Combustion Analysis: Videos & Practice Problems

Combustion analysis is a valuable method for determining the empirical formula of a compound, particularly through the use of combustion reactions. These reactions typically involve hydrocarbons, which are compounds made up solely of carbon and hydrogen, or compounds that include carbon, hydrogen, and oxygen. During combustion, these compounds react with oxygen gas (O₂), resulting in the formation of carbon dioxide (CO₂) and water (H₂O).

For instance, consider octane, a hydrocarbon with the molecular formula C₈H₁₈. When octane undergoes combustion with O₂, it produces CO₂ and H₂O. Similarly, glucose, which has the molecular formula C₆H₁₂O₆, also reacts with O₂ to yield CO₂ and H₂O. In both cases, the key focus is on the products formed—carbon dioxide and water—rather than the balancing of the chemical equations.

This analytical technique is particularly useful because it allows chemists to derive the empirical formula of a compound by analyzing the amounts of CO₂ and H₂O produced during combustion. By measuring the mass of these products, one can infer the original composition of the hydrocarbon or organic compound, thus providing insight into its empirical formula.

The combustion of carbon compounds can be analyzed to determine their empirical formulas. In this example, we will explore the empirical formula of 2,3-dihydroxytartaric acid, which consists of carbon, hydrogen, and oxygen. Given that the combustion of 12.01 grams of this acid produces 14.08 grams of carbon dioxide (CO₂) and 4.32 grams of water (H₂O), we can derive its empirical formula through a series of calculations.

First, we convert the grams of carbon dioxide into grams of carbon. The molar mass of CO₂ is 44.01 grams per mole, which allows us to find the moles of CO₂:

\[ \text{Moles of CO}_2 = \frac{14.08 \text{ g CO}_2}{44.01 \text{ g/mol}} \approx 0.3199 \text{ moles CO}_2 \]

Since each mole of CO₂ contains one mole of carbon, we find:

\[ \text{Grams of Carbon} = 0.3199 \text{ moles} \times 12.01 \text{ g/mol} \approx 3.8423 \text{ g C} \]

Next, we convert the grams of water into grams of hydrogen. The molar mass of H₂O is 18.016 grams per mole:

\[ \text{Moles of H}_2\text{O} = \frac{4.32 \text{ g H}_2\text{O}}{18.016 \text{ g/mol}} \approx 0.2400 \text{ moles H}_2\text{O} \]

Since each mole of H₂O contains two moles of hydrogen, we calculate:

\[ \text{Grams of Hydrogen} = 0.2400 \text{ moles} \times 2 \times 1.008 \text{ g/mol} \approx 0.4834 \text{ g H} \]

To find the grams of oxygen, we subtract the grams of carbon and hydrogen from the total mass of the acid:

\[ \text{Grams of Oxygen} = 12.01 \text{ g} - (3.8423 \text{ g C} + 0.4834 \text{ g H}) \approx 7.6843 \text{ g O} \]

Now, we convert the masses of each element into moles:

\[ \text{Moles of Carbon} = \frac{3.8423 \text{ g}}{12.01 \text{ g/mol}} \approx 0.3199 \text{ moles C} \]

\[ \text{Moles of Hydrogen} = \frac{0.4834 \text{ g}}{1.008 \text{ g/mol}} \approx 0.4796 \text{ moles H} \]

\[ \text{Moles of Oxygen} = \frac{7.6843 \text{ g}}{16 \text{ g/mol}} \approx 0.4803 \text{ moles O} \]

Next, we divide each mole value by the smallest mole value (0.3199) to obtain a ratio:

\[ \text{C: } \frac{0.3199}{0.3199} = 1, \quad \text{H: } \frac{0.4796}{0.3199} \approx 1.5, \quad \text{O: } \frac{0.4803}{0.3199} \approx 1.5 \]

Since we have 1.5 for both hydrogen and oxygen, we multiply all ratios by 2 to obtain whole numbers:

\[ \text{C: } 1 \times 2 = 2, \quad \text{H: } 1.5 \times 2 = 3, \quad \text{O: } 1.5 \times 2 = 3 \]

Thus, the empirical formula for 2,3-dihydroxytartaric acid is C₂H₃O₃.

