Question

A quantity of \(1 \mathrm{~g}\) of metallic carbonate \(\mathrm{XCO}_{3}\) is completely converted into a \(\begin{array}{lll}\text { chloride } \mathrm{XCl}_{2} \text { weighing } & 1.11 & \text { g. } & \text { The }\end{array}\) atomic mass of the element ' \(\mathrm{X}\) ' is (a) 10 (b) 20 (c) 30 (d) 40

Short answer

The atomic mass of the element 'X' is 40 g/mol.

Step-by-step solution

- Write the reaction formula

Express the conversion of the metallic carbonate \(XCO_3\) into its chloride \(XCl_2\) using a chemical equation. Given the complete conversion, the equation would be: \[ XCO_3 \rightarrow XCl_2 + CO_2 \]This shows that each mole of \(XCO_3\) yields one mole of \(XCl_2\) upon decomposition.

- Calculate the molar mass difference

Using the given masses, calculate the difference in molar mass between \(XCl_2\) and \(XCO_3\). \[\text{Molar mass difference} = \text{Mass of } XCl_2 - \text{Mass of } XCO_{3} = 1.11 g - 1 g = 0.11 g\]

- Determine the mass of \(CO_2\) released

Given that the \(CO_2\) has been completely released in the reaction, subtract the mass of \(CO_2\) from the initial mass of \(XCO_3\) to find the mass of \(X\) in \(XCO_3\). \[\text{Molar mass of } CO_2 = 12 + (2 \times 16) = 44 g/mol\text{Mass of } CO_2 = \frac{44}{\text{Molar mass of } XCO_3} \times 1g\text{Mass of } X = 1 g - \text{Mass of } CO_2\]

- Write the molar mass equation for \(XCl_2\) and \(XCO_3\)

The molar mass of \(XCO_3\) must equal the molar mass of \(X\) plus the molar mass of \(CO_3\), which is the same as the molar mass of \(CO_2\) plus the mass difference calculated in Step 2. \[M(XCO_3) = M(X) + M(CO_3)\M(XCl_2) = M(X) + 2 \times M(Cl)\]

- Calculate the atomic mass of 'X'

Using the molar mass of chlorine (35.5 g/mol for each atom of Cl), write an equation for the atomic mass of \(X\) based on \(XCl_2\)'s molar mass from the given mass difference, and solve for \(M(X)\). \[M(XCl_2) - 2 \times M(Cl) = M(X)\1.11 g - 2(35.5 g/mol) = M(X)\M(X) = 1.11 g - 71 g/mol \frac{M(X)\times \text{Molar mass of } XCO_3}{1 g} + 0.11 g = 44 g/mol\]

- Solve for the atomic mass

Isolate \(M(X)\) on one side of the equation to solve for the atomic mass of \(X\). \[ \frac{M(X)\times \text{Molar mass of } XCO_3}{1 g} = 44 g/mol - 0.11 g\M(X) = \frac{1 g \times (44 g/mol - 0.11 g)}{\text{Molar mass of } XCO_3}\] The Molar mass of \(XCO_3\) is \((M(X) + 60\) g/mol. After substituting and solving, we find that \(M(X)\) is approximately 40 g/mol. Therefore, the correct answer is (d) 40.

Key concepts

Chemical Reactions

Understanding chemical reactions is key to entering the realm of chemical equations and their transformations. At its core, a chemical reaction is a process where reactants transform into products through bonds breaking and forming. This transformation follows the law of conservation of mass, which states that in a closed system, mass cannot be created or destroyed. In the given exercise, metallic carbonate \(XCO_3\) reacts in such a way that it is completely converted into chloride \(XCl_2\) and releases carbon dioxide \(CO_2\) gas.

This chemical change can be expressed succinctly by a balanced chemical equation. The balanced equation allows us to see the stoichiometry of the reaction, meaning the proportional relationship between reactants and products. In the case of \(XCO_3\) becoming \(XCl_2\), we are dealing with a decomposition reaction where the compound breaks down into different substances. The point of the exercise improvement advice is to ensure students can visualize this transformation clearly and understand its significance in determining the stoichiometry of the reaction.

Molar Mass Calculation

The concept of molar mass is fundamental in stoichiometry as it connects the amount of substance to the mass that can be measured. Molar mass is defined as the mass of one mole of a substance and is expressed in grams per mole (g/mol). It allows chemists to work out chemical calculations using the atomic unit—the mole—which is based on the number of atoms in exactly 12 grams of carbon-12, amounting to Avogadro's number of atoms, \(6.022 \times 10^{23}\) atoms.

The molar mass calculation is used to convert between mass and moles in a substance and can be found by summing up the atomic masses of all atoms in the formula unit of the compound. Understanding molar mass is crucial in our exercise when we compare masses of \(XCO_3\) and \(XCl_2\) to infer the atomic mass of element X. Through the correct application of molar mass principles, the solution offers a detailed route to tackle this stoichiometric problem.

Atomic Mass Determination

Determining atomic mass is a fundamental skill in interpreting and predicting the outcomes of chemical reactions. Atomic mass, often confused with atomic weight, actually refers to the mass of a single atom, typically expressed in atomic mass units (amu) or unified atomic mass units (u). The atomic mass is roughly equivalent to the number of protons and neutrons in the atom's nucleus.

In stoichiometric calculations, specifically like the one in our exercise, we leverage the concept of atomic mass to find the mass of an unknown element X. By using the known molar masses of compounds and elements involved in the reaction, and seeing how they combine or decompose, we can isolate and calculate the atomic mass of X. The improvement advice in this context ensures that students grasp how every piece of data—molar masses of compounds and elements—plays a crucial role in unwrapping the atomic mass of an unknown entity within a reaction.