Problem 11 - Chemistry The Central Science

Question

A 1.0 -g sample of carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is fully decomposed into its elements, yielding 0.273 g of carbon and 0.727 \(\mathrm{g}\) of oxygen. (a) What is the ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?\) (b) If a sample of a different compound decomposes into 0.429 \(\mathrm{g}\) of carbon and 0.571 \(\mathrm{g}\) of oxygen, what is its ratio of the mass of \(\mathrm{O}\) to \(\mathrm{C} ?(\mathbf{c})\) According to Dalton's atomic theory, what is the empirical formula of the second compound?

Step-by-Step Solution

Answer

The mass ratio of O to C for the first compound is \( \frac{0.727}{0.273} \approx 2.66 \) and for the second compound is \( \frac{0.571}{0.429} \approx 1.33 \). The empirical formula of the second compound can be found using the mole ratio, resulting in the empirical formula as \( C_{1}O_{1} \) or simply CO.

Step 1: Calculation of mass ratio for the first compound

To calculate the mass ratio of O to C for the first compound, we have the following masses: Mass of Carbon (C) = 0.273 g Mass of Oxygen (O) = 0.727 g Now we can find the ratio of the mass of O to C: \( Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} \) \( Ratio_{O:C} = \frac{0.727}{0.273} \)

Step 2: Calculation of mass ratio for the second compound

For the second compound, we have the following masses: Mass of Carbon (C) = 0.429 g Mass of Oxygen (O) = 0.571 g Now we can find the ratio of the mass of O to C: \( Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} \) \( Ratio_{O:C} = \frac{0.571}{0.429} \)

Step 3: Calculating the empirical formula of the second compound

To find the empirical formula for the second compound, we will look at the ratio of O to C and compare it with the atomic masses of Oxygen and Carbon. The atomic mass of Oxygen (O) is 16 g/mol, and Carbon (C) is 12 g/mol. Now, we can calculate the mole ratio of O to C: \( Mole \ Ratio_{O:C} = \frac{\frac{0.571}{16}}{\frac{0.429}{12}} \) Simplify the ratio by finding the simplest whole number ratio: Now that we have the mole ratio in simplest whole number ratio, we can write the empirical formula for the second compound. Let the empirical formula is CXOY, where X and Y are the simplest whole-number quantities of C and O, respectively. Now we have the empirical formula for the second compound.

Key Concepts

Mass Ratio

In chemistry, understanding the mass ratio of elements in a compound is key to determining its composition. To find the mass ratio of oxygen to carbon in a compound, we divide the mass of oxygen by the mass of carbon. For example, in the decomposition of carbon dioxide

• Mass of Carbon ( \( C \)) = 0.273 g
• Mass of Oxygen ( \( O \)) = 0.727 g

Thus, the mass ratio of O to C is calculated as:\[Ratio_{O:C} = \frac{Mass_{O}}{Mass_{C}} = \frac{0.727}{0.273}\]This ratio helps to understand the proportion of each element in the compound. Similarly, for another compound decomposing into 0.429 g of carbon and 0.571 g of oxygen, the mass ratio of O to C would be:\[Ratio_{O:C} = \frac{0.571}{0.429}\]Knowing these ratios is crucial when comparing different compounds and understanding their chemical properties.

Dalton's Atomic Theory

Dalton's Atomic Theory is a fundamental concept in chemistry that provides insights into the nature of matter. According to this theory, elements are composed of tiny, indivisible particles called atoms. Here are some key points:

• Each atom of an element is identical in mass and properties.
• Atoms of different elements have distinct masses and properties.
• Atoms combine in fixed, simple ratios to form compounds.

In the context of calculating empirical formulas, Dalton's theory helps us understand how atoms combine in specific ratios. Using the mass ratios calculated, these values can be converted to mole ratios, helping us deterministically create the empirical formula of a compound. For example, from the mass ratio of O to C in the second compound, we can convert this to a simple whole number atom ratio, using atomic masses: - Atomic mass of Oxygen (O) = 16 g/mol - Atomic mass of Carbon (C) = 12 g/mol This reflects the fixed and simple way atoms combine, aligning with Dalton's idea about chemical compounds.

Chemical Decomposition

Chemical decomposition is the process where a single compound breaks down into two or more elements or new compounds. This can be thought of as "reverse synthesis." Understanding it is essential in chemistry because it helps us study the composition of compounds. When a compound like carbon dioxide decomposes, it can form elemental carbon and oxygen. - Example: CO₂ decomposing into carbon (C) and oxygen (O). The decomposition reaction provides a way to verify the Law of Conservation of Mass, which states that matter is neither created nor destroyed in a chemical reaction. When analyzing decomposition reactions, recognizing the resulting mass ratios of elements aids in determining the compound’s original composition. This is instrumental when deciding the empirical formula. By observing the decomposition processes and calculating mass ratios, like in the given problem, students can apply knowledge of chemical bonding and composition to broader chemical analyses.