Question

Reaction of a mixture of yttrium oxide, barium carbonate and copper oxide in a stream of oxygen produces A. Elemental analysis of the oxide A gave \(11.9 \% \mathrm{Y}, 36.8 \%\) barium and \(34.1 \%\) copper. (a) Calculate the empirical formula of \(\mathrm{A}\), assuming the rest of the molecular \% is oxygen. (b) Determine the average oxidation state of the copper. (c) If you brought a pellet of \(\mathrm{A}\) at very low temperature towards a permanent magnet, what would you expect to happen and why? (d) Draw the graph expected for the change in resistance with temperature for A. Explain the salient features of the graph, with the aid of diagrams where necessary. What type of superconductor would you expect \(\mathrm{A}\) to be?

Short answer

The empirical formula of A is \(YBa_2Cu_4O_8\), and the average oxidation state of Copper is \(\frac{13}{4}\). At very low temperatures, A will exhibit the Meissner effect, acting as a perfect diamagnet due to the presence of Cooper pairs. The resistance vs. temperature graph for A shows an exponentially decreasing resistance as temperature approaches 0 Kelvin, indicating its superconducting properties, and A can be classified as a high-temperature or Type II superconductor.

Step-by-step solution

Calculate Empirical Formula

First, we will assume a 100g sample containing Yttrium, Barium, Copper, and Oxygen. The mass percentages can be converted to masses in grams, and then into numbers of moles using the molar masses. Yttrium: \(11.9 \% \Rightarrow 11.9 g \Rightarrow \frac{11.9}{88.91} = 0.1338 \hspace{0.3cm}mol\) Barium: \(36.8 \% \Rightarrow 36.8 g \Rightarrow \frac{36.8}{137.33} = 0.2679 \hspace{0.3cm}mol\) Copper: \(34.1 \% \Rightarrow 34.1 g \Rightarrow \frac{34.1}{63.55} = 0.5368 \hspace{0.3cm}mol\) Oxygen: \((100 - 11.9 - 36.8 - 34.1) \% \Rightarrow 17.2 g \Rightarrow \frac{17.2}{16} = 1.0750 \hspace{0.3cm}mol\) Divide each value by the smallest value (0.1338) to get the approximate whole number ratio of the elements: \(Y:Ba:Cu:O = \frac{0.1338}{0.1338}:\frac{0.2679}{0.1338}:\frac{0.5368}{0.1338}:\frac{1.0750}{0.1338} \approx 1:2:4:8\) Thus, the empirical formula is \(YBa_2Cu_4O_8\).

Determine Average Oxidation State of Copper

The oxidation states of the elements in the compound are: Y (+3), Ba (+2), Cu (unknown) and O (-2). We can set up the equation to find the average oxidation state of Copper: \(1(+3)+2(+2)+4x+8(-2)=0\) Solving for x: \(3+4+4x-16=0\) \(4x=-3+16\) \(x=\frac{13}{4}\) The average oxidation state of Copper is \(\frac{13}{4}\).

Mixture A and a Permanent Magnet

When the pellet of A is brought towards a permanent magnet at very low temperatures, we can expect that the pellet will show no attraction or repulsion from the magnet. This phenomenon is known as the Meissner effect, where a superconducting material expels the magnetic field lines, making the material act as a perfect diamagnet. The pellet of A shows superconducting properties due to the specific composition and structure of the compound which allows electrons to form Cooper pairs, leading to the superconductivity.

Resistance vs. Temperature Graph and Superconductor Type

The graph for the change in resistance with temperature for A will have an exponentially decreasing resistance as the temperature approaches 0 Kelvin. The pellet of A becomes a superconductor at low temperatures as the material transitions from normal to a superconducting phase. Resistance will become zero or nearly zero in the superconducting state. The compound A is an example of a high-temperature superconductor (HTS), also known as a Type II superconductor. These superconductors have a higher critical temperature when compared to classical superconductors. They allow higher magnetic fields before becoming normal conductors, which makes them suitable for various applications, such as magnetically levitated trains and power transmission.