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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 \(2: 69.59 \%\) 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 M? (b) What is the empirical formula for Compound 2 and the identity of 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. \((\mathbf{e})\) Are compounds 1 and 2 ionic or molecular?

Step by step solution

01

Determine Molar Masses

To find the empirical formula, we first need the molar masses of carbon (C), oxygen (O), and the possible Group 2 metals: magnesium (Mg), calcium (Ca), strontium (Sr), or barium (Ba). These are approximated as: C = 12 g/mol, O = 16 g/mol, Mg = 24 g/mol, Ca = 40 g/mol, Sr = 88 g/mol, Ba = 137 g/mol.

02

Calculate Moles for Compound 1

For Compound 1, calculate moles for each element using given percentages: \( n(M) = \frac{40.04}{M_m} \), \( n(C) = \frac{12.00}{12} = 1 \), \( n(O) = \frac{47.96}{16} = 2.997 \approx 3 \). Accurate results require the mass of M to match the molar ratio with C (1 mol C to 3 mol O suggests a formula like MCO鈧�).

03

Identify M for Compound 1

By matching the calculated mole ratio 1:1:3 to the known carbonates, compute \( M_m \) for potential metals. Only calcium (Ca) with \( M_m = 40 \) matches the empirical formula MCO鈧�, identifying \( M \) as calcium (Ca). Thus, the empirical formula is CaCO鈧�.

04

Calculate Moles for Compound 2

For Compound 2, calculate moles using: \( n(M) = \frac{69.59}{M_m} \), \( n(C) = \frac{6.09}{12} = 0.5075 \approx 0.51 \), \( n(O) = \frac{24.32}{16} = 1.52 \approx 1.5 \). This suggests a formula like M鈧侰O鈧�.

05

Identify M for Compound 2

By observing the simplest ratio of 2:1:3 for met carb, a metallic molar mass approximating barium's (Ba) molar mass fits. So the empirical formula is BaCO鈧� with \( M \) as barium (Ba).

06

Identify Colorless Gas

Both compounds decompose upon heating, which is typical for carbonates to release carbon dioxide (CO鈧�), a colorless gas.

07

Write Decomposition Reactions

Write balanced chemical equations: Compound 1: \[ \text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2 \] Compound 2: \[ \text{BaCO}_3 \rightarrow \text{BaO} + \text{CO}_2 \]

08

Determine Ionic or Molecular Nature

Barium and calcium carbonates are ionic compounds because they consist of metal cations (Ca虏鈦�, Ba虏鈦�) and the carbonate anion (CO鈧兟测伝).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula

The empirical formula is the simplest way to represent a chemical compound using the smallest whole number ratios of atoms involved. When trying to determine this formula, it's important to find the correct mass ratios of the constituents and simplify those into integer ratios.
To calculate the empirical formula for a compound like Compound 1, the percentages by mass given for each element are crucial. For Compound 1, the lab analysis provides a composition of 40.04% metal, 12.00% carbon, and 47.96% oxygen. These percentages convert into moles per the molar masses of the individual elements: carbon (12 g/mol), oxygen (16 g/mol), and the unknown metal M.
By dividing these percentages by the molar mass of each element, we obtain the numbers of moles. For carbon and oxygen in Compound 1, we find approximately 1 mol of carbon and 3 moles of oxygen. These ratios suggest a carbonate-type constituent, commonly represented as MCO鈧�.
Similarly, for Compound 2, the ratios suggest a formula of M鈧侰O鈧� based on the approximate 2:1:3 ratio of M to C to O.

Molar Mass Calculation

Calculating molar mass is a fundamental step for identifying a compound's composition. The molar mass is the mass of one mole of a given substance. This is essential when turning the percentage mass of each element into moles.
For the task, let's consider the molar masses of Group 2 metals: magnesium (3 24 g/mol), calcium (40 g/mol), strontium (88 g/mol), and barium (137 g/mol). For the case of Compound 1, our calculation matched well with calcium. We used 40 g/mol for M and confirmed the empirical formula as CaCO鈧�.
In Compound 2, the analysis led us to consider Barium due to the higher metal percentage and its corresponding molar mass of 137 g/mol. This aligned with the calculated ratios and identified the compound as BaCO鈧�. Molar mass hence acts as a bridge between percentage composition and chemical identity.

Chemical Decomposition Reactions

Chemical decomposition reactions occur when a single compound breaks down into two or more simpler substances. This process uses energy, often in the form of heat, to break the bonds in the compound.
In our lab scenario, both compounds undergo decomposition instead of melting. This behavior is typical of carbonates. Normally, thermal decomposition of carbonates produces a metal oxide and releases carbon dioxide as a byproduct. Thus, for Compound 1 (CaCO鈧�), heating results in:

• CaCO鈧� 鈫� CaO + CO鈧�

For Compound 2 (BaCO鈧�), a similar reaction occurs:

• BaCO鈧� 鈫� BaO + CO鈧�

These reactions reinforce the understanding that, despite not changing state to a liquid, carbonates break down into consistent gaseous and solid products.

Ionic Compounds

Ionic compounds consist of positively charged cations and negatively charged anions. In this situation, the Group 2 metal carbonates are ionic due to the presence of metal cations and non-metal anions.
For Compounds 1 and 2, both contain metal cations鈥攃alcium (Ca虏鈦�) and barium (Ba虏鈦�) respectively鈥攁nd carbonate anions (CO鈧兟测伝). The strong electrostatic forces between these oppositely charged ions result in the formation of a stable ionic lattice.
Ionic compounds like these often have high melting points, but due to the thermal decomposition in our experiment, we notice that they decompose before melting. This highlights the robust nature of ionic bonds, yet underlines the often overlooked thermal instability of certain ionic compounds like metal carbonates.