Problem 143 - Chemistry The Molecular Science

Question

In a reaction, \(1.2 \mathrm{~g}\) element A reacts with exactly \(3.2 \mathrm{~g}\) oxygen to form an oxide, \(\mathrm{AO}_{x} ; 2.4 \mathrm{~g}\) element A reacts with exactly \(3.2 \mathrm{~g}\) oxygen to form a second oxide, \(\mathrm{AO}_{y}\) (a) Determine the ratio \(x: y\). (b) If \(x=2\), determine what the identity of element A might be.

Step-by-Step Solution

Answer

(a) The ratio \(x:y = 2:1\). (b) Element A is Carbon.

Step 1: Understand the Problem

We need to find the ratios of oxygen that react with element A in two different reactions forming two oxides, A\(_x\)O\(_x\) and A\(_y\)O\(_y\). Additionally, if \(x = 2\), we need to find the identity of element A.

Step 2: Analyze First Reaction

In the first reaction, \(1.2\, \text{g}\) of element A combines with \(3.2\, \text{g}\) of oxygen to form oxide AO\(_x\). The mass of the oxide is \(1.2 + 3.2 = 4.4\, \text{g}\).

Step 3: Analyze Second Reaction

In the second reaction, \(2.4\, \text{g}\) of element A combines with \(3.2\, \text{g}\) of oxygen to form oxide AO\(_y\). The mass of the oxide is \(2.4 + 3.2 = 5.6\, \text{g}\).

Step 4: Calculate Moles of Oxygen

The moles of oxygen in both reactions are the same. Use the mass of oxygen and molar mass of oxygen (\(16\, \text{g/mol}\) per oxygen atom) to find the number of moles: \(n = \frac{3.2}{16} = 0.2 \text{ mol O}_2\).

Step 5: Determine Ratios of Oxygen

For both oxides, since they use the same amount of oxygen (0.2 mol O\(_2\)), the ratio can be established: both oxides have \(x : y = 1 : 1\), but each uses \(0.4 \text{ mol}\) of individual O atoms (since 0.2 mol O\(_2\) is 0.4 mol O).

Step 6: Analyze Mass Ratio of A to Oxygen

Calculate the mass ratio of A to oxygen: First oxide is \(1.2 : 3.2\), simplifying to \(3 : 8\). Second oxide is \(2.4 : 3.2\), simplifying to \(3 : 4\). This indicates for AO\(_2\), 1 A to 2 O \(x = 2\); for AO, 1 A to 1 O \(y=1\). Thus, \(x:y = 2:1\).

Step 7: Identify the Element A

Assume \(x = 2\) for AO\(_2\): Use the atomic masses and the reaction to find the identity of A.From AO\(_2\) \(M_A + 2 \times 16 = 44\) yields \(M_A = 12\). The probable element with atomic mass of 12 is Carbon (C).

Key Concepts

Chemical Reactions

Chemical reactions explain the process where substances transform into new materials. In our exercise, element A reacts with oxygen to form two different oxides. Here, it is essential to understand the terms "reactants" and "products". The reactants in this scenario are element A and oxygen, while the products are the oxides, indicated as AO\(_x\) and AO\(_y\).
A chemical reaction is represented by a balanced equation, showing the conservation of mass. That means the total mass of the reactants always equals the total mass of the products. This principle is clearly demonstrated when 1.2 g of element A reacts with 3.2 g of oxygen to yield 4.4 g of oxide in the first reaction. Similarly, the second reaction uses 2.4 g of element A and 3.2 g of oxygen to give 5.6 g of oxide.
One key takeaway from the exercise is the formation of compounds with distinct ratios of elements, highlighting the stoichiometric nature of chemical reactions. Understanding this ensures accurate prediction of chemical changes and the quantities of substances involved.

Mole Concept

The mole concept is a fundamental chemical principle that relates the mass of a substance to the number of particles it contains. In our exercise, the crucial part is calculating the number of moles of oxygen involved in the reactions.
We know the mass of oxygen is 3.2 g. Given that the molar mass of an oxygen molecule (O\(_2\)) is 32 g/mol, the number of moles of oxygen is calculated using the formula:\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{3.2}{32} = 0.1 \text{ mol O}_2 \] This calculation shows that 0.1 moles of oxygen molecules react with element A in both reactions. However, since each oxygen molecule contains two oxygen atoms, we need to consider the stoichiometry: this corresponds to 0.2 moles of atomic oxygen.
Thus, the mole concept not only aids in understanding the quantities of substances but also in balancing the stoichiometric coefficients in chemical equations. This underlines the importance of moles in defining the proportions of elements in chemical reactions.

Empirical Formulas

Empirical formulas provide the simplest whole-number ratio of atoms in a compound, fundamental for understanding chemical compositions. In our exercise, we determine the empirical formulas of the oxides AO\(_x\) and AO\(_y\).
First, we examine the mass ratio of element A to oxygen in both reactions. For oxide AO\(_x\), the ratio 1.2 g (A) to 3.2 g (O) simplifies to 3:8. For oxide AO\(_y\), 2.4 g (A) to 3.2 g (O) simplifies to 3:4. These ratios help determine the number of atoms of each element in the formula.The empirical formula of AO\(_x\) indicates a simpler ratio, synonymous with A and two oxygens (e.g., AO\(_2\)). Here, x=2. Conversely, AO\(_y\) has a 1:1 ratio, meaning one atom of A to one atom of oxygen, giving the empirical formula AO, implicating y=1.
Understanding empirical formulas bridges the mass and mole aspect of stoichiometry, enabling us to recognize the simplest representation of a compound in reactions. In our case, confirming x and y as 2 and 1 helps identify element A, resting on established empirical relationships.