Question

Several samples of carbon dioxide are obtained and decomposed into carbon and oxygen. The masses of the carbon and oxygen are then weighed, and the results are tabulated as shown here. One of these results does not follow the law of constant composition and is therefore wrong. Which one? a. 12 grams of carbon and 32 grams of oxygen b. \(4.0\) grams of carbon and 16 grams of oxygen c. \(1.5\) grams of carbon and \(4.0\) grams of oxygen d. \(22.3\) grams of carbon and \(59.4\) grams of oxygen

Short answer

Option b) is incorrect because it has a mass ratio of 4.0g C to 16g O (0.25), which does not match the expected ratio for carbon dioxide (0.375).

Step-by-step solution

Understand the Law of Constant Composition

The Law of Constant Composition states that a given chemical compound will always contain its component elements in a fixed ratio by mass, regardless of its source or how it was prepared. For carbon dioxide (CO2), this means the mass ratio of carbon to oxygen should always be the same.

Calculate the Mass Ratios

Find the mass ratio of carbon to oxygen for each sample by dividing the mass of carbon by the mass of oxygen. For carbon dioxide, the correct mass ratio should match that of CO2's known composition, which is 1 part carbon (12 amu for a carbon atom) to 2 parts oxygen (16 amu each for an oxygen atom, hence 32 amu total for both oxygen atoms).

Identify the Incorrect Sample

Compare the calculated ratios to the expected ratio of carbon to oxygen in carbon dioxide: Expected ratio for CO2 = 12/32. Check if the calculated ratio for each sample matches this expected ratio. The one that does not match is incorrect.

Evaluate Each Option

For each option, calculate and compare the mass ratio of carbon to oxygen: a) 12g C / 32g O = 0.375; b) 4.0g C / 16g O = 0.25; c) 1.5g C / 4.0g O = 0.375; d) 22.3g C / 59.4g O = 0.375. Compare these ratios to the expected ratio of 12g C to 32g O (0.375).

Key concepts

Chemical Composition

Understanding the chemical composition of compounds is crucial in the study of chemistry. The chemical composition of a substance refers to the elements that make it up and the relative amounts of these elements. In the context of the exercise, carbon dioxide (\textbf{CO}\(_{2}\)) is composed of one carbon atom and two oxygen atoms. This fixed composition is why the Law of Constant Composition applies.

When exploring the Law of Constant Composition, students are shown that no matter the source or the amount of carbon dioxide, the ratio by mass of carbon to oxygen will always be constant, reflecting its chemical structure. Each molecule of carbon dioxide will always consist of one carbon atom and two oxygen atoms (thus, \textbf{CO}\(_{2}\)). This is why when decomposing carbon dioxide, the resulting masses of carbon and oxygen should exhibit a consistent ratio, essentially reaffirming the compound's defined chemical composition.
Additionally, to grasp the significance of this law, examples such as minerals with a variable composition like the amphibole group contrast with those that have a fixed composition. The amphibole group can show a range of compositions due to ionic substitution within their structure, while substances like water (H2O) will always consist of two hydrogen atoms for every oxygen atom.

Stoichiometry

Stoichiometry is the section of chemistry that pertains to the calculation of the quantities of reactants and products involved in chemical reactions. It's like a recipe for science, telling you how much of each ingredient you need. In relation to the exercise, stoichiometry allows us to understand the calculations behind the mass ratio of carbon to oxygen in carbon dioxide.

For stoichiometry to work, it relies on the conservation of mass and the Law of Constant Composition. To clarify, if we know the molecular formula of a certain compound, like \textbf{CO}\(_{2}\), stoichiometry is used to determine the proportions of carbon and oxygen in any given sample of carbon dioxide. The molecular mass of carbon dioxide is calculated by summing the masses of its constituent atoms, which we will explain in the next section about molecular mass.
Understanding the stoichiometric relationship in compounds helps students realize why the ratio should always be consistent, as shown in the stepwise approach to solving the given exercise. It is a foundational concept to predict the outcome of chemical reactions, maintain balance in equations, and to deduce the exact composition of substances based on their chemical formulas.

Molecular Mass

The concept of molecular mass can sometimes puzzle students, but it's essentially the sum of the mass of all atoms in a molecule. It is often expressed in atomic mass units (amu). For instance, the molecular mass of carbon dioxide (\textbf{CO}\(_{2}\)) is calculated by adding the mass of one carbon atom (12 amu) to the mass of two oxygen atoms (each 16 amu). This gives us \textbf{CO}\(_{2}\)'s molecular mass as 44 amu (12 amu for C + 2 x 16 amu for O).

In the exercise, once students understand the molecular mass, they can begin to see the practical application of this concept through the ratios calculated in the options given. For aid in visualization, it's helpful to consider molecular models or diagrams that illustrate the atom's individual masses, summing up to a molecule's total mass. By knowing the molecular mass and the ratio provided by the Law of Constant Composition, students can readily determine whether a sample matches the correct proportions, as the mass of an individual molecule is spread out consistently across its many instances.
Thus, through the analysis of molecular mass, students find that sample option (b), with a mass ratio of 0.25, doesn't follow the expected ratio of 0.375, based on the molecular mass of \textbf{CO}\(_{2}\). By understanding molecular mass, students learn it's not just the number of atoms, but also the type of atoms (with their respective atomic masses) that determine the mass of a molecule and, accordingly, allow us to interpret and predict patterns in chemical behavior.