Problem 45 - Chemistry The Central Science In Si Units

Question

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.052 \mathrm{~mol} \mathrm{C}, 0.103 \mathrm{~mol}\) \(\mathrm{H}\), and \(0.017 \mathrm{~mol} \mathrm{O} ;\) (b) \(2.10 \mathrm{~g}\) nickel and \(0.58 \mathrm{~g}\) oxygen; (c) \(26.56 \% \mathrm{~K}, 35.41 \% \mathrm{Cr}\), and \(38.03 \% \mathrm{O}\) by mass.

Step-by-Step Solution

Answer

Empirical formulas: (a) \(\text{C}_3\text{H}_6\text{O}\), (b) \(\text{NiO}\), (c) \(\text{K}_2\text{Cr}_2\text{O}_7\).

Step 1: Identify Moles in Compound (a)

For the compound with mol values given directly: - Moles of C = 0.052. - Moles of H = 0.103. - Moles of O = 0.017.

Step 2: Find Simplest Ratio for Compound (a)

Divide each of the moles by the smallest number of moles in the set:- C: \( \frac{0.052}{0.017} \approx 3.06 \approx 3 \)- H: \( \frac{0.103}{0.017} \approx 6.06 \approx 6 \)- O: \( \frac{0.017}{0.017} = 1 \)The ratio is approximately 3:6:1, so the empirical formula is \( \text{C}_3\text{H}_6\text{O} \).

Step 3: Convert Mass to Moles for Compound (b)

Given masses, convert to moles using atomic masses:- Atomic mass of Ni = 58.69 g/mol, O = 16.00 g/mol.- Moles of Ni = \( \frac{2.10}{58.69} \approx 0.0358 \).- Moles of O = \( \frac{0.58}{16.00} \approx 0.03625 \).

Step 4: Find Simplest Ratio for Compound (b)

Divide each number of moles by the smallest: - Ni: \( \frac{0.0358}{0.0358} = 1 \)- O: \( \frac{0.03625}{0.0358} \approx 1.012 \approx 1 \)The empirical formula is \( \text{NiO} \).

Step 5: Convert % Composition to Moles for Compound (c)

Convert mass percentages to moles based on 100 g total:- K: 26.56 g, Cr: 35.41 g, O: 38.03 g.- Moles of K = \( \frac{26.56}{39.10} \approx 0.679 \).- Moles of Cr = \( \frac{35.41}{52.00} \approx 0.681 \).- Moles of O = \( \frac{38.03}{16.00} \approx 2.377 \).

Step 6: Find Simplest Ratio for Compound (c)

Divide each mole value by the smallest number of moles:- K: \( \frac{0.679}{0.679} = 1 \)- Cr: \( \frac{0.681}{0.679} \approx 1 \)- O: \( \frac{2.377}{0.679} \approx 3.5 \)Approximate to the simplest whole number ratio: 2:2:7. The empirical formula is \( \text{K}_2\text{Cr}_2\text{O}_7 \).

Key Concepts

Mole Ratio

The mole ratio is an essential concept in chemistry that helps us understand the proportion of different elements in a compound. When you are given a set of measurements, either in moles or mass, you need to convert them into moles to simplify the comparison.
Mole ratios are derived by dividing each mole value by the smallest number of moles present. This allows you to establish whole number ratios between the elements, which form the basis of the empirical formula. For example, if you have 0.052 mol of carbon, 0.103 mol of hydrogen, and 0.017 mol of oxygen, you would calculate these ratios as follows:

• Carbon: \( \frac{0.052}{0.017} \approx 3.06 \approx 3 \)
• Hydrogen: \( \frac{0.103}{0.017} \approx 6.06 \approx 6 \)
• Oxygen: \( \frac{0.017}{0.017} = 1 \)

This results in a carbon to hydrogen to oxygen ratio of 3:6:1, which gives the empirical formula \( \text{C}_3\text{H}_6\text{O} \). A clear understanding of mole ratio helps in converting any complex data into simplified whole numbers necessary for further calculation.

Elemental Composition

Elemental composition refers to the percentage or fraction of each element within a compound. Understanding elemental composition is crucial for determining empirical formulas when elements are given in percentages by mass or as part of a compound's chemical structure.
For instance, if a sample consists of 26.56% potassium, 35.41% chromium, and 38.03% oxygen by mass, you assume a 100 g sample to find the moles of each element. You perform the conversion:

• Potassium: \( \frac{26.56}{39.10} \approx 0.679 \text{ mol} \)
• Chromium: \( \frac{35.41}{52.00} \approx 0.681 \text{ mol} \)
• Oxygen: \( \frac{38.03}{16.00} \approx 2.377 \text{ mol} \)

After determining these mole values, divide each by the smallest number to find a mole ratio, leading you to the empirical formula. Elemental composition is fundamental in deducing the simplest whole number ratios, helping to describe the basic structure of compounds.

Chemical Compounds

Chemical compounds are substances formed by the combination of two or more different elements bonded together in fixed proportions. Recognizing the empirical formula, which is the simplest ratio of elements in a compound, is vital for understanding chemical compounds.
An empirical formula gives insights into the nature of the chemical binding and composition. For example, \( \text{NiO} \) is derived from a sample that contains nickel and oxygen in such proportions that the smallest whole number ratio is reached. This formula signifies that in every one mole of nickel oxide, there exists one mole of nickel atom and one mole of oxygen atom bonded together.

• Empirical formulas provide the simplest description, ideal for basic level stoichiometric calculations.
• They help predict how compounds will react with others in chemical reactions.
• They differ from molecular formulas, which denote the actual number of atoms of each element in a molecule.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It uses principles like mole ratio and empirical formulas to predict the outcome of reactions. Understanding stoichiometry is necessary for planning chemical equations and ensuring that reactions occur with correct proportions.
By using the simplest forms of chemical equations (empirical formulas), stoichiometry lets you determine the amount of reactants needed and the amount of products formed.
Consider the stoichiometric calculations for a reaction involving \( \text{K}_2\text{Cr}_2\text{O}_7 \):

• Firstly, interpret the empirical formula to define input quantities.
• Secondly, use mole ratios from the equation to balance reactants and products.
• Finally, predict the yield of products, ensuring the reaction adheres to the law of conservation of mass.

Stoichiometry provides a deeper insight into how substances interact, transforming raw data about elemental compositions into practical chemical knowledge that guides various applications in chemistry.