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

Expert verified
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

01

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.
02

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.
03

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.
04

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.
05

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.
06

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.
07

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.
08

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.
09

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Potassium peroxide is composed of \(70.96 \% \mathrm{~K}\) and $29.04 \% \mathrm{O}\(. Each peroxide ion has a net charge of \)2^{-}$. (a) Given the molecular weight of the compound is 110.19 , determine the chemical formula of potassium peroxide. \((\mathbf{b})\) What is the number of valance electron pairs in the peroxide ion? (c) Draw the Lewis structure of the peroxide ion. (d) Compare the \(\mathrm{O}-\mathrm{O}\) bond distance in oxygen molecule and potassium peroxide.

Consider the formate ion, \(\mathrm{HCO}_{2}^{-}\), which is the anion formed when formic acid loses an \(\mathrm{H}^{+}\) ion. The \(\mathrm{H}\) and the two O atoms are bonded to the central C atom. (a) Draw the best Lewis structure(s) for this ion. (b) Are resonance structures needed to describe the structure? (c) Would you predict that the \(\mathrm{C}-\mathrm{O}\) bond lengths in the formate ion would be longer or shorter relative to those in \(\mathrm{CO}_{2}\) ?

List the individual steps used in constructing a Born-Haber cycle for the formation of BaI \(_{2}\) from the elements. Which of the steps would you expect to be exothermic?

In the following pairs of binary compounds, determine which one is a molecular substance and which one is an ionic substance. Use the appropriate naming convention (for ionic or molecular substances) to assign a name to each compound: (a) \(\mathrm{SiF}_{4}\) and \(\mathrm{LaF}_{3}\), (b) \(\mathrm{FeCl}_{2}\) and \(\mathrm{ReCl}_{6}\), (c) \(\mathrm{PbCl}_{4}\) and RbCl.

Which of the following statements about electronegativity is false? (a) Electronegativity is the ability of an atom in a molecule to attract electron density toward itself. (b) Electronegativity is the same thing as electron affinity. (c) The numerical values for electronegativity have no units. (d) Fluorine is the most electronegative element. (e) Cesium is the least electronegative element.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free