Problem 98 - Chemistry The Molecular Science

Question

In an experiment, \(1.056 \mathrm{~g}\) of a metal carbonate containing an unknown metal \(\mathrm{M}\) was heated to give the metal oxide and \(0.376 \mathrm{~g} \mathrm{CO}_{2}\) $$\mathrm{MCO}_{3}(\mathrm{~s}) \longrightarrow \mathrm{MO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g})$$ Determine the identity of the metal \(\mathrm{M}\). (a) \(\mathrm{Ni}\) (b) \(\mathrm{Cu}\) (c) Co (d) \(\mathrm{Ba}\)

Step-by-Step Solution

Answer

The metal M is copper (Cu).

Step 1: Determine the Molar Mass of CO2

First, calculate the molar mass of \( \mathrm{CO}_{2} \). The atomic mass of carbon (C) is approximately 12.01 g/mol and that of oxygen (O) is 16.00 g/mol. Therefore, the molar mass of \( \mathrm{CO}_{2} \) is \( 12.01 + 2 \times 16.00 = 44.01 \) g/mol.

Step 2: Calculate Moles of CO2 Released

Since we know the mass of \( \mathrm{CO}_{2} \) released is 0.376 g, we can find the moles of \( \mathrm{CO}_{2} \) by dividing the mass by its molar mass: \[ \text{Moles of } \mathrm{CO}_{2} = \frac{0.376 \text{ g}}{44.01 \text{ g/mol}} \approx 0.00855 \text{ mol}. \]

Step 3: Find the Mass of MO Produced

The mass of the metal oxide, \( \mathrm{MO} \), produced is the difference between the initial mass of the metal carbonate and the mass of \( \mathrm{CO}_{2} \) that was lost during the reaction. Thus, \[ \text{Mass of } \mathrm{MO} = 1.056 \text{ g} - 0.376 \text{ g} = 0.680 \text{ g}. \]

Step 4: Determine the Molar Mass of MO

The number of moles of \( \mathrm{MO} \) is the same as the number of moles of \( \mathrm{CO}_{2} \), which is 0.00855 mol. The molar mass of \( \mathrm{MO} \) is therefore \[ \text{Molar Mass of } \mathrm{MO} = \frac{0.680 \text{ g}}{0.00855 \text{ mol}} \approx 79.53 \text{ g/mol}. \]

Step 5: Determine the Molar Mass of M

The molar mass of \( \mathrm{MO} \) is the sum of the molar masses of \( \mathrm{M} \) and oxygen (16.00 g/mol). Therefore, the molar mass of \( \mathrm{M} \) is \[ \text{Molar Mass of } \mathrm{M} = 79.53 \text{ g/mol} - 16.00 \text{ g/mol} = 63.53 \text{ g/mol}. \]

Step 6: Identify the Metal M

Compare the calculated molar mass of \( \mathrm{M} \) to the given options: (a) \( \mathrm{Ni} \approx 58.69 \text{ g/mol} \), (b) \( \mathrm{Cu} \approx 63.55 \text{ g/mol} \), (c) \( \mathrm{Co} \approx 58.93 \text{ g/mol} \), and (d) \( \mathrm{Ba} \approx 137.33 \text{ g/mol} \). The molar mass of \( \mathrm{Cu} \) is the closest match to our calculated value.

Key Concepts

Molar Mass Calculation

Understanding how to calculate molar mass is essential in stoichiometry. The molar mass of a compound is the sum of the molar masses of all the elements in it, measured in grams per mole (g/mol). For instance, in this exercise, we need to determine the molar mass of carbon dioxide (\(\mathrm{CO}_2\)). Carbon has an atomic mass of approximately 12.01 g/mol, and oxygen has an atomic mass of 16.00 g/mol. Since \(\mathrm{CO}_2\) contains one carbon and two oxygen atoms, we calculate its molar mass as follows:
\[\text{Molar Mass of } \mathrm{CO}_2 = 12.01 + 2 \times 16.00 = 44.01 \text{ g/mol}.\]
This understanding allows us to further calculate the number of moles from the mass of a substance. Since moles serve as a bridge to translate mass into a reactive quantity, knowing the molar mass is integral for any stoichiometric calculation.

Chemical Reactions

In stoichiometry, analyzing chemical reactions involves understanding the interaction between reactants and products. The exercise given revolves around the decomposition of a metal carbonate, \(\mathrm{MCO}_3\), into a metal oxide, \(\mathrm{MO}\), and carbon dioxide, \(\mathrm{CO}_2\). This reaction illustrates a common type of chemical decomposition where a compound breaks down into simpler substances.
The balanced chemical equation is:
\[\mathrm{MCO}_3(\mathrm{s}) \longrightarrow \mathrm{MO}(\mathrm{s}) + \mathrm{CO}_2(\mathrm{g})\]
Studying such reactions helps us comprehend how matter is conserved and can transform from one form to another. For a successful stoichiometric analysis, you need to keep track of masses which remain constant and use these to infer unknowns as in this practical example.

Identification of Elements

One of the fascinating parts of studying chemistry is identifying unknown elements through experimental data. In the given problem, after performing a reaction and making some calculations, you can identify the unknown metal \( \mathrm{M} \) in the compound \( \mathrm{MCO}_3 \).
Here's how it works: after calculating the molar mass of the resulting metal oxide \( \mathrm{MO} \) and subtracting the known mass of oxygen, the remainder gives the molar mass of the metal \( \mathrm{M} \). This calculated value is then compared to known atomic masses to identify the element.
In this exercise, the molar mass of \( \mathrm{M} \) was calculated to be approximately 63.53 g/mol. Comparing this with the atomic masses of the given options, copper (\( \mathrm{Cu} \)) with a molar mass close to 63.55 g/mol emerges as the identity of the metal. This process demonstrates how chemical knowledge, combined with mathematical calculations, can uncover the properties and identities of unknown substances.