Problem 74 - Chemistry An Atoms First Approach

Question

In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen \((77 \mathrm{K})\) was discovered. The approximate formula of this substance is \(\mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}\) Calculate the percent composition by mass of this material.

Step-by-Step Solution

Answer

The percent composition by mass of the YBa\(_2\)Cu\(_3\)O\(_7\) compound is: Yttrium (Y) - \(13.50\%\), Barium (Ba) - \(41.22\%\), Copper (Cu) - \(28.61\%\), and Oxygen (O) - \(16.67\%\).

Step 1: Determine the molar mass of each element

To find the percent composition, we will need the molar mass of each element in the compound (Y, Ba, Cu, and O). Using the periodic table, we find the molar mass of each element: Yttrium (Y): \(89.905\, \mathrm{g/mol}\) Barium (Ba): \(137.33\, \mathrm{g/mol}\) Copper (Cu): \(63.546\, \mathrm{g/mol}\) Oxygen (O): \(16.00\, \mathrm{g/mol}\)

Step 2: Calculate the total molar mass of the compound YBa\(_2\)Cu\(_3\)O\(_7\)

Use the molar masses and their respective proportions in the compound to find the total molar mass: Total molar mass = (1 x molar mass of Y) + (2 x molar mass of Ba) + (3 x molar mass of Cu) + (7 x molar mass of O) Total molar mass = \[ (1 x 89.905) + (2 x 137.33) + (3 x 63.546) + (7 x 16.00) \mathrm{g/mol} \] Total molar mass = \( 666.262 \, \mathrm{g/mol} \)

Step 3: Calculate the percent composition by mass of each element in the compound

Use the individual molar mass of each element and divide it by the total molar mass of the compound, then multiply by 100% to obtain the percent composition of each element: Percent composition (Y) = \(\frac{89.905}{666.262} \times 100\% = 13.50\%\) Percent composition (Ba) = \(\frac{2 \times 137.33}{666.262} \times 100\% = 41.22\%\) Percent composition (Cu) = \(\frac{3 \times 63.546}{666.262} \times 100\% = 28.61\%\) Percent composition (O) = \(\frac{7 \times 16.00}{666.262} \times 100\% = 16.67\%\)

Step 4: Summarize the results

Finally, the percent composition by mass of YBa\(_2\)Cu\(_3\)O\(_7\) is as follows: Yttrium (Y): \(13.50\%\) Barium (Ba): \(41.22\%\) Copper (Cu): \(28.61\%\) Oxygen (O): \(16.67\%\)

Key Concepts

Superconductor Materials

Superconductors are fascinating materials that can conduct electricity without any resistance under certain conditions. This means they can carry electric currents without losing energy as heat. A significant breakthrough in this field came in 1987 when a substance capable of acting as a superconductor above the temperature of liquid nitrogen, 77 K, was discovered. The exciting part about this discovery is that it made superconducting materials more practical since liquid nitrogen is more easily available and cheaper than other cooling methods used for superconductors.

The formula for this particular superconductor material is \(\text{YBa}_2\text{Cu}_3\text{O}_7\), which contains Yttrium, Barium, Copper, and Oxygen. These superconductors have a unique arrangement of atoms that allow them to exhibit their remarkable properties. Understanding what goes into these materials, such as their chemical composition, is essential for further advancements in technology reliant on superconductors.

Molar Mass Calculation

Calculating molar mass is a crucial step in determining the percent composition of a compound. The molar mass gives the mass of one mole of a compound and is expressed in grams per mole (g/mol). To find the molar mass of a compound, you need to sum the molar masses of all the atoms present in the compound according to their proportions.

For the compound \(\text{YBa}_2\text{Cu}_3\text{O}_7\), the process involves:

• Identifying the molar mass of each element from the periodic table
• Multiplying the molar mass by the number of times each element appears in the formula

Using the molecular weights: Yttrium: 89.905 g/mol, Barium: 137.33 g/mol, Copper: 63.546 g/mol, and Oxygen: 16.00 g/mol, we calculate the total molar mass of the compound:\[(1 \times 89.905) + (2 \times 137.33) + (3 \times 63.546) + (7 \times 16.00) = 666.262 \, \text{g/mol}.\]

Chemical Formulas

Chemical formulas are the language of chemistry. They provide a way to represent the elements in a compound and the ratio of atoms bonded together. In the formula \(\text{YBa}_2\text{Cu}_3\text{O}_7\), each letter represents a different element: Yttrium (Y), Barium (Ba), Copper (Cu), and Oxygen (O).

The subscript numbers following each element symbol show how many atoms of that element are in a single molecule of the compound. No subscript means there is just one atom. For example, the formula indicates that there are two Barium atoms, three Copper atoms, and seven Oxygen atoms for every Yttrium atom. Understanding these ratios helps in calculating molar masses and percent compositions, which are essential when dealing with chemical reactions and material properties.

Periodic Table Usage

The periodic table is an indispensable tool in chemistry. It organizes all known elements based on their properties and makes it easy to extract vital data needed for various calculations. Each element's cell in the table provides crucial information, including the atomic number, element symbol, and atomic mass.

To find the molar mass of an element, we rely on the atomic mass shown under each element on the table, usually expressed in atomic mass units (u). For calculations like those performed for \(\text{YBa}_2\text{Cu}_3\text{O}_7\), this atomic mass is typically converted to grams per mole (g/mol), aligning with the units needed for molar mass. Periodic table usage is not only limited to identifying molar masses but also extends to understanding the arrangement of elements, predicting chemical properties, and guiding chemical reactions.