Calculating Ratios to Find New Quantities of Substances| Respiration | C6H12O6 | + 6O2 | —› 6H2O | + 6CO2 | + 36 ATP (670 kcal) |
glucose | + oxygen | —› water | + carbon dioxide | + energy
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Table 1. Calculating kilogram inputs and outputs of respiration | Measures | Compounds | ATP | Energy | | Glucose
A | Oxygen
B | Water C | Carbon dioxide D | (moles)
| (kcal)> E | | moles | 1 | 6 | 6 | 6 | 36 | 670 | | grams | 1 x 180.16 = 180.16 | 6 x 32.00 = 192.00 | 6 x 18.02 = 108.12 | 6 x 44.01 = 264.06 | 36 | 670 | | kilograms | 0.18016 | 0.19200 | 0.10812 | 0.26406 | 36 | 670 | | kg/day | | | | | | * |
| C6H12O6 | + 6O2 | —› 6H2O | + 6CO2 | + 36 ATP (670 kcal) | | A | + B | —› C | + D | + E kcal | | X (A | + B | —› C | + D | + E kcal) | | XA | + XB | —› XC | + XD | + XE kcal |
What is X? X is the FACTOR by which if one quantity increases,
all quantities must increase. X x E = New E where E is the original kcal quantity
X = New E / E The factor, X, is the same for all compounds and kcal. X = New E / E = New A / A = New B / B = New C / C = New D / D Thus, if any one new quantity is known, it can be used to find
the factor that is multiplied by all the original quantities to obtain
the other new quantities. |