You and a partner are asked to complete a lab entitled "Oxides of Ruthenium" 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. In the second lab, you are to determine melting points. Upon going to lab you find two unlabeled vials, one containing a soft yellow substance and the other a black powder. You also find the following notes in your partner's notebook-Compound 1: \(76.0 \%\) \(\mathrm{Ru}\) and \(24.0 \% \mathrm{O}\) (by mass), Compound 2: \(61.2 \% \mathrm{Ru}\) and \(38.8 \%\) O (by mass). (a) What is the empirical formula for Compound \(1 ?\) (b) What is the empirical formula for Compound \(2 ?\) (c) 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 of your apparatus, \(1200^{\circ} \mathrm{C}\). What is the identity of the yellow compound? What is the identity of the black compound? Be sure to use the appropriate naming convention depending on whether the compound is better described as a molecular or ionic compound.

Step by step solution

01

Finding the Empirical Formula for Compound 1

To find the empirical formula, we will do the following steps: 1. Assume a sample of 100g so that the percent becomes mass. 2. Convert the masses to moles. 3. Divide the number of moles of each element by the smallest number from step 2. 4. If any ratios are close to a whole number, round them up or down. 5. Use the whole number ratios as subscripts to form the empirical formula. For Compound 1, with 76.0% Ru and 24.0% O: 1. Assume a sample of 100g: 76.0g Ru and 24.0g O. 2. Convert to moles: \( \frac{76.0g\: Ru}{101.07g/mol} = 0.752\: mol\: Ru \) and \( \frac{24.0g\: O}{16.00g/mol} = 1.50\: mol\: O \) 3. Divide by the smallest number of moles: 0.752 mol Ru and 1.50 mol O. Both by 0.752, we get: 1 Ru : 2 O 4. The ratio is already in whole numbers. 5. Empirical Formula for Compound 1: RuO₂.

02

Finding the Empirical Formula for Compound 2

For Compound 2, with 61.2% Ru and 38.8% O: 1. Assume a sample of 100g: 61.2g Ru and 38.8g O. 2. Convert to moles: \( \frac{61.2g\: Ru}{101.07g/mol} = 0.606\: mol\: Ru \) and \( \frac{38.8g\: O}{16.00g/mol} = 2.43\: mol\: O \) 3. Divide by the smallest number of moles: 0.606 mol Ru and 2.43 mol O. Both by 0.606, we get: 1 Ru : 4 O 4. The ratio is already in whole numbers. 5. Empirical Formula for Compound 2: RuO₄. Now we will identify the compounds using their melting points:

03

Identification of Compound 1

The yellow compound melts at 25°C. Comparing the physical description of the compound to the two compounds, we find that Compound 1 matches this description. The empirical formula for Compound 1 is RuO₂. Given that Ruthenium is a metal and Oxygen is a non-metal, we can classify this compound as an ionic compound. Therefore, the correct naming convention for Compound 1 is Ruthenium(IV) Oxide.

04

Identification of Compound 2

The black compound does not melt up to a temperature of 1200°C. Comparing the physical description of the compound to the two compounds, we find that Compound 2 matches this description. The empirical formula for Compound 2 is RuO₄. Given that Ruthenium is a metal and Oxygen is a non-metal, we can also classify this compound as an ionic compound. Therefore, the correct naming convention for Compound 2 is Ruthenium(VIII) Oxide.

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