# Physical quantities conversion

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.

• What is $cell$? Like what are the units / what is the size of it? Commented Jul 5, 2020 at 3:45
• @SuperCiocia I'm guessing it's a number? As in "per cell" or "per person"? Commented Jul 5, 2020 at 4:01
• 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? Commented Jul 5, 2020 at 4:02
• @Philip yeah but you need to give that in units of cm. Commented Jul 5, 2020 at 4:02
• I read the book @Philip (the part with the matter of interest), it was very good. Thanks! Commented Jul 6, 2020 at 19:25

• 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: $$5.5\cdot 10^{-15}\frac{\mathrm{mol}}{\mathrm{cell}\cdot \mathrm{s}} \Rightarrow 5.5\cdot 10^{-15}\frac{0.032 \,\mathrm{kg}}{\alpha\, \mathrm{cm}\cdot \mathrm{s}} = \frac{1.76\cdot 10^{-16}}{\alpha}\frac{\mathrm{kg}}{\mathrm{cm}\cdot \mathrm{s}}.$$
• @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. Commented Jul 6, 2020 at 19:49