Problem 162 - 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) \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{3}\) (c) \(\mathrm{C}_{7} \mathrm{H}_{10} \mathrm{O}_{4}\) (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{5}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{5}\). Errors occur if ratios are not correctly simplified or normalized.

Step 1: Assume 100 grams of compound

To simplify calculations, assume the total mass of the compound is 100 grams. This allows us to directly convert the percent composition to grams: - Carbon: 54.0 grams - Hydrogen: 6.00 grams - Oxygen: 40.0 grams.

Step 2: Convert mass to moles

Use the molar mass of each element to convert grams to moles:- Moles of Carbon: \[ \frac{54.0 \text{ g}}{12.01 \text{ g/mol}} = 4.498 \text{ moles} \]- Moles of Hydrogen: \[ \frac{6.00 \text{ g}}{1.008 \text{ g/mol}} = 5.952 \text{ moles} \]- Moles of Oxygen: \[ \frac{40.0 \text{ g}}{16.00 \text{ g/mol}} = 2.500 \text{ moles} \]

Step 3: Determine the simplest ratio

Divide all the mole values by the smallest number of moles calculated to get the simplest whole number ratio:- Carbon: \[ \frac{4.498}{2.500} \approx 1.799 \]- Hydrogen: \[ \frac{5.952}{2.500} \approx 2.381 \]- Oxygen: \[ \frac{2.500}{2.500} = 1 \]

Step 4: Normalize to whole numbers

Multiply each of these numbers by an appropriate factor to convert them to the nearest whole number. Here, multiplying by approximately 5 helps clear fractions:- Carbon: \[ 1.799 \times 5 = 9.00 \approx 9 \]- Hydrogen: \[ 2.381 \times 5 = 11.905 \approx 12 \]- Oxygen: \[ 1 \times 5 = 5 \]

Step 5: Verify empirical formula

The resulting empirical formula is \(\mathrm{C}_{9}\mathrm{H}_{12}\mathrm{O}_{5}\). Compare this with the options given:- (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) \(\mathrm{C}_{9} \mathrm{H}_{12} \mathrm{O}_{5}\)Thus, option (d) matches the calculated empirical formula.

Key Concepts

Molar Mass Calculations

To determine the empirical formula of a compound, we first need to convert the percent composition of each element into moles. This process requires us to use the molar mass of each element. The molar mass is essentially the weight of one mole of an element, measured in grams per mole (g/mol). For example, the molar mass of carbon (C) is 12.01 g/mol, hydrogen (H) is 1.008 g/mol, and oxygen (O) is 16.00 g/mol. By dividing the mass of each element given in the compound by its molar mass, we find the number of moles present. This conversion is critical as it facilitates comparing elements in a common scale, allowing further steps like calculating the simplest ratio.

Stoichiometry

Stoichiometry provides the tools to relate quantities of reactants and products in chemical reactions. In the context of determining an empirical formula, stoichiometry involves converting masses calculated from percent composition into moles, and then using these moles to find the simplest ratio. Once moles are determined, the next step is to calculate the ratio between them. This is done by dividing each mole quantity by the smallest number of moles present. Stoichiometry helps ensure that these ratios reflect the simplest whole number relationships between the elements in the compound. It is a method based on the law of conservation of mass, which ensures that matter is not created or destroyed in a chemical reaction.

Percent Composition

Percent composition refers to the percentage by mass of each element in a compound. It is a crucial step in deducing empirical formulas. In practice, chemists express it in terms of the element's contribution to the overall mass of the compound. For example, in a compound where carbon constitutes 54.0% of the mass, hydrogen 6.00%, and oxygen 40.0%, these percentages effectively express how much of the 100 grams (if we assume the total compound weighs this much) each element contributes.

• Carbon contributes 54.0 grams
• Hydrogen contributes 6.0 grams
• Oxygen contributes 40.0 grams

By knowing these values, you can calculate the empirical formula by determining the actual moles of each element present.

Chemical Formula Calculation

Calculating the chemical or empirical formula involves the process of establishing the relative number of atoms for each element present in a compound. After determining the moles of each element, the next step is to find the simplest ratio by dividing each value by the smallest number. In this problem, dividing the number of moles of carbon, hydrogen, and oxygen by the smallest number of moles results in a mole ratio. However, these ratios are often not perfect integers. Multiplying each ratio by the same factor can help achieve whole numbers. For instance, if the ratio for carbon, hydrogen, and oxygen is approximately 1.799:2.381:1, we multiply by 5 to get close to whole numbers, resulting in a ratio of 9:12:5. Finally, we convert these numbers into subscripts in the chemical formula, thereby providing a clear, understandable empirical formula for the compound.