814 EXPERIMENT STATION RECORD. 



been derived from some substance other than protein — that is, from fat 

 and carbohydrates — is shown as follows : 



160 gm. C (excreted in breath) — 43.7 gm. (derived from protein) =116.3 gm. 

 (derived from fat and starch). 



500 gm. <> (amount consumed) —144.3 (amount necessary for oxidation of pro- 

 tein) =355.7 gm. O (amount for oxidation of carbon from fat and starch). 



The proportion of fat and starch burned in the body may be deter- 

 mined by studying the respiratory quotient. Ordinary animal fat has 

 the following percentage composition: 76.54 carbon, 12.01 hydrogen, 

 11.45 oxygen. When 100 gm. fat is burned 11.45 gm. oxygen and 1.43 

 gm. hydrogen unite to form water, and there remain 10.58 gm. hydrogen 

 and 7G.54 gm. carbon. These require 84.04 gm. and 204.11 gm. oxygen, 

 respectively, or a total of 288.75 gm. for complete combustion, 280.65 gm. 

 C0 3 being produced. This oxygen must be derived from the air. The 

 respiratory quotient would therefore be 



280.65 + 1.96633 _ 

 288.75 +■ 1.43003 ~ °' 70b9 



Starch has the formula (O G H l0 O 5 ) x . Its molecular weight is there- 

 fore some multiple of 162. The oxygen in the starch molecule is 

 sufficient for a complete combustion of the hydrogen; therefore only an 

 amount sufficient for the combustion of the carbon need be taken from 

 the air. The carbon in 162 gm. of starch would require 192 gm. of oxy- 

 gen for its combustion and produce 264 gm. carbon dioxid. Therefore 

 the carbon in 100 gm. would require 118.5 gm. oxygen, producing 163 

 gm. carbon dioxid. The respiratory quotient when starch is burned 

 is therefore 1. If a respiratory quotient of 1 is found in an experiment, 

 then starch alone is burned iu the body. A respiratory quotient of 

 0.7069 shows that fat only is bnrned. If the respiratory quotient is 

 between these values, it indicates the combustion of a mixture of fat 

 and starch. 



The amount of oxygen necessary for combustion may be computed as 

 follows : 



Let A = total O used, B = 0O 2 produced, and X = O necessary for 

 oxidation of fat. Then A — X = O available tor oxidation of starch; 

 and X x 0.7069 -4- ( A — X) x 1 = B (the C0 2 produced). 



In the concrete example cited above there were 116.3 gm. of carbon 

 (equal to 216.9 liters of carbon dioxid) and 355.7 gm. (equal to 248.7 

 liters) of oxygen remaining to be accounted for by combustion of fat and 

 starch. The respiratory quotient would therefore be 0.872. 



Substituting the proper values in the above equations would give 

 the following: 



0.7069 X+248.7— X=216.9. Then X=108.5, the oxygen necessary 

 for oxidation of fat; and 248.7—108.5=140.2, the oxygen necessary for 

 oxidation of starch. 



