Conversion from a PPB value to µg/m3 of Isobutylene

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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:

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