Question

You and a partner are asked to complete a lab entitled " \(\mathrm{Ox}\) ides of Ruthenium" that is scheduled to extend over two lal periods. The first lab, which is to be completed by your part ner, is devoted to carrying out compositional analysis. In th second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yel low substance and the other a black powder. You also find the following notes in your partner's notebook-Compounc \(1: 76.0 \% \mathrm{Ru}\) and \(24.0 \% \mathrm{O}\) (by mass), Compound \(2: 61.2 \% \mathrm{R}\) and \(38.8 \% \mathrm{O}\) (by mass). (a) What is the empirical formula for Compound 1 ? (b) What is the empirical formula for Compound 2? Upon determining the melting points of these two compounds, you find that the yellow compound melts at \(25^{\circ} \mathrm{C}\), while the black powder does not melt up to the maximum temperature o. your apparatus, \(1200^{\circ} \mathrm{C}\). (c) What is the identity of the yellow compound? (d) What is the identity of the black compound? (e) Which compound is molecular? (f) Which compound is ionic?

Short answer

The empirical formula for Compound 1 is RuO₂ and for Compound 2 is RuO₄. The yellow compound is Ruthenium dioxide (RuO₂) and the black compound is Ruthenium tetroxide (RuO₄). The molecular compound is RuO₄, and the ionic compound is RuO₂.

Step-by-step solution

Calculate moles of Ru and O in Compound 1

Assume that we have 100 grams of Compound 1. In this case, we have 76 grams of Ru and 24 grams of O. To find the moles of each element, we can use the formula: moles = mass / molar_mass The molar mass of Ru is 101.07 g/mol, and the molar mass of O is 16.00 g/mol. Moles of Ru = 76 g / 101.07 g/mol = 0.752 moles Moles of O = 24 g / 16.00 g/mol = 1.500 moles

Find the empirical formula for Compound 1

To find the empirical formula, we will divide the moles of each element by the smallest number of moles, in this case 0.752 moles, and then round to the nearest whole numbers: Ru = 0.752 / 0.752 = 1 O = 1.500 / 0.752 = 1.99 ≈ 2 So, the empirical formula for Compound 1 is RuO₂.

Calculate moles of Ru and O in Compound 2

Again, assume that we have 100 grams of Compound 2, which contains 61.2 grams of Ru and 38.8 grams of O. Using the same formula as in step 1: Moles of Ru = 61.2 g / 101.07 g/mol = 0.605 moles Moles of O = 38.8 g / 16.00 g/mol = 2.425 moles

Find the empirical formula for Compound 2

As we did in step 2, we will divide the moles of each element by the smallest number of moles, in this case 0.605 moles, and then round to the nearest whole numbers: Ru = 0.605 / 0.605 = 1 O = 2.425 / 0.605 = 4.01 ≈ 4 So, the empirical formula for Compound 2 is RuO₄.

Identify the yellow and black compounds

We are given that the yellow compound has a melting point of 25°C and the black compound doesn't melt up to 1200°C. We can use this information along with the known properties of ruthenium oxides to identify these compounds. The yellow compound with a melting point of 25°C is actually a known oxide of ruthenium called Ruthenium(IV) oxide or Ruthenium dioxide (RuO₂). The black compound that doesn't melt below 1200°C is another known oxide of ruthenium called Ruthenium(VIII) oxide or Ruthenium tetroxide (RuO₄). So, Compound 1 is RuO₂, and Compound 2 is RuO₄.

Determine the molecular and ionic compounds

RuO₂ (Ruthenium dioxide) is an ionic compound due to the transfer of electrons between Ruthenium and Oxygen, forming a crystalline structure. RuO₄ (Ruthenium tetroxide) is a molecular compound, as it forms covalent bonds between Ruthenium and Oxygen atoms. Therefore, the answers are: (c) The identity of the yellow compound is Ruthenium dioxide (RuO₂). (d) The identity of the black compound is Ruthenium tetroxide (RuO₄). (e) The molecular compound is RuO₄. (f) The ionic compound is RuO₂.

Key concepts

Compositional Analysis

Compositional analysis is a fundamental process used in chemistry to determine the percentage of each element present in a compound. This analysis provides valuable clues about the compound's identity and its empirical formula, which represents the simplest whole-number ratio of the elements involved.

In the given exercise, your partner conducted compositional analysis on two ruthenium oxides. The method involved determining the mass percentages of ruthenium (Ru) and oxygen (O) in each compound. For instance, in Compound 1, Ru constituted 76.0% of the mass, while O made up 24.0%. For Compound 2, the percentages were 61.2% Ru and 38.8% O.

To derive the empirical formula, you convert these percentages to moles by considering a hypothetical 100 g sample of the compound. Thus, the mass percentages translate directly into mass in grams. By dividing these masses by the respective molar masses (101.07 g/mol for Ru and 16.00 g/mol for O), you obtain the number of moles of each element.

Finally, by dividing the moles of each element by the smallest number of moles calculated, you simplify the ratio to whole numbers. This results in the empirical formula, such as RuO₂ for Compound 1 and RuO₄ for Compound 2.

Melting Points

Melting points are a key physical property often used to help identify and characterize substances. They are especially useful when working with unknown compounds, such as the yellow and black samples in the experiment. Generally, ionic compounds have higher melting points than molecular compounds due to the stronger forces holding their crystal lattice together.

In this exercise, the yellow compound had a melting point of 25°C, indicating it is likely a molecular compound, as these tend to melt at lower temperatures. This melting point is consistent with the known properties of Ruthenium(IV) oxide, or RuO₂, which matches this characteristic.

Conversely, the black compound did not melt even up to 1200°C, showcasing a significantly higher melting point typical of ionic or more robust compounds. Ruthenium tetroxide, or RuO₄, being a known black compound of ruthenium, does not match typical ionic compound characteristics by color but fits by bonding. Such observations make it crucial to use melting points alongside compositional analysis when identifying compounds.

Ionic and Molecular Compounds

Understanding the difference between ionic and molecular compounds is essential in interpreting chemical behaviors and properties. **Ionic compounds** are composed of positively and negatively charged ions held together by electrostatic forces known as ionic bonds. These compounds generally exhibit high melting points, are solid at room temperature, and can conduct electricity in molten or dissolved (aqueous) states because their ions are free to move.

In contrast, **molecular compounds** consist of molecules formed by covalent bonds, where electrons are shared between atoms. These compounds typically have lower melting points compared to ionic compounds and exist as gases, liquids, or solids at room temperature but do not conduct electricity in solid form.

In the experiment, RuO₂ was identified as the ionic compound. It forms a crystalline structure typical for ionic compounds, which aligns with its high melting point observation. RuO₄ is identified as the molecular compound. Although black in color, its physical properties, such as relatively low reactivity under testing conditions and sublimation, make it consistent with molecular characteristics. Knowing the compound type helps in predicting its physical properties and reactions, which is vital in chemical analysis and applications.