Problem 48 - Chemistry The Central Science In Si Units

Question

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(42.1 \% \mathrm{Na}, 18.9 \% \mathrm{P}\), and \(39.0 \% \mathrm{O}\) (b) \(18.7 \% \mathrm{Li}, 16.3 \% \mathrm{C},\) and \(65.0 \% \mathrm{O}\) (c) \(60.0 \% \mathrm{C}, 4.4 \% \mathrm{H},\) and the remainder \(\mathrm{O}\)

Step-by-Step Solution

Answer

(a) \(\mathrm{Na}_3\mathrm{PO}_4\); (b) \(\mathrm{Li}_2\mathrm{CO}_3\); (c) \(\mathrm{C}_5\mathrm{H}_4\mathrm{O}\).

Step 1: Assume 100 g sample

To simplify calculations, assume you have 100 grams of each compound. This makes the mass of each element equal to its percentage in grams, e.g., in part (a), you have 42.1 g of Na, 18.9 g of P, and 39.0 g of O.

Step 2: Convert masses to moles

Divide each mass by the molar mass of the respective element to find the number of moles. (a) For Na: \(\frac{42.1}{22.99} \approx 1.83\), P: \(\frac{18.9}{30.97} \approx 0.61\), O: \(\frac{39.0}{16.00} \approx 2.44\)(b) For Li: \(\frac{18.7}{6.94} \approx 2.69\), C: \(\frac{16.3}{12.01} \approx 1.36\), O: \(\frac{65.0}{16.00} \approx 4.06\)(c) For C: \(\frac{60.0}{12.01} \approx 4.99\), H: \(\frac{4.4}{1.008} \approx 4.37\), O: Assume 35.6% left for O: \(\frac{35.6}{16.00} \approx 2.23\)

Step 3: Find simplest mole ratio

Divide the number of moles of each element by the smallest number of moles in the set.(a) Na: \(\frac{1.83}{0.61} = 3\), P: \(\frac{0.61}{0.61} = 1\), O: \(\frac{2.44}{0.61} \approx 4\) leads to the ratio 3:1:4.(b) Li: \(\frac{2.69}{1.36} \approx 2\), C: \(\frac{1.36}{1.36} = 1\), O: \(\frac{4.06}{1.36} \approx 3\) gives the ratio 2:1:3.(c) C: \(\frac{4.99}{2.23} \approx 2.24\approx 5\), H: \(\frac{4.37}{2.23} \approx 1.96 \approx 4\), O: \(\frac{2.23}{2.23} = 1\) leads to the ratio 5:4:1.

Step 4: Write the empirical formula

Use the mole ratios to write the empirical formula for each compound. (a) The empirical formula is \(\mathrm{Na}_3\mathrm{PO}_4\).(b) The empirical formula is \(\mathrm{Li}_2\mathrm{CO}_3\).(c) The empirical formula is \(\mathrm{C}_5\mathrm{H}_4\mathrm{O}_1\) or \(\mathrm{C}_5\mathrm{H}_4\mathrm{O}\).

Key Concepts

Understanding Mole Ratios

Mole ratios are an essential concept in chemistry, acting as the backbone of empirical formula determination. They simplify how we express the relative amounts of different substances reacting or produced in a chemical reaction.

To understand mole ratios, start by converting the mass of each element in a compound to moles. This is accomplished using the formula \( \, \text{moles of element} = \frac{\text{mass of element}}{\text{molar mass of element}} \).

Once you have calculated the moles for each element, it is important to determine the simplest whole number ratio of these moles. This ratio is derived by dividing the mole amount of each element by the smallest number of moles calculated among them. For instance, if you have 1 mole of Carbon and 2 moles of Oxygen, the simplest ratio would be 1:2.

Understanding the significance of mole ratios helps in writing balanced chemical equations, deducing empirical formulas, and gaining insights into the stoichiometric relationships between reactants and products in chemical reactions.

Exploring Mass Composition

Mass composition is the percentage by mass of each element in a compound. When determining empirical formulas, knowing the mass composition of a substance is your first step. It's like creating a recipe where each ingredient represents an element, and the percentages are how much of each you have.

Typically, you begin by assuming you have 100 grams of a compound. This assumption is crucial because it converts the percentage composition directly into grams. For example, if a compound has 18% of an element, in a 100 g sample, that element contributes 18 g.

• This process makes calculations simpler and more intuitive.
• It provides a straightforward path to convert mass into moles.

Once the mass is known, use the molar mass of each element to convert grams to moles, facilitating the development of mole ratios. Thus, mastering mass composition serves as a bedrock for further chemical analysis.

The Significance of Chemical Formulas

Chemical formulas are the symbolic representation of a compound's composition. They provide insights into the ratios and arrangements of elements in chemicals, helping chemists understand and predict the behavior of substances.

The empirical formula, specifically, gives the simplest whole number ratio of atoms of each element present in a compound. To determine this, you use the mole ratios derived from mass composition calculations. For instance, given the mole ratio of 3:1:4 for a compound consisting of Sodium, Phosphorus, and Oxygen, the empirical formula would be \(\text{Na}_3\text{PO}_4\).

The critical component in these formulas is the subscript next to each chemical symbol, indicating the number of moles of each element in the simplest form.

• Empirical formulas aid in understanding fundamental material properties.
• They are foundational in stoichiometry and chemical reaction predictions.

Mastering empirical formulas cultivates a deeper comprehension of chemical composition, enabling diverse applications from basic laboratory work to complex industrial processes.