# 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.

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:

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