Gas Laws - How To Convert Ppm To G/m3? - Chemistry Stack Exchange

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Learn more about Teams How to convert ppm to g/m3? Ask Question Asked 7 years, 5 months ago Modified 3 years, 10 months ago Viewed 12k times 4 $\begingroup$

I'm not very comfortable with chemistry and I need to convert $\ce{CH4}$ concentration values in $\pu{ppm}$ to $\pu{g/m3}$. Is that possible?

I've already researched a bit and realised that for water you can assume $\pu{1 ppm}$ equals $\pu{1 mg/L} = \pu{1 g/m3}$. But since I'm measuring concentrations in the air this might not be correct.

I really appreciate any help. Thanks!

• gas-laws
• concentration
• units

Share Cite Improve this question Follow edited May 3, 2018 at 14:26 user7951 asked Jun 9, 2017 at 15:52 PatriciaPatricia 5711 gold badge11 silver badge44 bronze badges $\endgroup$ 4

• 4 $\begingroup$ 1 ppm is like you have 1 part , here 1 molecule of $\ce {CH_4} $ in a million molecules of air. If we assume air to be ideal gas then you may use the Ideal gas equation for knowing the volume of total air and then take out value in $g/m^3$ and don't forget the $g $ represent weight of methane $\endgroup$ – Physicsapproval Commented Jun 9, 2017 at 16:32
• $\begingroup$ @Physicsapproval Thanks for your help! I have estimated the volume using the Ideal Gas Law (assuming 1 mol of $CH_{4}$) but I'm unsure as to what to do next. Should I divide said volume per the molecular weight of $CH_{4}$? $\endgroup$ – Patricia Commented Jun 12, 2017 at 9:35
• $\begingroup$ I have tried a different approach. Knowing that: $ 1 ppm = 1 \frac{\mu g}{g}$; first, I multiplied the ppm values per the density of (in this case) methane ($656 g/m^{3}$) and them multiplied again by the factor $(10^{-6})$. Here's the units calculation: $\frac{\mu g}{g} \times \frac{g}{m^{3}} = \frac{\mu g}{m^{3}} \times (10^{-6}) = \frac{g}{m^{3}}$. What do you think? $\endgroup$ – Patricia Commented Jun 12, 2017 at 18:31
• $\begingroup$ okay is the methane gas in a mixture I believe assuming air then how have you calculated the density ? Again have you used ideal gas law here to find density ? $\endgroup$ – Physicsapproval Commented Jun 13, 2017 at 3:47

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I am trying to understand ppm, too. As far as I have understood there are different kinds of ppm, which is basically a ratio: it can be a ratio of amount of substance, masses, or volumes.

Assuming your ppm is a molar ratio I did this reasoning:

Indicating with $n$ the amount of substance, with $M$ the molar mass and with $V$ the volume, the concentration of your gas is: $$c = \frac{n_\mathrm{gas} \cdot M}{V},$$ and defining the $\mathrm{ppm}$ as: $$\mathrm{ppm} =\frac{n_\mathrm{gas}}{n_\mathrm{total}}\cdot 10^6.$$

Using the gas law: $$n_\mathrm{tot} = \frac{p\cdot V}{R \cdot T}, $$ where $T$ is the temperature in kelvin and $p$ the pressure in pascal, and substituting, you get: $$c = \frac{\mathrm{ppm} \cdot M \cdot p}{R\cdot T}\left[\frac{\mu \pu{g}}{\pu{m^3}}\right].$$

Share Cite Improve this answer Follow edited Jan 27, 2021 at 11:35 andselisk♦ 38.4k1414 gold badges133133 silver badges223223 bronze badges answered May 14, 2018 at 13:52 Francesco BoiFrancesco Boi 19911 silver badge77 bronze badges $\endgroup$ Add a comment | 1 $\begingroup$

You really don't need to overcomplicate things for this answer.

The key things worth knowing is that in an ideal gas (a good approximation for most at standard conditions (0°C and standard atmospheric pressure)) one mole of the gas will occupy 22.4 L of volume. A mixture of gases isn't any different and to know the weight of the gas you want you just need to multiply the molar mass of the gas with the proportion in the mixture (ppm is the proportion here).

So each ppm of methane will contribute about 16 / 1,000,000 g to each 22.4L of the gas mixture. Or (adjusting for the volume conversion to cubic metres which contain 1,000 L) 44.7 * 16 /1,000,000 g/cubic metre.

By this formula, a cubic metre of pure methane would weigh ~715g at STP so you could just work with that by multiplying by the ppm value.

It only gets more complicated if you need proportions by mass in the mixture: then you have to know the molar masses of all the other components. But, if you stick with volumes, gas laws keep things really simple.

PS if your conditions (pressure or temperature are different) the only thing you need to adjust is the volume an ideal gas under those conditions (molar volume is closer to 24.8 L at 25 °C, for example).

Share Cite Improve this answer Follow answered Apr 3, 2018 at 13:25 matt_blackmatt_black 37.7k44 gold badges9191 silver badges184184 bronze badges $\endgroup$ Add a comment |

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