J. A. CAMPBELL Harvey Mudd College Claremont, California 9171 1
Questions
41. The only known essential function of the huma n stomach is to accomplish absorption of vitamin B-12. How do you suppose this ~ v a sdiscovered? 42. In 1838 J. B. Boussingault in France planted seeds in a pot containing no nitrogen fertilizer and ohtairoed the following data. He got similar results in field c:xperiments. Interpret the results in terms of the law ol conservation of atoms.
'
; N in seed ; N in 3-month-old plant
wheat oats 57 .59 60 53
clover 114 156
pea
47 100
G. H. Rergold summarizes analytical data on complmition of the Bombyz mori (silkworm) virus. [Bioclhemistry of Insect Viruses in "The Viruses," nod W. RI. STANLEY F. M. BURNET (Editors), Academic Press, 1959.1 Calculate an empirical formula for the virus using the following average weight percents: C 51.10, N 14.9,H6.8.5, S 1.19,P0.243, C10.075,ash 0.30,lbalance oxygen (26.7). Neglect ash in calculating the empirical formula, Bergold gives 378,000 as the total molecular weight. Suggest a possible molecular formu[la. Degradation experiments suggest each virus partic:le may be split reversibly into six almost equal fragm~ents,each of which can be irreversibly split
43.
into three. Are your formulas consistent with these observations? Assume there are eightef :n identical fragments, adjust your molecular formula t l3 allow this, and calculate a theoretical percent composition. Compare to the experimental and comm;ent on the likelihood of 18 identical parts in light of all the data given here.
44. A glass of whiskey (90 proof or 4liyOethanol, volume 100 cm3, density 1 g cm-9 can produce a concentration of about 0.001 g ahou t 1 hr after consumption in a total blood volume of 7 1. What percent of the imbibed alcohol is not then i~1 the blood?
45. An astronaut drinks about 2 l of I1*0per day but excretes ahout 2.4 1 H 2 0 per day. Accsount for the difference. 46. The Apollo 11 lunar module was ~ropelledby Aerozine 50 consisting of a 50/50y0 by weigl~tfuel made up of hydrazine, N2H4, and 1,l-dimethy1 hydrazine (CH,),NNH,, plus dinitrogen tetroxide,, & 0 4 , as oxidizer. If 2200 kg of Aerozine 50 were required for lunar take-off, how much oxidizer should be carried? If water constituted one-third of the welight of the exhaust gases, how much hydrogen came (lut in other forms? Volume 49, Number I , Jonuor
ial .te a) LC-
er rd is Ju CY 1%
re
Pin rimental uncertainty ot about lux. Hilt clover and peas have much more nitrogen in t,he plant compared to the seed than could reasonably be attributed to experimental uncertainty. Since there were no nitrogen atoms in the soil, the plant must have ingested them from elsewherepresumably from the atmosphere, which is about 79% by volume nitrogen, N?. Calculate the atomic ratios (neglecting ash) and assuming the virus contains one atom of C1 (the least common element) per molocr~leof virus. The empirical formule.wcight A3.
The experimental values are given in parentheses. All the calculated values agree with the experimental values with an uncertainly to about *lYo of the value (which seems a reasonable range), except chlorine which is off by a factor of two and phosphorus off about 20%. These differences are so large as to rule against 18 equal fragments (assuming the analytical data are acceptable). The fact that the analytical data give a good formula (all numbers of atoms integral) for a molecular weight of 378,000 gives confidence in the analysis. Thus it appears that at least some of t,he six fragments are different in numbers of chlorine atoms per fragment. Note that, in general, tho most information (in terms of limitations) is obtained from those elements present in the least amounts. Yet it is just these which are most difficult to determine acnrrately. Good analytical data are a rare and valuable commodity. We conclude that the virus has a molecular weight of about 378,000 and contains 8CI, 30P, 1405, and about 16,200C, 63300, 4040N, 25,ROOH per molec~tle,that the six initial fragments are not idontical, nor are the three secondary fragments, at least in some cases. A4. Initid alcohol int,ake = 100 em3 .X. -1- ,elem3 X 0.45 = 45 g CZHSOHintake. In blood one how later = 0.001 g/ems X 7 X 1000 em3 = 7 g C2H60Hin blood. (7/45) X 100 = 15% of initial alcohol in blood, or 857?of alcohol is elsewhere. ~
~~~
AS. Excretion of 2.4 1 of water having drunk only 2 l indicates 0.4 ]/day of metabolic water plus water present in food. The thrce sources of the additional water are: (1) present in food ns H20, (2) present in food, e.g., carbohydrate, as H and 0 then metabolized to water, (3) produced when H in food is oxidized Lo H U by inhaled oxygen. A6.
2N&
+ N104 = 3Nz + 4HxO
F. wt NzH4= 32.0
1100 kg N2H4 = 1100 kg X 1000 (g/kg)/32.0 (glmole) = 34,400 moles N2H4 1100 kg (CHx)nN2Hn= 1100 kg X 1000 (g/kg)/60.0 (g/mole) = 18,300 moles (CHJ)%NIHZ 34,400 moles in NIHl require (112) X 34,400 = 17,200 moles N204 18,300 moles ( C H S ) ~ N ~ H ~ 2 X 18,300 = 36,600 moles N a , 17,200 36,600 = 53,800 moles NnOl total = 53,800 moles X 92.0 (g/mole)/l000 (g/kg) = 4,950 kg of NzOatotal for 2200 kg of Aerozine
+
Total exhaust gases = 2200 kg Aerazine
+ 4950 kg Nn04 =
7150 kg gases.
1100 kg N2H4= 14.03 (g H/mole)/32.0 (g Aerosine/mole)l X 1100 kg Aeroeine = 139 kg hydrogen in NzHl 1100 kg (CH&N2H2 = 18.06 (g H/mole/60.0 (g Aeroaine/ mole)] X 1100 kg Aeroeine = 148 kg hydrogen in (CHa)~NIH1 147
emicol Education
7150 X 1/3 X 2/18 = 264 kg hydrogen out as water 137 - 264 = 20 kg hydrogen out in some other form
+
