Question

Determine the empirical formulas of the compounds with the following compositions: (a) 2.1 percent \(\mathrm{H}\), 65.3 percent \(\mathrm{O}, 32.6\) percent \(\mathrm{S} ;\) (b) 20.2 percent \(\mathrm{Al}\), 79.8 percent \(\mathrm{Cl}\).

Short answer

(a) \(\text{H}_2\text{O}_4\text{S}\); (b) \(\text{AlCl}_3\).

Step-by-step solution

Convert Percentage to Grams

Assume you have 100 grams of each compound. This simplifies the conversion, as the percentage can directly be considered as grams. So, for (a), you have 2.1 g of H, 65.3 g of O, and 32.6 g of S. For (b), you have 20.2 g of Al and 79.8 g of Cl.

Convert Grams to Moles

Reactants are often better considered in terms of moles rather than mass. Use the molar masses: - For \text{H}\; (1.008 \text{g/mol}), O \; (16.00 \text{g/mol}), and S \; (32.07 \text{g/mol}) for (a).- For Al \; (26.98 \text{g/mol}) and Cl \; (35.45 \text{g/mol}) for (b).Calculate the moles:a) \[\text{Moles of H} = \frac{2.1}{1.008} = 2.08 \]\[\text{Moles of O} = \frac{65.3}{16.00} = 4.08 \]\[\text{Moles of S} = \frac{32.6}{32.07} = 1.02 \]b) \[\text{Moles of Al} = \frac{20.2}{26.98} = 0.748 \]\[\text{Moles of Cl} = \frac{79.8}{35.45} = 2.25 \]

Determine the Simplest Whole Number Ratios

For each compound, divide all numbers of moles by the smallest number of moles obtained to get the simplest ratio.a) Divide each by 1.02:\[\text{H: } \frac{2.08}{1.02} \approx 2.04 \approx 2 \]\[\text{O: } \frac{4.08}{1.02} \approx 4.00 \approx 4 \]\[\text{S: } \frac{1.02}{1.02} = 1.00 \approx 1 \]b) Divide each by 0.748:\[\text{Al: } \frac{0.748}{0.748} = 1.00 \approx 1 \]\[\text{Cl: } \frac{2.25}{0.748} \approx 3.00 \approx 3 \]

Write the Empirical Formula

Using the whole number ratios from Step 3, write the empirical formula for each compound.a) The empirical formula for compound (a) is \(\text{H}_2\text{O}_4\text{S}\).b) The empirical formula for compound (b) is \(\text{AlCl}_3\).

Key concepts

Percentage Composition

To understand the empirical formula of a compound, we start with its percentage composition. This representation shows the percentage by mass of each element in a compound. It's like breaking a recipe down to its main ingredients and their amounts.

• Consider a compound with percentages for hydrogen (H), oxygen (O), and sulfur (S): 2.1% H, 65.3% O, and 32.6% S.
• Another compound could have 20.2% aluminum (Al) and 79.8% chlorine (Cl).

For calculations, we usually convert these percentages to grams as it helps to simplify further steps by assuming a 100 g sample. This means the percentage value directly reflects the mass in grams of each element present in the sample. Understanding the mass proportion is crucial because the next step involves turning it into a countable number of atoms or moles.

Molar Mass

Molar mass is an important concept that helps us convert grams of each element to moles, which is a more natural way to compare amounts in chemistry. It is the mass of one mole of a substance and expressed in g/mol.
For example:

• Hydrogen (H) has a molar mass of 1.008 g/mol.
• Oxygen (O) has a molar mass of 16.00 g/mol.
• Sulfur (S) has a molar mass of 32.07 g/mol.
• Aluminum (Al) has a molar mass of 26.98 g/mol.
• Chlorine (Cl) has a molar mass of 35.45 g/mol.

When you have a sample, like from our earlier conversion of percentages to grams, you use these molar masses to determine the number of moles of each element. This foundational step enables you to see the elements in terms of number of particles, which paves the way to discovering the simplest formula or empirical formula.

Moles Conversion

Converting grams to moles is pivotal in forming the empirical formula, as it connects the amount of substance to the number of atoms involved. You do this using the molar mass of each element:

• For a substance with a certain percentage composition, assume a 100 g sample for simplicity.
• Then, convert the mass of each element to moles by dividing the mass by the element's molar mass.

For example:

• For 2.1 grams of H: Moles of H = \( \frac{2.1}{1.008} \) = 2.08 moles.
• For 65.3 grams of O: Moles of O = \( \frac{65.3}{16.00} \) = 4.08 moles.
• For 32.6 grams of S: Moles of S = \( \frac{32.6}{32.07} \) = 1.02 moles.

Understanding how many moles of each element are present is crucial for deducing the simplest ratio, or empirical formula, which is the goal of such calculations.

Chemical Compounds

Chemical compounds are substances made from two or more elements in a specific ratio.
The empirical formula of a compound is derived by comparing the relative amounts of each element based on proportion and moles.

• After converting grams to moles, divide each value by the smallest number of moles calculated to find the simplest whole-number ratio.
• For compound with 2.1% H, 65.3% O, 32.6% S, these calculations give us a ratio, leading to the empirical formula \( \text{H}_2\text{O}_4\text{S} \).
• For compound with 20.2% Al and 79.8% Cl, the empirical formula is \( \text{AlCl}_3 \).

The empirical formula doesn't reveal the exact number of atoms, but it does provide the simplest ratio of elements within a compound, essential for understanding its basic composition and properties.