19 A 1724 mg sample of a naturally occurring dental anesthet

19. A 172.4 mg sample of a naturally occurring dental anesthetic was placed in an evacuated flask with a volume of 500.0 mL at 280.0°C. The pressure that the compound exerted in the flask under those conditions was found to be 48.3 mm Hg. In a combustion experiment, 18.80 mg of the compound was burned to give 50.39 mg of carbon dioxide and 12.36 mg of water. What is the molecular formula of the compound?

Solution

Ans. Step 1: Determine molar mass of the compound:

Given

            Volume, V = 500.0 mL = 0.500 L

            Temperature, T = 280.0⁰C = 553.15 K

            Pressure, P = 48.3 mm Hg = (48.3 / 760) atm = 0.0635526 atm

Using Ideal gas equation:    PV = nRT      - equation 1

            Where, P = pressure in atm

            V = volume in L                   

            n = number of moles

            R = universal gas constant= 0.0821 atm L mol^-1K^-1

            T = absolute temperature (in K) = (⁰C + 273.15) K

Putting the values in above equation-

            0.0635526 atm x 0.500 L = n x (0.0821 atm L mol^-1K^-1) x 553.15 K

            Or, n = 0.0317763 atm L / (45.413615 atm L mol^-1)

            Hence, n = 6.9971 x 10^-4 mol

# Therefore, moles of the compound in flask = 6.9971 x 10^-4 mol

# Given, mass of sample = 172.4 mg = 0.1724 g

Now,

            Molar mass of the compound = Mass / number of moles

                                                = 0.1724 g / (6.9971 x 10^-4 mol)

                                                = 246.39 g/ mol

# Step 2: Determine simple molar ratios of C, H and O:

# Moles of CO₂ produced = Mass / Molar mass

= 0.05039 g / (44.0098 g/ mol)

= 0.00114497 mol

Moles of H₂O produced = 0.01236 g / (18.01528 g/ mol) = 0.00068608 mol

# Note that-

I. 1 mol CO_2­ contains 1 mol C.

Hence, number of moles of C in sample = 0.00114497 mol

II. 1 mol H₂O contains 2 mol H-atoms.

So,

Moles of H in sample = 2 x moles of H₂O produced

                                    = 2 x 0.00068608 mol

                                    = 0.00137216 mol

# Total mass of C- and H in the sample = (Moles of C x MW) + (Moles of H x MW)

            = (0.00114497 mol x 12.011 g mol^-1) + (0.00137216 mol x 1.00794 g mol^-1)

            = 0.0151352896204 g

            = 15.135 mg

# We have, total mass of sample taken = 18.80 mg

Remaining un-combusted mass = 18.80 mg – 15.135 mg = 2.965 mg

The un-combusted mass of oxygen because if it were any other element, respective oxide would have been produced.

So,

            Moles of O = 0.00295 g / (15.9994 g/ mol) = 0.00018532 mol

# Smallest molar ratio of C : H : O =

(0.00114497 mol : 0.00137216 mol : 0.00018532 mol) / 0.00018532 mol       = 6.18 : 7.40 : 1

Therefore, molar ratio of elements in the compound = C_6.18H_7.40O

# Step 3: Determine molecular formula:

# Molar mass of (C_6.18H_7.40O) = 97.28296 g/ mol

# We have calculated molar mass of the compound = 246.39 g/ mol

# Let the actual formula of compound be n(C_6.18H_7.40O)

So,

            n = Actual molar mass / Empirical molar mass

            Or, n = (246.39 g/ mol) / (97.28296 g/ mol)

            Hence, n = 2.53

# So actual formula of the compound = 2.53 x (C_6.18H_7.40O)

                                    = C_15.6H_18.72O_2.53

                                    = C₁₆H₁₉O₃ (nearest whole number ration)

Therefore, molecular formula of the compound = C₁₆H₁₉O₃

Note: Round off the stoichiometric ratio to nearest whole number in the final step only.