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?

  • 2
    $\begingroup$ Given the details you provided, your method is correct. $\endgroup$
    – Jori
    May 31, 2014 at 8:43

1 Answer 1


For the sake of correct formalism, here the answer as it should be given in the textbook:

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.