Chapter 24, Problem 24.54QP

The combustion of 3.795 mg of liquid B, which contains only C, H, and O, with excess oxygen gave 9.708 mg of CO₂ and 3.969 mg of H₂O. In a molar mass determination, 0.205 g of B vaporized at 1.00 atm and 200.0°C and occupied a volume of 89.8 mL. Derive the empirical formula, molar mass, and molecular formula of B and draw three plausible structures.

Interpretation Introduction

Interpretation:

Empirical formula, molar mass and molecular formula of B have to be derived; three plausible structures have to be drawn.

Concept introduction:

Steps to calculate empirical formula:

• Convert the mass of elements into moles.
• Divide each mole value by the smallest number of moles calculated.
• Round to the nearest whole number.

Number of moles = Molarity $\text{×}$ volume

$\text{Number of moles}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Explanation of Solution

Calculate moles of each given elements:

$\text{C:}\left(\text{9}{\text{.708×10}}^{\text{-3}}\text{g}{\text{CO}}_{\text{2}}\right)\frac{\text{1}\text{mol}{\text{CO}}_{\text{2}}}{\text{44}\text{.01 g}{\text{CO}}_{\text{2}}}\text{×}\frac{\text{1}\text{mol}\text{C}}{\text{1}\text{mol}{\text{CO}}_{\text{2}}}\text{=2}\text{.206}{\text{×10}}^{\text{-4}}\text{mol}\text{C}$

$\text{H:}\left(\text{3}{\text{.969×10}}^{\text{-3}}\text{g}{\text{H}}_{\text{2}}\text{O}\right)\frac{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}{\text{18}\text{.02 g}{\text{H}}_{\text{2}}\text{O}}\text{×}\frac{\text{2}\text{mol}\text{H}}{\text{1}\text{mol}{\text{H}}_{\text{2}}\text{O}}\text{=}\text{4}\text{.405}{\text{×10}}^{\text{-4}}\text{mol}\text{H}$

The mass of oxygen is found by difference:

Convert moles of carbon into mass in mg: $\text{Mass}\text{=}\left(\text{2}\text{.206}{\text{×10}}^{{}^{\text{-4}}}\text{mol}\text{C}\right)\left(\text{12}\text{g}\text{/mol}\right)\text{=}\text{2}\text{.647mg}$

Convert moles of hydrogen into mass in mg: $\text{Mass}\text{=}\left(\text{4}\text{.405}{\text{×10}}^{{}^{\text{-4}}}\text{mol}\text{H}\right)\left(\text{1}\text{g}\text{/mol}\right)\text{=}\text{0}\text{.44mg}$

$\text{3}\text{.795}\text{mg}\text{compound}\text{-}\left(\text{2}\text{.649}\text{mg}\text{C}\text{+}\text{0}\text{.440}\text{mg}\text{H}\right)\text{=}\text{0}\text{.706}\text{mg}\text{O}$

$\text{O:}\left(\text{0}{\text{.706×10}}^{\text{-3}}\text{g}\text{O}\right)\text{×}\frac{\text{1}\text{mol}\text{O}}{\text{16}\text{.00}\text{g}\text{O}}\text{=}\text{4}{\text{.41×10}}^{\text{-5}}\text{mol}\text{O}$

This gives the formula ${\text{C}}_{\text{2}{\text{.206×10}}^{\text{-4}}}{\text{H}}_{\text{4}{\text{.405×10}}^{\text{-4}}}{\text{O}}_{\text{4}{\text{.41×10}}^{\text{-5}}}$. Dividing the smallest number of moles gives the empirical formula, $C_5{H}_{10}O$

Now, let’s calculate moles using the ideal gas equation, and then calculate the molar mass.

$\text{n}\text{=}\frac{\text{PV}}{\text{RT}}\text{=}\frac{\left(\text{1}\text{.00}\text{atm}\right)\left(\text{0}\text{.0898}\text{L}\right)}{\left(\text{0}\text{.0821}\text{L}\text{.atm/K}\text{.mol}\right)\left(\text{473}\text{K}\right)}\text{=}\text{0}\text{.00231mol}$

$\text{Molar mass = }\frac{\text{g}\text{of}\text{substance}}{\text{mol}\text{of}\text{substance}}\text{=}\frac{\text{0}\text{.205}\text{g}}{\text{0}\text{.00231}\text{mol}}\text{=}\text{88}\text{.7}\text{g/}\text{mol}$

The formula mass of $C_5{H}_{10}O$ is $\text{86}\text{.13}\text{g}$, so is the molecular formula. Three possible structures are,