How to calculate what mass of halothane is inspired in one minute given the partial pressure and temperature?

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?

• It looks you have two errors. You used 1L instead of .1 and in your algebra you multiplied where you should have divided. – Tyberius Jul 27 '17 at 4:54

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.