Question

What is the empirical formula of a neutral compound containing \(58.6 \%\) oxygen, \(39 \%\) sulfur, \(2.4 \%\) hydrogen by mass? A. \(\mathrm{HSO}_3^{-}\) B. \(\mathrm{HSO}_3^{-}\) \(\mathrm{C}_{+} \quad \mathrm{H}_2 \mathrm{SO}_3\) D. \(\mathrm{H}_2 \mathrm{SO}_4\)

Short answer

Empirical formula: \( \mathrm{H}_2 \mathrm{SO}_3 \)

Step-by-step solution

Convert Percentages to Masses

Assume a sample size of 100 grams to make calculations easier. Then, the masses are: Oxygen = 58.6 g, Sulfur = 39 g, Hydrogen = 2.4 g.

Calculate Moles of Each Element

Use the molar masses to convert the masses to moles: Moles of Oxygen: \( \frac{58.6 \, \text{g}}{16 \, \text{g/mole}} = 3.6625 \, \text{moles} \), Moles of Sulfur: \( \frac{39 \, \text{g}}{32 \, \text{g/mole}} = 1.21875 \, \text{moles} \), Moles of Hydrogen: \( \frac{2.4 \, \text{g}}{1.01 \, \text{g/mole}} = 2.376 \, \text{moles} \)

Simplify Moles Ratio

Divide all moles by the smallest number of moles: Oxygen: \( \frac{3.6625}{1.21875} \approx 3 \), Sulfur: \( \frac{1.21875}{1.21875} = 1 \), Hydrogen: \( \frac{2.376}{1.21875} \approx 2 \)

Write the Empirical Formula

Using the simplified ratios, the empirical formula is \( \mathrm{H}_2 \mathrm{SO}_3 \).

Key concepts

Molar Mass Conversion

To start with molar mass conversion, you need to understand what molar mass is. It is the mass of one mole of a substance, usually measured in grams per mole (g/mole). Each chemical element has a different molar mass. For example, the molar mass of oxygen (O) is 16 g/mole, sulfur (S) is 32 g/mole, and hydrogen (H) is 1.01 g/mole.

When you convert mass percentages to moles, you use the molar mass to divide the mass of each element. This step helps us understand how many moles of each element are present in the sample. As shown in the exercise, you will use the masses given (like 58.6 g of oxygen) and divide them by their molar masses, like this: \ \[ \text{Moles of Oxygen:} \frac{58.6 \text{g}}{16 \text{g/mole}} = 3.6625 \text{moles} \ \]

This process is crucial for all the subsequent calculations and ensures that the amount of substance is accurate.

Mass Percentage to Moles

In chemistry, converting mass percentages to moles is a crucial step for empirical formula determination. Assume a sample size of 100 grams. This simplifies calculations because percentages can directly be interpreted as masses. For example, if a compound is 39% sulfur, it means there are 39 grams of sulfur in a 100 grams sample.

With the mass converted to grams, the next step is to find moles. Divide the mass of each element by its molar mass. For instance:

• Sulfur: \ \[\frac{39 \text{g}}{32 \text{g/mole}} = 1.21875 \text{moles}\ \]
• Hydrogen: \ \[\frac{2.4 \text{g}}{1.01 \text{g/mole}} = 2.376 \text{moles}\ \]

These mole values will be used in the next stages to simplify ratios and determine the empirical formula.

Simplifying Mole Ratios

Simplifying mole ratios is pivotal for finding the empirical formula. After converting mass percentages to moles, you'll have a set of mole values for each element. However, these values often aren't whole numbers.

To simplify these ratios, divide all moles by the smallest number of moles among the elements. This helps normalize the moles to the smallest amount, making it easier to find the ratios. For example:

• Smallest number of moles: 1.21875 (Sulfur)
• Oxygen: \ \[\frac{3.6625}{1.21875} \rightarrow 3\ \]
• Hydrogen: \ \[\frac{2.376}{1.21875} \rightarrow 2\ \]
• Sulfur: \ \[\frac{1.21875}{1.21875} \rightarrow 1\ \]

These simplified ratios will be used to determine the empirical formula.

Empirical Formula Determination

The simplified mole ratios now provide the subscripts for each element in the empirical formula. In our example, the ratios are approximately 3 for Oxygen, 2 for Hydrogen, and 1 for Sulfur. These ratios become the subscripts in the empirical formula.

Therefore, based on the steps above, the empirical formula is written by placing these subscripts next to their corresponding elements: \ \[\text{Empirical Formula: } \text{H}_2\text{SO}_3\ \]

Thus, the neutral compound with 58.6% oxygen, 39% sulfur, and 2.4% hydrogen by mass corresponds to the empirical formula \ \[\text{H}_2\text{SO}_3\ \]. Understanding this process is crucial for solving similar problems and mastering empirical formula calculations. Remember, each subscript in the empirical formula reflects the simplest whole-number ratio of moles of each element in the compound.