Answer: A certain anesthetic contains 64.9 percent C, 13.5

Step 1 of 2

The goal of the problem is to find the molecular formula of the anesthetic compound.

Given:

T = 120°C = 120 + 273 = 393 K

P = 750 mmHg

V = 1.00 L

Mass of compound \(=2.30 \mathrm{~g}\)

Mass percent of \(\mathrm{C}=64.9 \%\)

Mass percent of \(\mathrm{H}=13.5 \%\)

Mass percent of \(\mathrm{O}=21.6 \%\)

First, let's convert pressure in mmHg to atm:

\(1 \mathrm{~atm}=760 \mathrm{mmHg}\)

Therefore, \(750 \mathrm{mmHg}\) in atm is:

\(=750 \mathrm{mmHg} \times 760 \mathrm{mmHg}\)

\(=0.986 \mathrm{~atm}.\)

We know, the ideal gas equation:

\(\mathrm{PV}=\mathrm{nRT}\)

First, let's find the number of moles:

\(\begin{aligned}\mathrm{n}=& \frac{P V}{R T} \\=& \quad\left[\mathrm{R}=\text { gas constant }=0.0821 \mathrm{~L} \cdot \mathrm{atm} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\right] \\&=0.986 \mathrm{~atm} \times 1.0 \mathrm{~L} \\&=0.0305 \text { moles }\end{aligned}\)

Now, let's calculate the molar mass:

We know the formula to calculate the number of moles:

\(\text { Number of moles }(\mathrm{n})=\frac{\operatorname{mass}(\text { in } g)}{\operatorname{Molarmass}(M)}\)

This can be rewritten as:

\(\begin{aligned}\mathrm{M} &=\frac{\operatorname{mass}(\operatorname{in} g)}{n \text { mberof moles }} \\&=\frac{2.30 \mathrm{~g}}{0.0305 \mathrm{~mol}} \\&=75.40 \mathrm{~g} / \mathrm{mol}\end{aligned}\)