Question

When analyzed, an unknown compound gave these experimental results: \(\mathbf{C}, 54.0 \% ; \mathbf{H}, \mathbf{6} .00 \% ;\) and \(\mathbf{O}, 40.0 \%\) 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) \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{3}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{O}_{4}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{5}\)

Short answer

The correct answer is (d) \(\mathrm{C}_9\mathrm{H}_{12}\mathrm{O}_5\) by converting percentages to moles and simplifying the ratios.

Step-by-step solution

Determine the number of moles of each element

First, we need to convert the percentage composition to grams. Assume you have a 100 g sample of the compound. This means you have 54.0 g of Carbon (C), 6.00 g of Hydrogen (H), and 40.0 g of Oxygen (O).Now, convert these masses to moles using the molar mass of each element:- Molar mass of C = 12.01 g/mol- Molar mass of H = 1.008 g/mol- Molar mass of O = 16.00 g/molCalculate the moles of each:\[ \text{moles of C} = \frac{54.0 \text{ g}}{12.01 \text{ g/mol}} \approx 4.50 \text{ mol} \]\[ \text{moles of H} = \frac{6.00 \text{ g}}{1.008 \text{ g/mol}} \approx 5.95 \text{ mol} \]\[ \text{moles of O} = \frac{40.0 \text{ g}}{16.00 \text{ g/mol}} = 2.50 \text{ mol} \]

Determine the mole ratio

Next, we find the simplest whole number ratio by dividing the number of moles of each element by the smallest number of moles calculated in Step 1, which is 2.50 mol (for Oxygen).Calculate the ratio:\[ \text{Ratio of C}: \frac{4.50}{2.50} = 1.80 \]\[ \text{Ratio of H}: \frac{5.95}{2.50} = 2.38 \]\[ \text{Ratio of O}: \frac{2.50}{2.50} = 1.00 \]

Adjust the mole ratio to whole numbers

Since the ratios are not whole numbers, multiply each mole ratio by a common factor to convert them to the nearest whole numbers. Here we can multiply each by 5:\[ \text{Adjusted Ratio of C}: 1.80 \times 5 = 9 \]\[ \text{Adjusted Ratio of H}: 2.38 \times 5 \approx 12 \]\[ \text{Adjusted Ratio of O}: 1.00 \times 5 = 5 \]

Write the empirical formula

Using the whole number ratios obtained in Step 3, the empirical formula is formed as:\[ \text{Empirical formula}: \mathrm{C}_9\mathrm{H}_{12}\mathrm{O}_5 \]

Verify the correct empirical formula answer

The correct empirical formula from the options given is (d) \(\mathrm{C}_9\mathrm{H}_{12}\mathrm{O}_5\). It accurately reflects the simplified whole number ratios calculated from the given percentages.

Key concepts

Mole Ratio

The mole ratio is a crucial step in determining the empirical formula of a compound. It involves comparing the number of moles of each element present based on given data, often expressed as percentages. To calculate this ratio, you first need to convert the percentage of each element into grams, assuming a 100 g sample for simplicity. Then, you calculate moles by dividing the mass of each element by its molar mass.
Once you have the moles for all elements, the mole ratio is determined by dividing each mole value by the smallest number of moles present among the elements. This helps to break down the amounts into their simplest form, allowing you to derive a basic comparison of how much more an element is present in relation to others. By adjusting these ratios to whole numbers, you identify the simplest formula depicting the atom count from each element. The mole ratio of each element in the empirical formula should reflect these whole numbers.

Percentage Composition

Percentage composition gives insight into the proportion of each element within a compound. It's represented as the percentage by mass of each element. For example, in a compound with 54.0% carbon, 6.00% hydrogen, and 40.0% oxygen, these percentages highlight the mass contribution of each element to the total mass of the compound.
To use percentage composition in empirical formula calculations, you convert these percentages to grams (assuming a 100 g sample, which makes calculation easier). Each percentage directly translates into grams (e.g., 54.0% becomes 54.0 g), which is then used to find the mole ratio for the elements in the compound.
By understanding percentage composition, you are effectively using it to dissect the compound into its elemental make-up, thus enabling further calculations like finding mole ratios and eventually deducing the empirical formula.

Molar Mass Calculation

Calculating molar mass is a foundational step in analyzing a compound's elemental composition. The molar mass of an element is the mass of one mole of its atoms, typically expressed in grams per mole (g/mol). Knowing this allows us to convert from mass to moles, bridging the gap between experimental data and theoretical mole ratios.
To calculate the molar mass, one must use atomic weights from the periodic table: for example, carbon has a molar mass of 12.01 g/mol, hydrogen is 1.008 g/mol, and oxygen is 16.00 g/mol. By dividing the mass of each element in the sample by its molar mass, you determine the moles of each element present.

• Moles of an element = Mass of the element (g) / Molar mass (g/mol)

This conversion is essential for obtaining the mole ratio, a subsequent step in finding the empirical formula of the compound.

Chemical Analysis

Chemical analysis involves applying quantitative techniques to decipher the composition of an unknown compound. In this context, it usually means determining the empirical formula, which requires converting experimental findings into useful data.
This process is grounded in accurate measurements of elemental percentages followed by conversion into grams. Assuming a 100 g sample is a common strategy, simplifying percentage conversion into mass directly. Next, using the periodic table's molar masses, these masses are turned into moles, which are the fundamental part of further calculations.
Through chemical analysis, you determine the smallest whole number ratio between elements by converting mole results. If necessary, they are adjusted to achieve whole numbers, reflecting the actual number of atoms in an empirical formula. Such analysis bridges practical experimentation with theoretical models, aiding in predicting compound behavior and reactions.