0-0815 gm. C. 



44 II 



0*1092 gm. H 2 O = 0-1092 x 2- gm. H = 0-1092 x - gm. H = 0-01213 gm. H. 

 io 9 



15-2 c.c. moist N at 767 mm. and 16-5 C. = * 5 2 x 753 x 2 ? 3 c.c. at o and 760 mm. 



760 x 289-5 



s= 14-2 C.C. 



28 



= 14-2 x gm. 



22,400 

 = 0-01775 g m - N. 



/.percentage of C = ' o8l 5 XIO = 40*56. 

 0-2009 



H - ' 01213 x I0 - 6-04. 



0*2009 

 N== 0-01775x100^ 



0-1887 



O by difference = 44-00. 

 Total = 100-00 



The formula of the compound is obtained by dividing the percent- 

 ages by the atomic weights of the elements ; the ratio of the number 

 of atoms to each other is then obtained by dividing by the lowest 

 value : 



C 12^ = 3-38*0*67 = 5. 

 H ^f = 6*o 4 ^o-6 7 = 9. 

 N O? = 0-67 -f 0*67 = i. 

 O li^ = 2-75 -r 0*67 = 4-1. 



The formula of the compound is therefore C 5 H 9 NO 4 . 



In any estimation only a difference of o*2-O'3 per cent, is allowed 

 between the values found and those calculated from the formula. The 

 calculated values are 



C = 40*81. diff. = - 0*25. 

 H = 6*12. = + 0-08. 



N = 9*52. = - 0*12. 



The analysis was therefore sufficiently accurate. 



* Vapour pressure of water at 16*5 C. = 14*0 mm. .*. pressure on gas = 767 - 14 = 753 



