Question

Write the balanced chemical equation for the complete combustion of adipic acid, an organic acid containing \(49.31 \% \mathrm{C}, 6.90 \% \mathrm{H},\) and the remainder \(\mathrm{O},\) by mass.

Short answer

The balanced equation for the combustion of adipic acid is: \[ 2\text{C}_6\text{H}_{10}\text{O}_4 + 15 \text{O}_2 \rightarrow 12 \text{CO}_2 + 10 \text{H}_2\text{O} \]

Step-by-step solution

Determine the empirical formula

First, assume a 100 g sample of adipic acid. This means it contains 49.31 g of C, 6.90 g of H, and 100 - (49.31 + 6.90) = 43.79 g of O. Convert these masses to moles: \[\text{Moles of C} = \frac{49.31}{12.01} = 4.106 \, \text{mol} \]\[\text{Moles of H} = \frac{6.90}{1.008} = 6.845 \, \text{mol} \]\[\text{Moles of O} = \frac{43.79}{16.00} = 2.737 \, \text{mol} \]

Find the mole ratio

Determine the smallest mole value and divide all moles by this smallest value to find the ratios:\[\text{Smallest mole value} = 2.737\]\[\text{Ratio for C} = \frac{4.106}{2.737} \approx 1.5 \]\[\text{Ratio for H} = \frac{6.845}{2.737} \approx 2.5 \]\[\text{Ratio for O} = \frac{2.737}{2.737} = 1 \]

Adjust the ratios to whole numbers

To convert these ratios to the smallest whole numbers, we multiply each by 2:\[\text{C: } 1.5 \times 2 = 3 \]\[\text{H: } 2.5 \times 2 = 5 \]\[\text{O: } 1 \times 2 = 2 \]The empirical formula is \( \mathrm{C_3H_5O_2} \).

Determine the molecular formula

Using the known molecular formula of adipic acid \( \mathrm{C_6H_{10}O_4} \), compare it to the empirical formula \( \mathrm{C_3H_5O_2} \). The molecular formula is twice the empirical formula, verifying \( \mathrm{C_6H_{10}O_4} \).

Write the balanced combustion equation

In the combustion of adipic acid: \[ \text{C}_6\text{H}_{10}\text{O}_4 + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O} \]Balance the equation:1. \( \text{C}_6\text{H}_{10}\text{O}_4 + 7 \text{O}_2 \rightarrow 6 \text{CO}_2 + 5 \text{H}_2\text{O} \)2. Simplifying the equation gives:\[ \text{C}_6\text{H}_{10}\text{O}_4 + \frac{15}{2} \text{O}_2 \rightarrow 6 \text{CO}_2 + 5 \text{H}_2\text{O} \]Multiply through by 2 for whole numbers:\[ 2\text{C}_6\text{H}_{10}\text{O}_4 + 15 \text{O}_2 \rightarrow 12 \text{CO}_2 + 10 \text{H}_2\text{O} \]

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of atoms in a compound. It's a foundational concept in chemistry that helps us understand the basic composition of compounds. To determine the empirical formula, you first need to convert the percentage composition of each element into grams, assuming you have a 100g sample. This exercise used 49.31% of carbon, 6.90% of hydrogen, and the remainder oxygen to calculate the grams of each element. Next, these grams are converted into moles using the molar mass (C: 12.01 g/mol, H: 1.008 g/mol, O: 16.00 g/mol). To find the empirical formula, identify the smallest number of moles calculated and divide all molecular numbers by this smallest value. This provides the ratio of each element, which can be adjusted to whole numbers, resulting in the empirical formula. Here, it's found to be \( \text{C}_3\text{H}_5\text{O}_2 \). This step is crucial for understanding the fundamental composition of organic molecules.

Molecular Formula

The molecular formula of a compound provides the actual numbers of atoms of each element in a molecule. While similar to the empirical formula, the molecular formula isn't necessarily the simplest form. It's essentially a multiple of the empirical formula. Knowing the molar mass of the compound, you can compare this with the empirical formula's molar mass to find the molecular formula. In adipic acid, the empirical formula is \( \text{C}_3\text{H}_5\text{O}_2 \), which needs to match the molecular mass of adipic acid. If the molecular mass is known, you can determine how many times the empirical formula fits into this mass. For adipic acid, the empirical formula was multiplied by two to yield the known compound formula, \( \text{C}_6\text{H}_{10}\text{O}_4 \). This verification confirms the identity and structure of the compound in question.

Combustion Reaction

Combustion reactions are fundamental in understanding how organic compounds react. In a combustion reaction, a compound typically combines with oxygen (\( \text{O}_2 \)) to produce carbon dioxide (\( \text{CO}_2 \)) and water (\( \text{H}_2\text{O} \)). These reactions must be balanced to obey the law of conservation of mass. Balancing involves ensuring the number of atoms of each element on the reactants side matches those on the products side. In the given exercise, adipic acid combustion is balanced as: 2\( \text{C}_6\text{H}_{10}\text{O}_4 \) + 15\( \text{O}_2 \) \( \rightarrow \) 12\( \text{CO}_2 \) + 10\( \text{H}_2\text{O} \). Ensuring that every component is accounted for helps us predict the products of combustion reactions and understand their broader chemical implications.

Organic Compounds

Organic compounds are the backbone of chemistry and biological systems. Characterized by containing carbon atoms, often bonded with elements like hydrogen, oxygen, nitrogen, and others, organic compounds range from simple molecules like methane (\( \text{CH}_4 \)) to complex proteins found in living organisms. Key features of organic compounds include:

• Carbon-to-carbon bonding forming chains or rings
• The presence of functional groups that define chemical properties
• Varied molecular structures, which allow for large diversity

Adipic acid, the subject of this exercise, is an organic acid utilized in the production of nylon. This organic compound highlights how diverse carbon compounds are employed in industry and nature. Understanding the properties and behaviors of organic compounds enriches our ability to create materials and understand biological systems.