Question

A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products have cooled to 20.0 °C at 1 bar, there are 4.41 L of CO2 and 3.26 mL of H2O . The density of water at 20.0 °C is 0.998 g/mL.

Answer 1

Explanation:

First, calculate the moles of $CO_{2}$ using ideal gas equation as follows.

                PV = nRT

or,          n = $\frac{PV}{RT}$

                = $\frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}$      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = $\frac{mass}{volume}$

Hence, mass of water will be as follows.

                Density = $\frac{mass}{volume}$

             0.998 g/ml = $\frac{mass}{3.26 ml}$    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = $\frac{mass}{\text{molar mass}}$

                              =  $\frac{3.25 g}{18.02 g/mol}$            

                              = 0.180 mol

Moles of hydrogen = $0.180 \times 2$ = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = $\frac{mass}{\text{molar mass}}$

          0.183 mol =  $\frac{mass}{12 g/mol}$            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = $\frac{mass}{\text{molar mass}}$

                               =  $\frac{0.95 g}{16 g/mol}$            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is $C_{3}H_{6}O$.

Thus, we can conclude that empirical formula of the given compound is $C_{3}H_{6}O$.