searching for experimental values of the consumption rate of $O_2$ by tumor cells I found an article that measures the rate in the units of measure

$$\frac{mol}{cell\cdot s}.$$

The actual measurement was

$$5.5\cdot 10^{-15}\frac{mol}{cell\cdot s},$$

I precise to convert the expression in terms of $kg,\;cm,\;s$ units, but I'm not sure of how to do that. When I tried the result where far away from the order of values that I was expecting. So, I appreciate any help.

  • $\begingroup$ What is $cell$? Like what are the units / what is the size of it? $\endgroup$
    – SuperCiocia
    Commented Jul 5, 2020 at 3:45
  • $\begingroup$ @SuperCiocia I'm guessing it's a number? As in "per cell" or "per person"? $\endgroup$
    – Philip
    Commented Jul 5, 2020 at 4:01
  • $\begingroup$ Oxygen has a molar mass of 32 g, and there are 1000 g in a kg. Your answer should end up being g/cell-s or kg/cell-s, but chemists would normally go with g/cell-s. By the way, why would you want to convert to something besides mole/cell-s? $\endgroup$ Commented Jul 5, 2020 at 4:02
  • $\begingroup$ @Philip yeah but you need to give that in units of cm. $\endgroup$
    – SuperCiocia
    Commented Jul 5, 2020 at 4:02
  • 1
    $\begingroup$ I read the book @Philip (the part with the matter of interest), it was very good. Thanks! $\endgroup$
    – Fernando
    Commented Jul 6, 2020 at 19:25

1 Answer 1


You have to write the old units in terms of the new units.

  • 1 mole of $0_2$ is an Avogradro's number ($N_A = 6.02214076 \times 10^{23}$) of particles. Each particle here is an $O_2$ molecule, so the mass of a single particle is $m_{O_2} = 2\times m_0$ (technically minus the binding energy, but it's small so let's ignore it). $m_0$ is the number of nucleons in an oxygen atom (neglecting the electrons since their mass contribution is so small) times by the mass of a nucleon. So $m_0 = 16\times u$ (for $^{16}O$) where $u$ is the atomic mass unit $1.66 \times 10^{-27}$ kg.

    Putting all together: $$ 1\,\mathrm{mol} = N_A\cdot m_{0_2} = N_A\cdot 2\cdot m_0 = N_A \cdot 2 \cdot 16 \cdot u \approx 0.032 \,\mathrm{kg}. $$

  • Whatever the dimension of a cell is, let's say it's $1 \, \mathrm{cell} = \alpha\, \mathrm{cm}$, where $\alpha$ is the dimension of the cell in centrimetres.

So, all together: $$ 5.5\cdot 10^{-15}\frac{\mathrm{mol}}{\mathrm{cell}\cdot \mathrm{s}} \Rightarrow 5.5\cdot 10^{-15}\frac{0.032 \,\mathrm{kg}}{\alpha\, \mathrm{cm}\cdot \mathrm{s}} = \frac{1.76\cdot 10^{-16}}{\alpha}\frac{\mathrm{kg}}{\mathrm{cm}\cdot \mathrm{s}}.$$

  • $\begingroup$ @Fernando I am so sorry, $87$ is something i had in my head from something else. For oxygen, it should be $16$ (twice the $Z$ number) unless you are dealing with isotopes. Edited the answer. $\endgroup$
    – SuperCiocia
    Commented Jul 6, 2020 at 19:49
  • $\begingroup$ Thanks, I was looking a lot and always appear to be 16, then I decided to ask, jajajaja $\endgroup$
    – Fernando
    Commented Jul 6, 2020 at 19:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.