# How to calculate the amount of chlorine emitted yearly by a leak in the air conditioning system?

A leak in the air conditioning system of an office building releases $12~\mathrm{kg}$ of $\ce{CHF2Cl}$ per month. If the leak continues, how many kilograms of $\ce{Cl}$ will be emitted into the atmosphere each year?

My working out is as follows:

\begin{align} 12~\mathrm{kg} &= \mathrm{CHF}_2\mathrm{Cl} \\ \text{Cl mass in compound} &= \frac{35.45}{86.448}\times100\% = 41\% \\ \end{align}

$41\%$ of $12~\mathrm{kg}$ is $4.92~\mathrm{kg}$, so I multiplied it by $12$ to get the annual result.

$$4.92~\mathrm{kg} \times 12(\text{months}) = 59.04~\mathrm{kg\ year}^{-1}$$

I don't have access to the answer so I'm not sure whether or not my answer is correct. Also, this process seems a little to simple to be correct. I say this because the question prior to it was very difficult. Can someone please tell me if I am right, and if I'm not, where did I go wrong?

• Given the details you provided, your method is correct.
– Jori
May 31 '14 at 8:43

The mass flow rate of the leaking substance, under the assumption of linear scaling, is $$\dot{m}_{\ce{CHF2Cl}} = 12 ~\mathrm{kg\, month^{-1}} = 144~\mathrm{kg\, a^{-1}}\; .$$
The mass flow rate of chlorine, as contained in the leaking substance, is then calculated as follows: $$\dot{m}_{\ce{Cl}} = \dot{m}_{\ce{CHF2Cl}} \cdot \frac{M_{\ce{Cl}}}{M_{\ce{CHF2Cl}}} = 144~\mathrm{kg\, a^{-1}} \cdot \frac{35.453~\mathrm{g\,mol^{-1}}}{86.47~\mathrm{g\,mol^{-1}}} = 59.04~\mathrm{kg\, a^{-1}}$$
For the end result to the question "how many kilograms of chlorine will be emitted into the atmosphere each year?", please note that the mass flow rate still has a dependency on time $$\dot{m} = \frac{\mathrm{d} m}{\mathrm{d} t}\; ,$$ which means that we have to do an integration so we can calculate the mass:
$$m(\tau) = \int_0^\tau \dot{m}\,\mathrm{d}t = \dot{m}\,\tau$$
For one year, we get $$m_{\ce{Cl}}(1~\mathrm{a}) = \dot{m}_{\ce{Cl}}\cdot1~\mathrm{a} = 59.04~\mathrm{kg}\; ,$$ which means that your attempt at answering your own question were very good, on spot and totally valid.