Combustion Apparatus: Videos & Practice Problems

Combustion analysis is a method used to determine the empirical formula of a compound by analyzing the products of its combustion. In this process, a sample is vaporized in a combustion apparatus consisting of multiple chambers. Oxygen gas is introduced as a reactant, essential for combustion. As the sample transitions to a gaseous state, any hydrogen present is converted to water, while nonmetals form gases such as:

• CO₂ for carbon
• NO₂ for nitrogen
• SO₂ for sulfur
• X₂ for halogens

Excess oxygen may also be present post-reaction. Understanding the function of each chamber in the apparatus is crucial for interpreting results and solving related analytical problems in chemistry.

Combustion Apparatus

Combustion analysis is a vital analytical technique used to determine the empirical formula of a compound. This process involves a specialized combustion apparatus designed to facilitate the combustion of a sample, allowing for the identification of its elemental composition. The apparatus typically consists of multiple chambers, labeled A through D, each serving a specific function in the analysis.

Initially, the sample is placed in chamber A, where it is vaporized in the presence of oxygen gas (O₂), a crucial reactant in combustion reactions. Once vaporized, the sample moves to chamber B. Here, if the sample contains hydrogen, it is converted into water (H₂O). Additionally, any nonmetal components of the sample are transformed into gaseous products. For instance, if carbon is present, it will typically be converted into carbon dioxide (CO₂); if nitrogen is present, it becomes nitrogen dioxide (NO₂); sulfur converts to sulfur dioxide (SO₂); and halogens form diatomic molecules.

In chamber C, the water produced is collected, while the gases generated from the nonmetals are also captured. The identity of these gases depends on the nonmetal present in the original sample. After the combustion process, any excess oxygen that remains may also be released, particularly in scenarios where the combustion occurs in an oxygen-rich environment.

Understanding the function of each chamber in the combustion apparatus is essential for interpreting the results of combustion analysis. This method not only reveals the empirical formula of the compound but also provides insights into its elemental composition, making it a fundamental technique in analytical chemistry.

Empirical Formula via Combustion Example

The combustion analysis of an experimental jet fuel, composed of carbon, hydrogen, and oxygen, allows us to derive its empirical formula through a systematic approach. Initially, we measure the mass of carbon dioxide (CO₂) and water (H₂O) produced during combustion, accounting for the initial amounts present in the combustion apparatus. By subtracting the initial masses from the final masses, we find that 22.0 grams of CO₂ and 13.5 grams of H₂O were generated.

Next, we convert the mass of CO₂ into moles to determine the amount of carbon. Using the molar mass of CO₂ (44.01 g/mol), we calculate:

\[\text{Moles of CO}_2 = \frac{22.0 \text{ g CO}_2}{44.01 \text{ g/mol}} \approx 0.4999 \text{ moles CO}_2\]

Since each mole of CO₂ contains one mole of carbon, we find that there are approximately 0.4999 moles of carbon, which corresponds to 6.0036 grams of carbon (C).

For the water, we perform a similar conversion. The molar mass of water is 18.016 g/mol, leading to:

\[\text{Moles of H}_2\text{O} = \frac{13.5 \text{ g H}_2\text{O}}{18.016 \text{ g/mol}} \approx 0.7493 \text{ moles H}_2\text{O}\]

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

\[\text{Moles of H} = 2 \times 0.7493 \approx 1.4987 \text{ moles H}\]

Next, we determine the mass of oxygen in the original sample. The total mass of the liquid is 11.5 grams, so we subtract the masses of carbon and hydrogen:

\[\text{Mass of O} = 11.5 \text{ g} - (6.0036 \text{ g C} + 1.5107 \text{ g H}) \approx 3.9857 \text{ g O}\]

Converting this mass into moles using the molar mass of oxygen (16 g/mol) gives us:

\[\text{Moles of O} = \frac{3.9857 \text{ g O}}{16 \text{ g/mol}} \approx 0.2491 \text{ moles O}\]

To find the empirical formula, we divide the moles of each element by the smallest number of moles obtained, which is for oxygen:

\[\text{C: } \frac{0.4999}{0.2491} \approx 2, \quad \text{H: } \frac{1.4987}{0.2491} \approx 6, \quad \text{O: } \frac{0.2491}{0.2491} \approx 1\]

Thus, the empirical formula of the unknown liquid is C₂H₆O. This formula indicates the simplest whole-number ratio of the elements present in the compound, which is essential for understanding its chemical composition and potential applications in jet fuel technology.

The following results are obtained from burning 5 g of an organic molecule (MW = 233.07 g/mol), which contains C, H, and O, in a combustion chamber:

Determine the molecular formula of linolenic acid.

• A. C₉H₁₅O
• B. C₃H₅O₁₂
• C. C₆H₅O
• D. C₁₈H₄₅O₃

Answer: (B)

Here’s what students ask on this topic:

What is combustion analysis and how is it used to determine the empirical formula of a compound?

Combustion analysis is an analytical method used to determine the empirical formula of a compound by analyzing the products of its combustion. In this process, a sample is vaporized in a combustion apparatus consisting of multiple chambers. Oxygen gas is introduced as a reactant, which is essential for combustion. As the sample transitions to a gaseous state, any hydrogen present is converted to water, while nonmetals form gases such as CO₂ for carbon, NO₂ for nitrogen, SO₂ for sulfur, and X₂ for halogens. By measuring the amounts of these products, the empirical formula of the original compound can be determined.

What are the key components of a combustion apparatus and their functions?

A combustion apparatus typically consists of multiple chambers, each with a specific function. Chamber A is where the sample is placed and vaporized, with oxygen gas introduced as a reactant. In Chamber B, hydrogen in the sample is converted to water, and nonmetals are converted to gases such as CO₂, NO₂, SO₂, or diatomic halogens. Chamber C traps the water produced, while the remaining gases continue to the next chamber. Excess oxygen may also be present post-reaction. Understanding the function of each chamber is crucial for interpreting the results of combustion analysis.

How does the presence of different nonmetals in a sample affect the products of combustion analysis?

The presence of different nonmetals in a sample affects the products of combustion analysis by forming specific gases. For example, if the nonmetal is carbon, it will form CO₂. If the nonmetal is nitrogen, it will form NO₂. If the nonmetal is sulfur, it will form SO₂. If the nonmetal is a halogen, it will form a diatomic molecule (e.g., Cl₂ for chlorine). These products are then measured to determine the empirical formula of the original compound.

Why is oxygen gas essential in a combustion reaction within a combustion apparatus?

Oxygen gas is essential in a combustion reaction within a combustion apparatus because it acts as a reactant that facilitates the combustion process. Combustion is a chemical reaction that typically involves the rapid combination of a substance with oxygen to produce heat and light. In the context of combustion analysis, oxygen helps to vaporize the sample and convert its components into measurable products such as CO₂, H₂O, NO₂, SO₂, and diatomic halogens. Without oxygen, the combustion reaction would not occur, and the analysis would not be possible.

What happens to excess oxygen in a combustion reaction within a combustion apparatus?

In a combustion reaction within a combustion apparatus, excess oxygen may remain after the reaction has taken place. This is because the reaction can occur in an environment that is rich in oxygen, leading to some oxygen not being consumed. The excess oxygen exits the apparatus along with the other gaseous products. It is important to account for this excess oxygen when analyzing the results of the combustion analysis to ensure accurate determination of the empirical formula of the compound.