Problem 113 - Chemistry and Chemical Reactivity

Question

When analyzed, an unknown compound gave these experimental results: \(\mathrm{C}, 54.0 \% ; \mathrm{H}, 6.00 \% ;\) and \(\mathrm{O}, 40.0 \%\) Four different students used these values to calculate the empirical formulas shown here. Which answer is correct? Why did some students not get the correct answer? (a) \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{O}_{4}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{5}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(\mathrm{C}_9\mathrm{H}_{12}\mathrm{O}_5\). Some students likely made calculation errors.

Step 1: Convert Percentages to Grams

Assume you have a 100g sample of the compound. This means the mass of carbon (C) is 54.0g, hydrogen (H) is 6.00g, and oxygen (O) is 40.0g.

Step 2: Convert Grams to Moles

Use the molar mass of each element to convert the grams to moles:\[ \text{For C: } \frac{54.0 \text{ g}}{12.01 \text{ g/mol}} = 4.50 \text{ moles} \]\[ \text{For H: } \frac{6.00 \text{ g}}{1.008 \text{ g/mol}} = 5.95 \text{ moles} \]\[ \text{For O: } \frac{40.0 \text{ g}}{16.00 \text{ g/mol}} = 2.50 \text{ moles} \]

Step 3: Determine the Mole Ratio

Divide each of the calculated moles by the smallest number of moles obtained:\[ \frac{4.50}{2.50} = 1.80 \] for C\[ \frac{5.95}{2.50} = 2.38 \] for H\[ \frac{2.50}{2.50} = 1 \] for O

Step 4: Simplify the Ratios to Small Whole Numbers

Multiply each value by the smallest factor that results in whole numbers:\[ \left(1.80 \right) \times 5 \approx 9 \] for C\[ \left(2.38 \right) \times 5 \approx 12 \] for H\[ 1 \times 5 = 5 \] for OThis results in the empirical formula \(\mathrm{C}_9\mathrm{H}_{12}\mathrm{O}_5\).

Step 5: Identify the Correct Answer

Upon comparing with the given choices, option (d) \(\mathrm{C}_9\mathrm{H}_{12}\mathrm{O}_5\) matches our calculated empirical formula.

Step 6: Conclusion on Incorrect Answers

Some students may not have converted grams to moles correctly, may have used incorrect atomic masses, or failed to simplify molecular ratios accurately leading to any of the incorrect options: (a) \(\mathrm{C}_4\mathrm{H}_5\mathrm{O}_2\), (b) \(\mathrm{C}_5\mathrm{H}_7\mathrm{O}_3\), or (c) \(\mathrm{C}_7\mathrm{H}_{10}\mathrm{O}_4\).

Key Concepts

Molar Mass Conversion

To get started with calculating the empirical formula, we first need to convert the percentage of each element into grams. For simplification, assume a 100g sample of the compound. This makes the percentages numerically equal to the mass in grams.

• For carbon (C), with 54.0% mass, you have 54.0g.
• For hydrogen (H), with 6.00% mass, you have 6.00g.
• For oxygen (O), with 40.0% mass, you have 40.0g.

Now, convert these masses into moles by using the molar mass of the respective elements. The molar mass is the weight of one mole of any element and is unique to each.

• Carbon has a molar mass of 12.01 g/mol, thus 54.0g equals 4.50 moles.
• Hydrogen's molar mass is 1.008 g/mol, thus 6.00g equals 5.95 moles.
• Oxygen's molar mass is 16.00 g/mol, thus 40.0g equals 2.50 moles.

Mole Ratio Determination

To determine the mole ratio, compare the number of moles for each element by dividing the amount of moles by the smallest number of moles. This step helps establish a ratio that links the elements together.

• For carbon, divide 4.50 moles by the smallest value, 2.50 moles, resulting in 1.80.
• For hydrogen, divide 5.95 moles by 2.50 moles, resulting in approximately 2.38.
• For oxygen, divide 2.50 moles by 2.50 moles, giving you exactly 1.

These ratios represent the relative number of each element present in the empirical formula. They demonstrate how each element's atoms relate to one another in the compound.

Simplifying Mole Ratios

Next, you need to simplify the mole ratios to the nearest whole numbers, as the empirical formula should consist of the smallest set of whole number ratios.
In this example, we saw non-whole number ratios like 1.80 for carbon and 2.38 for hydrogen. To convert these to whole numbers, multiply each ratio by the smallest factor that will eliminate the decimal.

• 1.80 multiplied by 5 results in approximately 9 for Carbon.
• 2.38 multiplied by 5 results in approximately 12 for Hydrogen.
• 1 multiplied by 5 remains 5 for Oxygen.

Now, the empirical formula can clearly be stated as \( \mathrm{C}_9 \mathrm{H}_{12} \mathrm{O}_5\).
This simplification process ensures that the ratios reflect the true proportionality of the elements in the simplest form, which is critical for accurate formula determination.

Error Analysis in Empirical Formulas

Empirical formula calculation involves a series of steps where small errors can lead to incorrect answers. Several factors can cause mistakes:

• Incorrect conversion from grams to moles due to wrong molar mass values.
• Failure to accurately calculate or convert the mole ratios, leading to floating decimal errors.
• Not multiplying by the correct factor to simplify ratios, hence arriving at fractional or improper ratios.

Errors at any point in this process can cause a student's outcome to be option (a) \(\mathrm{C}_4 \mathrm{H}_5 \mathrm{O}_2\), (b) \(\mathrm{C}_5 \mathrm{H}_7 \mathrm{O}_3\), or (c) \(\mathrm{C}_7 \mathrm{H}_{10} \mathrm{O}_4\) instead of the correct \(\mathrm{C}_9 \mathrm{H}_{12} \mathrm{O}_5\).
Paying attention to detail at every step is crucial to avoid these common pitfalls. Conducting a peer review or a secondary check can also help catch potential errors early in the process.