Problem 91 - Chemistry The Science in Context

Question

Adenine (135.14 g/mol; 44.44\% \(\mathrm{C}, 3.73 \% \mathrm{H}, \text { and } 51.84 \% \mathrm{N})\) was detected in mixtures of HCN, ammonia, and water under conditions that simulate early Earth. This observation suggests a possible origin for one of the bases found in DNA. What are the empirical and molecular formulas for adenine?

Step-by-Step Solution

Answer

Answer: The empirical formula for adenine is CHN, and the molecular formula is C5H5N5.

Step 1: Calculate the moles of each element in one mole of adenine.

We are given the percentage composition of adenine. Let's assume we have 100 g of adenine, so the mass of each element in this quantity will be: - Mass of Carbon (C) = 44.44% * 100 g = 44.44 g - Mass of Hydrogen (H) = 3.73% * 100 g = 3.73 g - Mass of Nitrogen (N) = 51.84% * 100 g = 51.84 g Now, we'll convert the mass of each element to moles using their respective molar masses: - Moles of Carbon (C) = \(\frac{44.44}{12.01}=3.70\) - Moles of Hydrogen (H) = \(\frac{3.73}{1.01}=3.69\) - Moles of Nitrogen (N) = \(\frac{51.84}{14.01}=3.70\)

Step 2: Determine the simplest whole number ratio

Divide the number of moles of each element by the smallest number of moles calculated in step 1 (which is 3.69) to get the mole ratio: - Carbon (C) ratio = \(\frac{3.70}{3.69}\approx1\) - Hydrogen (H) ratio = \(\frac{3.69}{3.69}=1\) - Nitrogen (N) ratio = \(\frac{3.70}{3.69}\approx1\) The empirical formula is therefore C\(_1\)H\(_1\)N\(_1\) or simply CHN.

Step 3: Calculate the molar masses of the empirical and molecular formulas

Calculate the molar mass of the empirical formula (CHN): - Molar mass of CHN = 12.01 (C) + 1.01 (H) + 14.01 (N) = 27.03 g/mol Now, compare the molar mass of the empirical formula with the given molar mass of adenine (135.14 g/mol): - Molecular formula multiplier = \(\frac{135.14}{27.03}\approx5\)

Step 4: Determine the Molecular Formula

Multiply the empirical formula by the molecular formula multiplier calculated in step 3 to obtain the molecular formula: C\(_{(1\times5)}\)H\(_{(1\times5)}\)N\(_{(1\times5)}\) = C\(_5\)H\(_5\)N\(_5\) So, the molecular formula for adenine is C\(_5\)H\(_5\)N\(_5\).

Key Concepts

Empirical Formula Calculation

When calculating the empirical formula, you are essentially finding the simplest whole number ratio of atoms in a compound. You start with the percentage composition of each element in the compound.
For adenine, the given percentages are:

• Carbon (C): 44.44%
• Hydrogen (H): 3.73%
• Nitrogen (N): 51.84%

Assume you have 100 grams of the compound. This makes it easier to convert percentages directly into grams.
The next step is to convert these masses into moles. To do this, divide the mass of each element by its atomic weight:

• Moles of Carbon (C): \( \frac{44.44\text{ g}}{12.01\text{ g/mol}} = 3.70 \text{ moles} \)
• Moles of Hydrogen (H): \( \frac{3.73\text{ g}}{1.01\text{ g/mol}} = 3.69 \text{ moles} \)
• Moles of Nitrogen (N): \( \frac{51.84\text{ g}}{14.01\text{ g/mol}} = 3.70 \text{ moles} \)

Finally, establish the simplest whole number ratio by dividing each mole value by the smallest one calculated. Here, 3.69 (moles of Hydrogen) is the smallest value. Dividing gives the ratios for each element:

• Carbon: 1
• Hydrogen: 1
• Nitrogen: 1

Therefore, the empirical formula for adenine is CHN, which represents the basic atomic composition in terms of the smallest whole numbers.

Molecular Formula Calculation

The molecular formula is essentially the true number of each type of atom in a compound's molecules. After determining the empirical formula CHN for adenine, we need to calculate the molecular formula based on the compound's given molar mass.
The empirical formula has a molar mass of 27.03 g/mol (since 12.01 for Carbon, 1.01 for Hydrogen, and 14.01 for Nitrogen total 27.03).
The molar mass of adenine is given as 135.14 g/mol.
To find out how many times the empirical formula repeats in the molecular formula, divide the molecular molar mass by the empirical molar mass:
\[\text{Multiplier} = \frac{135.14 \text{ g/mol}}{27.03 \text{ g/mol}} \approx 5\]This tells us that the empirical formula CHN occurs 5 times in the actual molecule. Hence, we multiply each subscript in the CHN formula by 5:

• Carbon: \(1 \times 5 = 5\)
• Hydrogen: \(1 \times 5 = 5\)
• Nitrogen: \(1 \times 5 = 5\)

Therefore, the molecular formula of adenine is \(\text{C}_5\text{H}_5\text{N}_5\). This formula indicates the exact number of each type of atom within a molecule of adenine.

Percentage Composition

Percentage composition helps in understanding a compound's structure by providing the percentage by mass of each element in the compound. To really grasp this for adenine, it is crucial to relate it back to its chemical formula.
For instance, when we're given a percentage composition:

• 44.44% Carbon (C)
• 3.73% Hydrogen (H)
• 51.84% Nitrogen (N)

This tells us how much of each element is present in a 100 gram sample of the compound. The values align with the mass calculations done for determining the empirical formula. Understanding the percentage composition allows us to set a foundational block for further calculations such as empirical and molecular formulas.
These percentages provide a direct pathway to finding the weight proportions, ensuring we start our empirical calculations with accurate approximations!
Grasping percentage composition is key to getting the whole picture of the compound's makeup. It acts as the starting blueprint on which further chemical formula calculations depend.