In the study of combustion reactions, it is essential to differentiate between hydrocarbons and non-hydrocarbons. Hydrocarbons are compounds composed solely of carbon and hydrogen, while non-hydrocarbons contain additional elements such as sulfur, nitrogen, or halogens. Halogens, which are found in Group 7A of the periodic table, include fluorine (F), chlorine (Cl), bromine (Br), and iodine (I).

When conducting combustion analysis on non-hydrocarbons, the products can vary significantly from those of hydrocarbons. Instead of only producing carbon dioxide (CO₂) and water (H₂O), the combustion of non-hydrocarbons can yield gaseous products such as sulfur dioxide (SO₂) and nitrogen dioxide (NO₂). Additionally, halogens in their diatomic forms (F₂, Cl₂, Br₂, I₂) may also be present in the products.

This variation in products highlights the complexity of combustion reactions involving non-hydrocarbons. It is crucial to recognize that the introduction of elements like sulfur and nitrogen leads to the formation of new types of final products, which differ from the typical outcomes observed with hydrocarbons. Understanding these differences is vital for accurately predicting the results of combustion reactions involving non-hydrocarbons.

To determine the molecular formula of a solvation agent with a molar mass of 147 grams per mole, which contains carbon, hydrogen, and chlorine, we start by analyzing the combustion products of a 0.250 grams sample. The combustion yields 0.451 grams of carbon dioxide and 0.0617 grams of water. The process involves several steps to convert these masses into moles of each element.

First, we convert the grams of carbon dioxide to grams of carbon. The molar mass of carbon dioxide (CO₂) is 44.01 grams, which consists of 1 carbon atom and 2 oxygen atoms. Using the relationship that 1 mole of CO₂ contains 1 mole of carbon, we find:

Mass of carbon = (0.451 g CO₂) × (1 g C / 44.01 g CO₂) × (12.01 g C / 1 g C) = 0.1231 g C.

Next, we convert the grams of water to grams of hydrogen. The molar mass of water (H₂O) is 18.016 grams, comprising 2 hydrogen atoms and 1 oxygen atom. Thus, we calculate:

Mass of hydrogen = (0.0617 g H₂O) × (2 g H / 18.016 g H₂O) × (1.008 g H / 1 g H) = 0.0069 g H.

Now, we subtract the masses of carbon and hydrogen from the total sample mass to find the mass of chlorine:

Mass of chlorine = 0.250 g sample - (0.1231 g C + 0.0069 g H) = 0.1200 g Cl.

Next, we convert the masses of each element into moles by dividing by their respective atomic masses:

• Moles of carbon = 0.1231 g C / 12.01 g/mol = 0.01025 moles C.
• Moles of hydrogen = 0.0069 g H / 1.008 g/mol = 0.00685 moles H.
• Moles of chlorine = 0.1200 g Cl / 35.45 g/mol = 0.00339 moles Cl.

To obtain whole number ratios, we divide each mole value by the smallest mole value (0.00339 moles Cl):

• Carbon: 0.01025 / 0.00339 = 3.02 ≈ 3
• Hydrogen: 0.00685 / 0.00339 = 2.02 ≈ 2
• Chlorine: 0.00339 / 0.00339 = 1

This gives us the empirical formula C₃H₂Cl. To find the molecular formula, we need to calculate the empirical mass:

Empirical mass = (3 × 12.01 g) + (2 × 1.008 g) + (1 × 35.45 g) = 73.496 g/mol.

Now, we determine the value of n by dividing the given molar mass (147 g/mol) by the empirical mass:

n = 147 g/mol / 73.496 g/mol = 2.

Finally, we multiply the empirical formula by n to find the molecular formula:

Molecular formula = (C₃H₂Cl)₂ = C₆H₄Cl₂.

Thus, the molecular formula of the solvation agent is C₆H₄Cl₂.