Problem 55 - Chemistry The Central Science

Question

(a) Combustion analysis of toluene, a common organic solvent, gives 5.86 \(\mathrm{mg}\) of \(\mathrm{CO}_{2}\) and 1.37 \(\mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\) . If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O} . \mathrm{A}\) 0.1005 -gsample of mentholis combusted, producing 0.2829 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) and 0.1159 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) What is the empirical formula for menthol? If menthol has a molar mass of 156 \(\mathrm{g} / \mathrm{mol}\) what is its molecular formula?

Step-by-Step Solution

Answer

The empirical formula for toluene is CH. For menthol, the empirical formula is C6H13O, and the molecular formula is C12H26O2.

Step 1: Convert mass of CO2 and H2O to moles.

First, let's convert the mass of CO2 and H2O to moles using their molar masses: Molar mass of CO2 = \(44.01\ \text{g/mol}\), Molar mass of H2O = \(18.02\ \text{g/mol}\) \[ \text{moles of CO}_2 = \frac{5.86\ \text{mg} \cdot (1\ \text{g} / 1000\ \text{mg})}{44.01\ \text{g/mol}} = 1.33 \times 10^{-4}\ \text{mol} \] \[ \text{moles of H}_2\text{O} = \frac{1.37\ \text{mg} \cdot (1\ \text{g} / 1000\ \text{mg})}{18.02\ \text{g/mol}} = 7.61 \times 10^{-5}\ \text{mol} \]

Step 2: Convert moles of CO2 and H2O to moles of C and H.

Since each mole of CO2 has 1 mole of carbon and each mole of H2O has 2 moles of hydrogen, we can obtain the moles of C and H as follows: \[ \text{moles of C} = 1.33 \times 10^{-4}\ \text{mol} \] \[ \text{moles of H} = 2 \times (7.61 \times 10^{-5}\ \text{mol}) = 1.52 \times 10^{-4}\ \text{mol} \]

Step 3: Determine the empirical formula of toluene.

Now divide both quantities by the smaller value (moles of C) to get the molar ratio: \[ \frac{\text{moles of H}}{\text{moles of C}} = \frac{1.52 \times 10^{-4}\ \text{mol}}{1.33 \times 10^{-4}\ \text{mol}} \approx 1.14 \approx 1 \] Since the molar ratio is approximately 1:1, the empirical formula for toluene is CH. #For Menthol (Part b):#

Step 4: Follow Steps 1-2 for menthol.

Following the same steps as for toluene, first calculate the moles of CO2 and H2O and then the moles of C and H for menthol: \[ \text{moles of CO}_2 = \frac{0.2829\ \text{g}}{44.01\ \text{g/mol}} = 0.00643\ \text{mol} \] \[ \text{moles of H}_2\text{O} = \frac{0.1159\ \text{g}}{18.02\ \text{g/mol}} = 0.00643\ \text{mol} \] \[ \text{moles of C} = 0.00643\ \text{mol} \] \[ \text{moles of H} = 2 \times (0.00643\ \text{mol}) = 0.0129\ \text{mol} \]

Step 5: Calculate moles of O in menthol.

To find the moles of O in menthol, subtract the mass of C and H from the total mass of menthol and divide it by the molar mass of O (16.00 g/mol): \[ \text{mass of O} = 0.1005\ \text{g} - (0.00643\ \text{mol} \times 12.01\ \text{g/mol}) - (0.0129\ \text{mol} \times 1.008\ \text{g/mol}) = 0.0161\ \text{g} \] \[ \text{moles of O} = \frac{0.0161\ \text{g}}{16.00\ \text{g/mol}} = 0.00101\ \text{mol} \]

Step 6: Determine the empirical formula of menthol.

Divide the moles of C, H, and O by the smallest value (moles of O) to get the molar ratio: \[ \text{C}:\text{H}:\text{O} = \frac{0.00643}{0.00101}:\frac{0.0129}{0.00101}:\frac{0.00101}{0.00101} \approx 6:13:1 \] The empirical formula of menthol is C6H13O.

Step 7: Determine the molecular formula of menthol.

Now calculate the molar mass of the empirical formula C6H13O: \[ \text{Molar mass} = 6(12.01\ \text{g/mol}) + 13(1.008\ \text{g/mol}) + 16.00\ \text{g/mol} = 100.18\ \text{g/mol} \] Divide the given molar mass of menthol (156 g/mol) by the molar mass of the empirical formula (100.18 g/mol) to get the multiplier: \[ \text{Multiplier} = \frac{156\ \text{g/mol}}{100.18\ \text{g/mol}} \approx 1.56 \approx 2 \] The molecular formula of menthol is thus twice the empirical formula: C12H26O2.

Key Concepts

Combustion Analysis

Combustion analysis is a laboratory method used to determine the elemental composition of a compound, particularly organic compounds containing carbon, hydrogen, and sometimes other elements such as sulfur or nitrogen. The process involves burning a small amount of the substance to produce combustion products, which are then analyzed.

For example, in a typical combustion analysis for carbon and hydrogen, the compound is burned in an excess of oxygen. The carbon in the compound is converted into carbon dioxide (CO₂), and the hydrogen is converted into water (H₂O). By measuring the mass of CO₂ and H₂O produced, and knowing their molar masses, we can calculate the amount of carbon and hydrogen in the original compound. This calculation involves converting the mass of the combustion products to moles, which then leads to determination of the empirical formula of the compound.

In our exercise, the compound toluene underwent combustion analysis resulting in specific masses of CO₂ and H₂O. These measures were critical to finding the ratio of carbon to hydrogen atoms, thus revealing the empirical formula.

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the mass of substances to the number of particles they contain. A mole is defined as the quantity of substance that contains as many elementary entities (atoms, molecules, ions, electrons, etc.) as there are atoms in exactly 12 grams of carbon-12. This number is known as Avogadro's number and is approximately 6.022 x 10²³ entities per mole.

In our exercise, the mole concept was used to convert the measured mass of CO₂ and H₂O to the number of moles. This conversion is key to understanding the relationship between the mass of a substance and its elemental composition. As often seen in textbook exercises, the stoichiometry of the reaction between elements and compounds is analyzed through the lens of the mole concept, giving us a clear and accurate understanding of chemical proportions.

Stoichiometry

Stoichiometry is the study of the quantitative relationships, or ratios, in chemical reactions and compositions. It is derived from the Greek words 'stoicheion' (element) and 'metron' (measure). Stoichiometry calculates the amounts of reactants and products involved in a chemical reaction, based on the balanced chemical equation.

In the context of our exercise involving combustion analysis, stoichiometry is used to calculate the number of moles of carbon and hydrogen from the moles of CO₂ and H₂O produced. For every mole of CO₂, we can infer there is one mole of carbon, and for every mole of H₂O, there are two moles of hydrogen. Using these stoichiometric ratios, we can then determine the empirical and molecular formulas of the compound by comparing the molar amounts of each element.

Molar Mass

Molar mass is the mass of one mole of a substance measured in grams per mole (g/mol). It numerically equals the average atomic or molecular weight of a substance in unified atomic mass units (u). The molar mass allows chemists to convert between mass and amount of substance using the mole concept.

In our solution, the molar mass was essential in converting the mass of CO₂ and H₂O to moles, and also in the final step of determining the molecular formula. Once the empirical formula was found, the molar mass of this empirical unit was calculated and compared to the given molar mass of menthol. The ratio of the molar masses provided the multiplier needed to convert the empirical formula to the molecular formula. This final piece connects the stoichiometric 'puzzle' allowing the discernment between empirical and molecular formulas based on mass data.