Question

You and a partner are asked to complete a lab entitled "Carbonates of Group 2 metal" that is scheduled to extend over two lab periods. The first lab, which is to be completed by your partner, is devoted to carrying out compositional analysis and determine the identity of the Group 2 metal (M). In the second lab, you are to determine the melting point of this compound. Upon going to lab you find two unlabeled vials containing white powder. You also find the following notes in your partner's notebook-Compound 1: \(40.04 \% \mathrm{M}\) and \(12.00 \%\) C, \(47.96 \%\) O (by mass), Compound \(2: 69.59 \% \mathrm{M}\), \(6.09 \% \mathrm{C},\) and \(24.32 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 and the identity of \(\mathrm{M} ?(\mathbf{b})\) What is the empirical formula for Compound 2 and the identity of \(\mathrm{M}\) ? Upon determining the melting points of these two compounds, you find that both compounds do not melt up to the maximum temperature of your apparatus, instead, the compounds decompose and liberate colorless gas. (c) What is the identity of the colorless gas? (d) Write the chemical equation for the decomposition reactions of compound 1 and 2. (e) Are compounds 1 and 2 ionic or molecular?

Short answer

The empirical formula for Compound 1 is MCO3, and the identity of M is calcium (Ca). The empirical formula for Compound 2 is MC2O6, and the identity of M is magnesium (Mg). The colorless gas produced during decomposition is carbon dioxide (CO2). The chemical equations for the decomposition of compound 1 and 2 are: 1) CaCO3 → CaO + CO2 2) MgC2O6 → MgO + 2 CO2 Both compounds 1 and 2 are ionic.

Step-by-step solution

Calculate the mole ratio of each element in Compound 1

To determine the empirical formula of Compound 1, first, we must find the moles of each element (M, C, and O) based on their mass percentages. Assume we have 100 g of the compound, we can calculate the number of moles as follows: For M: \( \frac{40.04 g}{molar \: mass \: of \: M} \) For C: \( \frac{12.00 g}{12.01\: g/mol} \) For O: \( \frac{47.96 g}{16.00\: g/mol} \) Next, we need to divide each mole value by the smallest mole value of the elements.

Determine the empirical formula of Compound 1

Divide all the mole values of Step 1 by the smallest mole value to get the mole ratio, and round them to the nearest whole number to get the empirical formula of Compound 1.

Identity of the Group 2 metal M for Compound 1

To find the identity of M, compare the mole ratio calculated in step 2 with the mole ratio of elements in known group 2= metal carbonates, then find the match.

Calculate the mole ratio of each element in Compound 2

Use the mass percentages given for Compound 2 to calculate the moles for M, C, and O in the same way as in Step 1.

Determine the empirical formula of Compound 2

Divide all the mole values of Step 4 by the smallest mole value to get the mole ratio, and round them to the nearest whole number to get the empirical formula of Compound 2.

Identity of the Group 2 metal M for Compound 2

To find the identity of M, compare the mole ratio calculated in step 5 with the mole ratio of elements in known group 2 metal carbonates, then find the match.

Identify the colorless gas produced during decomposition

The colorless gas produced during the decomposition of both compounds is a common byproduct of the decomposition of carbonates. In this case, the colorless gas is CO2.

Write the chemical equations for the decomposition of compound 1 and 2

Based on the empirical formulas and the gas produced, write the balanced chemical equations for the decomposition of the two compounds.

Determine if the compounds are ionic or molecular

Based on the nature of group 2 metal carbonates and the empirical formulas found, determine whether compounds 1 and 2 are ionic or molecular.

Key concepts

Empirical Formula

The empirical formula is the simplest representation of the relative number of atoms in a compound. It provides a basic insight into the proportion of different elements present. In chemistry, finding this formula is a crucial step when you've been given information about the percentages of elements, as in the case of Compound 1 and Compound 2 from the lab exercise.
To determine the empirical formula:

• Assume you have a 100 g sample to convert percentage data to grams.
• Calculate moles of each element using their respective atomic masses.
• Identify the smallest number of moles calculated, and then divide all mole quantities by this value to achieve the smallest whole number ratios.

Using this approach, for Compound 1, based on the given percentage composition, we can simplify the proportions of each element to understand the compound's basic structure.

Decomposition Reaction

A decomposition reaction involves breaking down a compound into simpler substances. When carbonates decompose, they typically produce an oxide and carbon dioxide gas. In our lab exercise, both compounds failed to melt and decomposed, releasing a colorless gas.
This gas is identified as carbon dioxide ( CO_2 ), a common byproduct of thermal decomposition of carbonates. Such reactions are crucial in understanding the chemical stability and properties of carbonate compounds.
The general form of a decomposition reaction for metal carbonates is:

• Metal Carbonate ( CaCO_3 ) → Metal Oxide ( CaO ) + CO_2

This helps to interpret experimental results and predict the behavior of similar compounds under heat.

Mole Ratio

Mole ratio represents the ratios of moles of each element in a compound compared to others. It's a vital concept in determining the empirical formula. In our exercise, finding the mole ratio of elements in Compounds 1 and 2 helps in understanding their compositions.
To calculate the mole ratio:

• Compute moles for each element using mass percentages and atomic weights.
• Select the smallest mole value and divide all mole quantities by this number.
• The resulting numbers may need to be rounded to the nearest whole number to reflect the simplest ratio.

This ratio doesn't just assist in empirical formula calculation but also in identifying unknown metals or validating theoretical results by comparing with known compounds.

Carbonate Compounds

Carbonate compounds contain the carbonate ion ( CO_3^{2-} ), which is pivotal in various chemical reactions. They are common in natural minerals and have significant industrial importance. Typically, when these compounds are subjected to heat, they decompose.
This decomposition reaction (as noticed in the lab exercise) under heat produces a metal oxide and releases carbon dioxide gas.
Knowing the nature of carbonate compounds is helpful for:

• Architectural studies, where many building stones are carbonate based.
• Understanding geological formations.
• Industrial processes like cement manufacturing.

The behaviors documented in this exercise help in applying theoretical knowledge to real-life scenarios.

Ionic Compounds

Ionic compounds consist of positively charged and negatively charged ions. Group 2 metals form ionic compounds with carbonate ions, leading to the substances you examined in the exercise. The exchange of electrons forms these ionic bonds, resulting in compounds with distinct physical properties like high melting points.
In the provided lab scenario, both compounds are likely ionic. This is evident from their behavior:

• They decompose rather than melt under heat, a trait common in ionic substances.
• The chemistry of Group 2 metals, known for forming ionic compounds, reaffirms this.

Understanding the ionic nature of compounds can assist in anticipating their chemical reactions and behavior under different environmental influences.