Problem 46 - Chemistry The Central Science In Si Units

Question

Determine the empirical formula of each of the following compounds if a sample contains (a) \(3.92 \mathrm{~mol} \mathrm{C}, 5.99 \mathrm{~mol} \mathrm{H}\) and \(2.94 \mathrm{~mol} \mathrm{O} ;(\mathbf{b}) 12.0 \mathrm{~g}\) calcium and \(2.8 \mathrm{~g}\) nitrogen; \((\mathbf{c})\) \(89.14 \%\) Au and \(10.86 \%\) O by mass.

Step-by-Step Solution

Answer

(a) C₄H₆O₃; (b) Ca₃N₂; (c) Au₂O₃.

Step 1: Identify Moles of Each Element

For part (a), the moles of each element are given: carbon (3.92 mol), hydrogen (5.99 mol), and oxygen (2.94 mol). Thus, start over competently with these values.

Step 2 for (a): Calculate Mole Ratios

Divide each number of moles by the smallest number of moles amongst the elements. Here, divide 3.92 mol C, 5.99 mol H, and 2.94 mol O all by 2.94 mol (the smallest value). This gives C: 1.33, H: 2.04, O: 1.

Step 3 for (a): Adjust Ratios to Whole Numbers

Multiply each ratio by an integer that converts all ratios to whole numbers. In this case, multiply each by 3: C: 1.33*3 = 4, H: 2.04*3 = 6, O: 1*3 = 3. Thus, the empirical formula is C₄H₆O₃.

Step 4: Calculate Moles of Each Element for (b)

Find the moles of calcium and nitrogen. For calcium: moles = 12.0 g / 40.08 g/mol = 0.299 mol. For nitrogen: moles = 2.8 g / 14.01 g/mol = 0.200 mol.

Step 5 for (b): Calculate Mole Ratios

Divide both mole values by the smaller of the two: calcium (0.299 mol) and nitrogen (0.200 mol) divided by 0.200. This gives Ca: 1.495 and N: 1.

Step 6 for (b): Adjust Ratios to Whole Numbers

Multiply each ratio by 2 to convert to whole numbers: Ca: 1.495*2 = 3, N: 1*2 = 2. Thus, the empirical formula is Ca₃N₂.

Step 7: Calculate Moles of Each Element for (c)

Use the given mass percentages to determine mass ratios. Assume a 100 g sample: 89.14 g of Au and 10.86 g of O.

Step 8 for (c): Convert Mass to Moles

Convert the masses to moles: Au = 89.14 g / 196.97 g/mol = 0.452 mol; O = 10.86 g / 16.00 g/mol = 0.679 mol.

Step 9 for (c): Calculate Mole Ratios

Divide each mol value by 0.452 (the fewer mol value): Au: 0.452/0.452 = 1, O: 0.679/0.452 = 1.501.

Step 10 for (c): Adjust Ratios to Whole Numbers

Multiply each ratio by 2 to eliminate the fraction: Au: 1*2 = 2, O: 1.501*2 ≈ 3. Thus, the empirical formula is Au₂O₃.

Key Concepts

Mole Ratio

A mole ratio is a comparison of the number of moles of each element in a compound. It is crucial when you're determining the empirical formula of a compound. In simple terms, the mole ratio allows you to see how many units of each element are present relative to each other.
For example, if you have a sample with 3.92 moles of carbon, 5.99 moles of hydrogen, and 2.94 moles of oxygen, first identify the smallest number of moles. Here, it is oxygen with 2.94 moles.

• Divide the moles of each element by 2.94, resulting in the following ratios: Carbon: 1.33, Hydrogen: 2.04, Oxygen: 1.
• If these numbers aren't whole numbers, multiply them by a common factor to convert them to the smallest possible whole numbers.

This process results in more accurate and standardized formulas which are much easier to use in further chemistry calculations.

Elemental Composition

Understanding the elemental composition of a compound involves knowing the percentage or amount of each element present within the compound. This knowledge is fundamental in determining empirical formulas.
Let's consider a sample containing 12.0 g of calcium and 2.8 g of nitrogen. To convert this information into useful data, you first need to convert the mass of each element into moles using their molar masses:

• Calcium's molar mass is around 40.08 g/mol. Therefore, the moles of calcium in the sample are calculated as follows: \( \text{moles of Ca} = \frac{12.0 \, \text{g}}{40.08 \, \text{g/mol}} = 0.299 \text{ mol} \).
• Nitrogen's molar mass is 14.01 g/mol, resulting in: \( \text{moles of N} = \frac{2.8 \, \text{g}}{14.01 \, \text{g/mol}} = 0.200 \text{ mol} \).

These calculations help to determine the mole ratio and, subsequently, the empirical formula of the compound.

Chemistry Problem Solving

Chemistry problem-solving often involves multi-step processes, especially when determining empirical formulas. The key is to break down the problem into simple, manageable steps.
Consider you are given the percentages of elements in a compound. The first step is to assume you have a 100 g sample, which simplifies the conversion of percentages to grams directly. For instance, if you have a compound containing 89.14% gold (Au) and 10.86% oxygen (O):

• The mass of gold is directly 89.14 g, and oxygen is 10.86 g in a 100 g sample.
• Convert these masses to moles: \( \text{Au} = \frac{89.14 \, \text{g}}{196.97 \, \text{g/mol}} = 0.452 \text{ mol} \), and \( \text{O} = \frac{10.86 \, \text{g}}{16.00 \, \text{g/mol}} = 0.679 \text{ mol} \).

With these mole values, calculate the mole ratios and refine them to get the simplest whole number ratio. This systematic approach ensures accuracy in determining the empirical formula, like Au₂O₃ in this example. Problems in chemistry often follow these logical steps, requiring practice and attention to detail for success.