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I was recently asked the following question

When breathing, approximately 5 percent of each exhaled breath is carbon dioxide. Neglecting any difference in water-vapour content, estimate the typical difference in mass between an inhaled breath and an exhaled breath. Assume that one’s lung capacity is about half a litre and that 20% of the air that is breathed in is oxygen. Note that in the process of breathing, approximately ¼ of the O inhaled is replaced by CO2. Air is 80% nitrogen and 20% oxygen.

given the molar mass of oxygen is 32 grams per mole, CO2 is 44 grams per mole and nitrogen is 28 grams per mole

I know you can use min = density of air * volume of 2 lungs.

And then work out 5% of this mass ( A = 0.05min)

Then use NCO2 = the 5% mass / molar mass of oxygen

This gives us the number of carbon dioxide moles and we can use

dm = change in mass = NCO2*(44-32)

using this method I get about 23 mg

However, before using this method, I used another which is provided below (my original method):

prior method

Why is it that these two methods disagree? In my original method, which statement causes the difference between the two methods answers?

(Which answer is more correct and why)

(Surely my original method follows from logic e.g. The first statement, where the total number of moles = the sum of moles of oxygen, nitrogen and CO2, and that 1 part in 5 of that sum is oxygen (when breathing in) etc.)

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  • $\begingroup$ I suspect the discrepancy is due to a misuse of units. Composition percentages are often by mass, by volume, or occasionally by moles. These expressions are not generally equivalent. According to [Wikipedia][1], the commonly cited percentages such as 78% $N_2$ are by volume. Thus to properly use this percentage in your computations, you should formally think of it as: $78% N_2 (by V) = \frac{78 L N_2}{100 L air} \neq \frac{78 g N_2}{100 g air} \neq \frac{78 mol N_2}{100 mol air}$ [1]: en.wikipedia.org/wiki/Atmosphere_of_Earth?wprov=sfla1 $\endgroup$
    – electronpusher
    Mar 21, 2017 at 19:15
  • $\begingroup$ If 80% (approx.) of the air is made up of Nitrogen by volume, surely it follows 80% of the moles you would find in a unit volume of air is Nitrogen e.g. say you have a volume of 1 cubic meter, with 100 moles; 20 moles of oxygen and 80 moles of Nitrogen, 80% of the mole count is nitrogen and 80% of that volume is Nitrogen ( of course, atoms are small and their individual volumes don't add to 1 cubic meter, but within that volume, you would say 80% of it was occupied by Nitrogen (even though most of it is occupied by nothing) therefore it should be the same independent of volume or mole count $\endgroup$
    – Think
    Mar 22, 2017 at 13:01
  • $\begingroup$ Keep in mind the ideal gas law PV=nRT. The temperature may be the same for all molecular species in air, but each gas has a different partial pressure. So if n, V, and P all vary, I don't think one can assume x% by mol = x% by volume. $\endgroup$
    – electronpusher
    Mar 22, 2017 at 13:37
  • $\begingroup$ surely PV = nRT can be applied to the gas as a whole (n = number of oxygen moles + number of CO2 moles +number of nitrogen moles). Also, why does each gas have a different Pressure $\endgroup$
    – Think
    Mar 22, 2017 at 14:08
  • $\begingroup$ en.wikipedia.org/wiki/Dalton%27s_law?wprov=sfla1 $\endgroup$
    – electronpusher
    Mar 22, 2017 at 16:38

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