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 \(1 : 76.0 \%\) Ru and 24.0\(\%\) O (by mass), Compound \(2 : 61.2 \%\) 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\(?\) 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, \(1200^{\circ} \mathrm{C}\) . (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?

Step by step solution

01

(a) Empirical formula of Compound 1:

To find the empirical formula of Compound 1, we need to start by determining the moles of each element present in the compound. First, let's assume we have 100 grams of compound 1; this assumption allows us to use the given percentage compositions directly as masses of the elements present. Ru: \(\frac{76.0\text{ g}}{101.07\text{ g/mol}} = 0.752\text{ mol}\) O: \(\frac{24.0\text{ g}}{16.00\text{ g/mol}} = 1.50\text{ mol}\) Now, divide both values by the smallest value: Ru: \(\frac{0.752}{0.752} = 1\) O: \(\frac{1.50}{0.752} \approx 2\) The empirical formula of Compound 1 is RuO\( _{2}\).

02

(b) Empirical formula of Compound 2:

Similarly, let's calculate the empirical formula for compound 2, again assuming we have 100 grams of the compound: Ru: \(\frac{61.2\text{ g}}{101.07\text{ g/mol}} = 0.606\text{ mol}\) O: \(\frac{38.8\text{ g}}{16.00\text{ g/mol}} = 2.43\text{ mol}\) Now, divide both values by the smallest value: Ru: \(\frac{0.606}{0.606} = 1\) O: \(\frac{2.43}{0.606} \approx 4\) The empirical formula of Compound 2 is RuO\( _{4}\).

03

(c) Identity of yellow compound:

The yellow compound has a melting point of \(25^{\circ}\mathrm{C}\). This is a relatively low melting point, which is characteristic of molecular compounds. Thus, the yellow compound is likely Compound 2 (RuO\( _{4}\)).

04

(d) Identity of black compound:

The black compound does not melt up to \(1200^{\circ} \mathrm{C}\), which is a very high melting point. This is characteristic of ionic compounds. Thus, the black compound is likely Compound 1 (RuO\( _{2}\)).

05

(e) Molecular compound:

From our discussion in part (c), the molecular compound is the yellow compound, which is compound 2 or RuO\( _{4}\).

06

(f) Ionic compound:

From our discussion in part (d), the ionic compound is the black compound, which is compound 1 or RuO\( _{2}\).

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