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Wednesday, 13 August 2008

How to convert ppm to mg/m3

According to BIPM(Bureau International des Poids et Mesures) “ppm” is “The term "ppm", meaning 10–6 relative value, or 1 in 106, or parts per million, is also used. This is analogous to the meaning of percent as parts per hundred”

According to Wikipedia “ppm” is “"Parts-per" notation is used, especially in science and engineering, to denote relative proportions in measured quantities; particularly in low-value (high-ratio) proportions at the parts-per-million (ppm), parts-per-billion (ppb), and parts-per-trillion (ppt) level.”

I think one example will be more clear. These days when everyone talks for ecology we in EU know that there is a limits of the sulfur contains in the liquid fuels. In the EU since 2007 the contain of sulfur shall not exceed 50 ppm(Euro IV). And from 2009 it must be 10 ppm.

What this digits mean actually – if you burn one litter of diesel then you will burn and 50 ppm of sulfur or 50 parts from all million. This means that you will burn 50 milliliters of sulfur! Of course the sulfur is not liquid – the stuff sulfur is a yellow crystal and it melt on 115.21 degrees. So it is impossible to be liquid in the diesel. This shows us that the ppm measurement is not so correct in when we talk about stuffs with different states – liquid and crystal.

Let see how the ppm is used in gases from chimneys and how it is used. Years ago when I have started to learn about measuring of pollutions in chimney gases everything was in mg/m3 or milligrams per cubic meter. This mean mass in volume which is also strange because we talk about gases and their masses differ and depends from temperatures and pressure. Many of the limits are given in these measurement units and if you saw the limits in EU, UK or even Bulgaria every limits are in mg/m3. For example in UK the limits of NOx in chimney gases on appliances which are new and work with liquid fuels is 450 mg/m3.

So how we should work to transfer ppm to mg/m3 at first let look for example when we have concentration of 35 ppm SO2. This mean that in one cubic meter we have 1 000 000 cubic centimeters. Then if we have 35 ppm this means that in one cubic meter there is 35 cubic centimeters of sulfur dioxide – 35 cm3/m3. Now we must find how the weight of these 35 cm3 is. And we have 25 degrees and 1 atm. pressure. From equation of Klapeyron

ρ = (M/22.4)*(ToP/PoT) – this give us a density of the gases in different temperature and pressure.
M – molecular mass of the stuff
To – normal temperature – 273K
Po – normal pressure 1.03 kg/cm2
T – measured temperature
P – measured pressure

Then in 25 Celsius degrees and 1 atm. the equitation will be

ρ = (64/22.4)*(273*1.03/1.03*298)
ρ = 2.85*0.916 = 2.617 kg/m3

So we have found the density of the 35 cubic centimeters of sulfur dioxide and it is 2.617 kg/m3. Now we could find how is the mass of sulfur dioxide in this 35 cm3:

1.35 cm3/m3 can be present as 35 x 10-6 m3/m3 = 0.000035 m3/m3
2.2.617 kg/m3 = 2 617 000 mg/m3
3.Then 2 617 000 mg/m3 * 0.000035 m3/m3= 91.56 mg/m3 sulfur dioxide

There is a simple way to calculate this here is the formula:

Xppm=(Ymg/m3*24.45)/(molecular weight)
Xmg/m3=(Xppm*molecular weight)/(24.45)
You should change the needed values.

And if you are so lazy try online converter here.


Michael said...

Try this--

mjekov said...

Thanks michael but mainly reason of mine post is to understand how to convert this and what is the mechanism of conversion.

Natalija said...

I am student in Faculty of Mechanical Engineering -Skopje on Environmental Management with master work "Decision making for Life cycle assessment and risk assessment in industrial process". I have finished Industrial Engineering and Management so I have forget chemistry staff even if was my best side in gymnasium.

So I found your side and saw that you are so good in your profession as chemical engineer, so could you help me with debts in converting emissions, please?

I have information from a company for their air and water releases in mg/m3 and mg/l and I need them in kg/kg Gross Product. It's about SO2 and CO emissions from steam petroleum jelly (мазут) boiler (we steal use it for heating:))and Cu and O2. It's a herbal tea industry.

From the formula that you have posted on the blog "Chemical Engineer"
Xppm=(Ymg/m3*24.45)/(molecular weight)
for the air emissions, I will have the kg of SO2 in the moment, (won't I?)and I need kg of SO2 and CO per year or er product, but I need only the average monthly bases of concentrations (SO2=1.569, CO=116 mg/m3, and dust 0.96mg/m3),
and for water emissions in the formula, witch density I should take as a parameter? 1kg/l or other?

Sorry for my 1 page questions, thank you in advance,

mjekov said...

Hi Natalija,
Could you tell me how much jelly you burn at year?
If you know the amount of jelly the answer is near to you. For example if you burn 1 000 kg per hour this will gave you 13 000 cubic meters chimney gases(this is a approximately and is given in CORINAIR).

1000 -> 13 000

Then you will receive from formula 13 000 m3 with concentration of SO2 = X. Let give the X just for the example 1000 mg/m3 concentration. Then we will have
13 000 m3 x 1000 mg/m3 =
13 000 000 mg SO2 or 13 kg of SO2 per hour.
So now you should put your concentration at the place of mine X or 1000.
I hope this is answer to your question about SO2.
For CO I am not completely sure what you mean. Do you mean CO2?
At the end water emissions. Do you ask me about the emissions which are IN the water or this ones who are in the water vapour?
In the first situation when the emissions are in the water then the density of 1kg/l is good choose but you should check the temperature of the water. Sometimes cooling water come out with 30 degrees and then the density falls.
In the situation when the SO2 is in the vapour then the things going totally different.
Please if you have other questions about this do not hesitate to ask me on my e-mail.

najla said...


I am a student of Chemical Engineering. Your solution for converting ppm to mg/m3 at elevated temperature and pressure seems valid.

But I am not sure about the equation of Klapeyron. Do u mind to share with me about your reference regarding the equation?

mjekov said...

Sorry najla,

Currently I am little busy right now if you wish contact with me on

#naymin said...

if you burn one litter of diesel then you will burn and 50 ppm of sulfur or 50 parts from all million. This means that you will burn 50 milliliters of sulfur!

I think 50 micro liter that should not be milliliter.

#naymin said...

if you burn one litter of diesel then you will burn and 50 ppm of sulfur or 50 parts from all million. This means that you will burn 50 milliliters of sulfur!

I think 50 micro liter that should not be milliliter.

angelitoidavid said...

I have a unique situation regarding the conversion of mg/cu m to ppm.

Could you pls help on how to convert said unit of volatile matters coming from a mixture of gasoline-diesel and water.
We are currently involved in the remediation of spilt diesel and gasoline product and in the remediation phase we are extracting the mixture gas-diesel and water. our standard is mg/cu.m.for the volatile matters generated by the mixture. what is the nearest conversion to ppm for this type of liquid.
Engr. Angelito I. David
Waste Solutions and management Services Inc.
Makati, Philippines

mjekov said...

Hello Angelito I. David ,

Please write me on

mjekov74 @

Just remove the space.

cielle said...

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