Problem 38 - General Chemistry: Principles and Modern Applications

Question

Adenine, a component of nucleic acids, has the mass percent composition: \(44.45 \%\) C \(, 3.73 \%\) H, \(51.82 \%\) N. Its molecular mass is 135.14 u. What is its molecular formula?

Step-by-Step Solution

Answer

The molecular formula of Adenine is \(CHN\)

Step 1: Convert mass percent into grams and find moles

First, convert the mass percentage into grams. Assume that the total mass is 100 g, thus there are \(44.45 g\) of Carbon, \(3.73 g\) of Hydrogen, and \(51.82 g\) of Nitrogen. Then, divide each mass by the atomic mass to find the number of moles: \(C: \frac{44.45 g}{12.01 g/mol} = 3.70 mol\) , \(H: \frac{3.73 g}{1.01 g/mol} = 3.69 mol\), \(N: \frac{51.82 g}{14.01 g/mol} = 3.70 mol\)

Step 2: Find element ratio

Find the ratio of each element to the smallest number of moles. In this case, the smallest number of moles is \(3.69 mol\) (Hydrogen). So the ratio of C, H, and N to H are as follows: \(C: \frac{3.70 mol}{3.69 mol} \approx 1\), \(H: \frac{3.69 mol}{3.69 mol} = 1\), \(N: \frac{3.70 mol}{3.69 mol} \approx 1\)

Step 3: Convert to whole numbers

Since the ratios found are all approximate to 1 and are very close to whole numbers, there is no need for a multiplier. However, if the ratio is a fraction, a suitable multiplier (2, 3, etc.) should be used to convert the ratio into whole numbers.

Step 4: Write the molecular formula

Combine the results from the previous steps to write the molecular formula. Use the symbols of the elements (C for Carbon, H for Hydrogen, and N for Nitrogen), followed by the subscript of the number ratio which in this case is all 1. The molecular formula is therefore \(C_1H_1N_1\) or simply \(CHN\)

Key Concepts

Mass Percent Composition

When determining the molecular formula of a compound, understanding mass percent composition is crucial. It tells us how much of each element is present in a compound by mass. For example, if you know that a compound contains 44.45% Carbon, 3.73% Hydrogen, and 51.82% Nitrogen, you understand the portion of each element's mass relative to the compound. To utilize these percentages in calculations, assume a total mass of 100 grams. This allows you to convert the percentages directly into grams:

• 44.45 grams of Carbon
• 3.73 grams of Hydrogen
• 51.82 grams of Nitrogen

This simplifies calculations and gives a clearer picture of the element distribution in the compound, which is essential for further steps in molecular formula determination.

Elemental Analysis

Elemental analysis involves determining the elemental composition of a compound. Once you know the mass of each element (in grams, from the mass percentages), you use elemental analysis to convert these masses into moles. This conversion is done by dividing the mass of each element by its respective atomic mass. For example:

• Carbon's atomic mass is 12.01 g/mol. Therefore, for 44.45 grams of Carbon: \( \frac{44.45}{12.01} = 3.70 \text{ moles} \).
• Hydrogen's atomic mass is 1.01 g/mol. Therefore, for 3.73 grams of Hydrogen: \( \frac{3.73}{1.01} = 3.69 \text{ moles} \).
• Nitrogen's atomic mass is 14.01 g/mol. Therefore, for 51.82 grams of Nitrogen: \( \frac{51.82}{14.01} = 3.70 \text{ moles} \).

Convert the mass to moles helps in comparing the amounts of different elements in a compound, paving the way for deducing their ratio, which is a key step in establishing the molecular formula.

Moles Calculation

After finding the moles of each element from their respective masses, the next step is to establish their ratios. The smallest number of moles acts as the baseline for this ratio comparison. For example, if the smallest mole count is 3.69 from Hydrogen, divide all mole quantities by 3.69. This yields:

• Carbon: \( \frac{3.70}{3.69} \approx 1 \)
• Hydrogen: \( \frac{3.69}{3.69} = 1 \)
• Nitrogen: \( \frac{3.70}{3.69} \approx 1 \)

If the ratios are near whole numbers without requiring further conversion, it suggests a simple and direct formula for the compound. Any fractional ratio should be multiplied by the smallest possible integer to obtain whole numbers.Thus, from these ratios, the molecular formula derived in this case is \(CHN\), showing that each element is present in approximately equal amounts, aligning with the calculated mole ratios.