If the anesthetic mixture is inspired at the rate of $\pu{100 mL/min}$, what mass of halothane, $\ce{CHCIBrCF3}$, molecular weight $M = \pu{197.4 g/mol}$, is inspired in one minute if the partial pressure of halothane is $\pu{7.6 torr}$ and the temperature is $\pu{21 ^\circ C}$.

  1. $\pu{0.08g}$
  2. $\pu{0.80g}$
  3. $\pu{1.80g}$
  4. $\pu{3.36g}$

My working: \begin{align} P &= \pu{0.01 atm} &&(\pu{1 atm} = \pu{760 torr})\\ T &= \pu{294 K} &&(273 + 21) \\ V &= \pu{1 L} &&(\pu{1000 mL} = \pu{1 L})\\ R &= 0.08 &&(\text{L-atm})\\ \end{align} Equation: \begin{align} 0.01 \times 1 &= n \times 0.08 \times 294\\ 0.01 &= n \times 23.52\\ n &= 4.3 \end{align}

Which suggest $\pu{4.3 mol}$ of $\ce{CHCIBrCF3}$ which would give the mass $\approx\pu{849g}$.

That is way off any of the answers. Can someone see what I have done wrong?

  • 5
    $\begingroup$ It looks you have two errors. You used 1L instead of .1 and in your algebra you multiplied where you should have divided. $\endgroup$
    – Tyberius
    Jul 27, 2017 at 4:54

1 Answer 1


Volumetric Flow Rate (VFR) is given to you. By applying ideal gas equation to the gas, in terms of volumetric flow rate, $$\frac{PV}{t} = \frac{nRT}{t}$$

And in an ideal gas mixture (the assumption), the partial pressure of a gas is what is exerts alone in same volume in same temperature as of the mixture. $$\frac{P_\mathrm{gas}\cdot V_\mathrm{tot}}{t} = \frac{n_\mathrm{gas}\cdot R\cdot T_\mathrm{total}}{t}$$

amount of substance = mass of the compound/molar mass (or molecular weight of the compound) $$n = \frac{m}{M}$$

$$\frac{P_\mathrm{gas}\cdot V_\mathrm{total}}{t} = \frac{(\frac{m_\mathrm{gas}}{M})\cdot R\cdot T_\mathrm{total}}{t}$$

Apply the values, mind the units: \begin{align} V &=\pu{0.1 L}\\ T &= (273+21)\pu{ K}\\ P_\mathrm{gas} &= \pu{0.01 atm}\\ R &=\pu{0.082057 L atm mol^-1K^-1}\\ M_\mathrm{gas} &=\pu{197.4 g mol^-1}\\ t &= \pu{1 min}\\ \end{align}

$$\frac{\pu{0.01 atm}\cdot\pu{0.1 L}}{\pu{1 min}} = \frac{ \frac{m_\mathrm{gas}}{\pu{197.4 g mol^-1}} \cdot\pu{0.082057 L atm mol^-1 K^-1}\cdot\pu{294 K}}{\pu{1 min}}$$

For $m_\mathrm{gas}$ I can see the value in your multiple choice.


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.