Combustion Analysis: Videos & Practice Problems

Combustion analysis is a method used to determine the empirical formula of a compound by analyzing the products of combustion reactions. Typically, compounds like hydrocarbons (containing only carbon and hydrogen) or those with carbon, hydrogen, and oxygen react with oxygen gas to produce carbon dioxide (CO₂) and water (H₂O). For example, octane (C₈H₁₈), a hydrocarbon, and glucose (C₆H₁₂O₆), which contains oxygen in addition to carbon and hydrogen, both yield CO₂ and H₂O upon combustion. However, non-hydrocarbons, which may include elements like sulfur, nitrogen, or halogens (fluorine, chlorine, bromine, iodine), can produce additional gaseous products such as sulfur dioxide (SO₂), nitrogen dioxide (NO₂), and diatomic molecules (F₂, Cl₂, Br₂, I₂) during combustion. Understanding the typical products of combustion reactions is crucial for performing accurate combustion analysis and determining the empirical formulas of unknown compounds.

Empirical Formula via Combustion

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.

Combustion of Carbon Compounds Example

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.

An unknown compound is composed of carbon, hydrogen and oxygen. A 4.30 g sample is ignited and creates 8.59 g CO₂ and 3.52 g H₂O. If the molar mass is 176.22 g/mol, what is the molecular formula?

• A. C₃H₅O₃
• B. C₂H₄O₄
• C. C₈H₁₆O₄
• D. C₁₂H₁₆O₄

Answer: (C)

In the presence of a small amount of oxygen a combustion reaction will not only produce carbon dioxide, but also carbon monoxide. The incomplete combustion of naphthalene, a hydrocarbon used in many dyes, produced 2.80 g CO, 4.40 g CO₂ and 1.44 g H₂O. Determine its empirical formula.

• A. C₃H₃O₃
• B. C₅H₄
• C. C₅H₄O
• D. C₁₀H₈

Answer: (B)

Combustion of Non-Hydrocarbons

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.

Molecular Formula via Combustion Example

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.

A compound composed of carbon, hydrogen and nitrogen undergoes a combustion reaction to produce 264.21 g CO₂, 63.06 g H₂O and 46.4 g NO₂. Determine its empirical formula.

• A. C₂H₁₁N₃
• B. C₆H₇N
• C. C₆H₁₂N₄
• D. CHN

Answer: (B)

The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO₂ and 2.52 g H₂O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?

The analysis of a compound containing carbon, hydrogen, oxygen, and fluorine begins with the combustion of a sample weighing 4.16 grams, which produces 7.7 grams of carbon dioxide and 2.52 grams of water. From these products, we can determine the amounts of carbon and hydrogen in the original compound. The mass of carbon is calculated using the molar mass of carbon dioxide, where 1 mole of CO₂ weighs 44.01 grams, and 1 mole of carbon weighs 12.01 grams. This results in approximately 2.1012 grams of carbon.

For hydrogen, we analyze the water produced. Since 1 mole of water (H₂O) weighs 18.016 grams and contains 2 moles of hydrogen, we find the mass of hydrogen to be about 0.2820 grams by using the molar mass of hydrogen (1.008 grams per mole).

Next, to find the mass of oxygen and fluorine in the compound, we utilize the law of definite proportions. A second sample of the compound weighing 3.63 grams contains 0.58 grams of fluorine. By setting up a proportion based on the mass of fluorine in both samples, we can determine that the 4.16-gram sample contains approximately 0.6647 grams of fluorine.

With the masses of carbon, hydrogen, and fluorine known, we can calculate the mass of oxygen by subtracting the total mass of these elements from the original sample mass. This yields about 1.1121 grams of oxygen.

Now, we convert the masses of each element into moles by dividing by their respective atomic masses: carbon (12.01 g/mol), hydrogen (1.008 g/mol), oxygen (16.00 g/mol), and fluorine (19.00 g/mol). This results in approximately 0.1750 moles of carbon, 0.2798 moles of hydrogen, 0.0695 moles of oxygen, and 0.0350 moles of fluorine.

To find the empirical formula, we divide the number of moles of each element by the smallest number of moles calculated (0.0350). This gives us the ratios: 5 for carbon, 8 for hydrogen, 2 for oxygen, and 1 for fluorine. Thus, the empirical formula of the compound is C₅H₈O₂F.

Here’s what students ask on this topic:

What is combustion analysis?

Combustion analysis is a laboratory procedure used to determine the elemental composition of a compound, particularly organic compounds containing elements like carbon and hydrogen. In this process, a sample of the substance is burned in the presence of excess oxygen to ensure complete combustion. The combustion products typically include carbon dioxide and water, and sometimes nitrogen oxides if nitrogen is present in the original compound.

By measuring the masses of these combustion products, you can calculate the ratios of the elements within the original compound. For example, the mass of carbon dioxide produced can be used to determine the amount of carbon in the sample, while the mass of water can be used to find the amount of hydrogen. This technique is particularly useful for determining the empirical formulas of unknown organic compounds. It's a fundamental method in chemical analysis and has applications in fields like environmental science, pharmaceuticals, and materials science.

How to do combustion analysis?

Combustion analysis is a method used to determine the elemental composition of a compound, particularly hydrocarbons. To perform a combustion analysis, follow these steps:

1. Weigh the Sample: Accurately weigh a known mass of the compound being analyzed.
2. Burn the Sample: The compound is combusted in a stream of oxygen, which converts the carbon in the compound to carbon dioxide (CO2) and the hydrogen to water (H2O).
3. Absorb the Products: The combustion gases are passed through substances that absorb the CO2 and H2O. Anhydrous calcium chloride can be used to absorb water, and sodium hydroxide can be used to absorb carbon dioxide.
4. Weigh the Absorbents: After the combustion process, the increase in mass of the absorbents is measured. This increase corresponds to the amount of water and carbon dioxide produced.
5. Calculate the Composition: Use the masses of CO2 and H2O to calculate the number of moles of carbon and hydrogen in the original sample. From the moles, you can determine the percentage composition of carbon and hydrogen.
6. Determine Other Elements: If there are elements other than carbon and hydrogen, such as sulfur or nitrogen, additional steps are required to measure the products of their combustion (e.g. sulfur dioxide).

How do you calculate the empirical formula from combustion analysis?

To calculate the empirical formula from combustion analysis, follow these steps:

1. Combustion Data: Obtain the mass of carbon dioxide (CO₂) and water (H₂O) produced during the combustion.
2. Moles of Carbon and Hydrogen: Convert these masses to moles by dividing by their respective molar masses (44.01 g/mol for CO₂ and 18.02 g/mol for H₂O).
3. Elemental Moles: Calculate the moles of carbon and hydrogen. Each mole of CO₂ contains one mole of carbon, and each mole of H₂O contains two moles of hydrogen.
4. Oxygen Moles: If the mass of the original compound is known, subtract the mass of carbon and hydrogen from the total mass to find the mass of oxygen. Then convert this mass to moles by dividing by the molar mass of oxygen (16.00 g/mol).
5. Simplest Ratio: Determine the simplest whole number ratio of moles of each element by dividing all the moles by the smallest number of moles calculated.
6. Empirical Formula: Use this ratio to write the empirical formula, with each element symbol followed by its corresponding subscript indicating the number of atoms in the simplest ratio.

Remember, the empirical formula represents the simplest whole number ratio of elements in a compound.