# Conversion from a PPB value to µg/m3 of Isobutylene

I have a quite simple problem, found a lot of information about it, but I am not sure anymore if I do my calculations right.

I own a sensor, which reports measurements in isobutylene units as PPB. So if I understand this right, the measurement would be the number of isobutylene molecules per one billion (x/1'000'000'000).

Therefore to convert this value into the more common µg/m3 unit, I would just use this formula:

$$\text{Concentration}\ \frac{µg}{m3} = \frac{\text{Concentration}\ \text{PPB}\times\text{Molecular Mass}\ \frac{g}{mol}}{\text{MolarVolume}\ l}$$

Therefore to convert 400 PPM of isobutylene, with a molecular mass of 56.106 g/mol or 0.0005879 g/l, in the molar volume of a gas at STP with 22.4 l, the calculation would be:

$$\frac{400\times56.106}{22.4} = 1001.9 \frac{µg}{m^3}$$

This seemed very high to me, and the Range of the sensor is 0-1056 PPM.

Is this the correct formula for the conversion?

This is much simpler than you think, even though the value you've obtained looks reasonable to me. Part per billion ($$\pu{1 ppb} = \pu{1e-9}$$) on its own is meaningless. Based on a context, it can refer to anything: mass fraction, mole fraction, particles etc. The thing is, for gaseous mixtures $$\pu{ppb}$$ always refers to the volume fraction $$\phi_i$$:

$$\varphi_i = \frac{V_i}{V}$$

where $$V_i$$ is the volume of the $$i$$th gas (here, isobutylene) and $$V$$ is the total volume. In order to convert volume fraction expressed in $$\pu{ppb}$$ to desired mass concentration $$\rho_i$$ expressed as

$$ρ_i = \frac{m_i}{V}$$

where $$m_i$$ is the mass of $$i$$th component in a mixture, you have to use density $$d_i$$:

$$ρ_i = \frac{m_i}{V} = \frac{m_i}{V_i}\varphi_i = d_i\varphi_i$$

To conform with your units of choice ($$\pu{μg m-3}$$) we need to juggle the units a bit as the density is usually given in $$\pu{kg m-3}$$, e.g. for gaseous isobutylene at NTP $$d(\ce{C4H8}) = \pu{2.3959 kg m-3}$$ (Source):

\begin{align} \pu{1 g} &= \pu{1e6 μg} \\ d(\ce{C4H8}) &= \pu{2.3959e6 μg m-3} \end{align}

Now, using this density value you can multiply your volume fraction in $$\pu{ppb}$$ or $$\pu{ppm}$$ and get an answer, e.g. for $$\pu{400 ppm}$$:

$$ρ_i = d(\ce{C4H8})\varphi (\ce{C4H8}) = \pu{2.3959e6 μg m-3}\cdot\pu{400e-6} = \pu{958 μg m-3}$$

Note that all notations involving "part per something" are deprecated and should be avoided. It also wouldn't hurt if you ask manufacturer whether $$\pu{1 ppb}$$ the sensor reports is indeed $$\pu{1e-9}$$ and not $$\pu{1e-12}$$ (just in case) if you think the reported values are too high.

• @downvoter I'd be grateful if you leave a comment about what's wrong with the answer and how it can be improved. – andselisk Feb 17 '19 at 20:09

400ppm means if you have million molecules, 400 of them would be isobutylene in this case. To convert ppm into µg/m3:

1. Find the number of gas molecule in $$1m^3$$ (using PV=nRT and Avogardo's const.)
2. Find the number of molecule of Isobutylene in $$1m^3$$ (400ppm)
3. Convert the number of Isobutylene molecules into moles to find the mass (Using molar mass)
4. Change the mass unit from g to µg

By the way, your formula is directly multiplying grams/liter with ratio of number of molecules, which gives non-physical number.

According to Environmental Science and Technology Briefs for Citizens (USA), concentrations in soil can be reported as parts per million (ppm) or parts per billion (ppb) where $$\pu{1 ppm} = \pu{1000 ppb}$$. For concentrations in soil, $$\pu{1 ppm} = \pu{1 mg/kg}$$ of contaminant in soil, and $$\pu{1 ppb} = \pu{1 \mu g/kg}$$.

For concentrations in water, $$\pu{1 ppm} \approx \pu{1 mg/L}$$ (also written as $$\pu{mg/L}$$) of contaminant in water, and $$\pu{1 ppb} \approx \pu{1 \mu g/L}$$ (also written as $$\pu{\mu g/L}$$).

Since, $$\pu{1 L} = \pu{1 dm^3}$$ and $$\pu{1 m^3} = \pu{1 dm^3} \times \frac{\pu{1 m^3}}{\pu{10^3 dm^3}}$$, your conversion for $$\pu{400 ppb}$$ should have been:

$$\pu{400 ppb} = \pu{400 \mu g/L} = 400 \times \frac{\pu{1 \mu g}}{\pu{1 dm^3}} \times \frac{\pu{10^3 dm^3}}{\pu {1 m^3}} = \pu{400 \times 10^3 \frac{\mu g}{m^3}}= \pu{400 \frac{mg}{m^3}}$$