Question

In \(1774,\) C. Scheele obtained a gas by reacting pyrolusite \(\left(\mathrm{MnO}_{2}\right)\) with sulfuric acid. The gas, which had been obtained that same year by Joseph Priestley by a different method, was an element, A. (a) What is the element isolated by Scheele and Priestley? (b) Element A combines with almost all other elements. For example, with cesium it gives a compound in which the mass percent of \(\mathrm{A}\) is \(19.39 \%\) The element combines with hydrogen to give a compound with a mass percent of element A of 94.12\%. Determine the formulas of the cesium and hydrogen compounds. (c) The compounds of cesium and hydrogen with element A react with one another. Write a balanced equation for the reaction.

Short answer

(a) Oxygen (O); (b) Cesium oxide: Cs2O, Water: H2O; (c) Cs2O + H2O → 2CsOH.

Step-by-step solution

Determine Element A

C. Scheele and Joseph Priestley both isolated the same gas using different methods in 1774. The element is oxygen, denoted by the symbol O. Scheele obtained it by reacting manganese dioxide (MnO2) with sulfuric acid, while Priestley did so by heating mercuric oxide.

Calculate Formula for Cesium Compound

Element A is oxygen. We need to determine the formula of the cesium compound where oxygen makes up 19.39% by mass. Let the formula of the compound be Cs_xO_y:- Mass of Cs = 132.91 g/mol; Mass of O = 16 g/mol.- Molecular mass of Cs_xO_y is given by: \[ 132.91x + 16y \]- Percent by mass of O is: \[ \frac{16y}{132.91x + 16y} \times 100 = 19.39 \% \approx y=1, x=2 \]- The formula is thus Cs2O.

Calculate Formula for Hydrogen Compound

Element A also combines with hydrogen to form a compound where oxygen constitutes 94.12% by mass. Let the formula of the compound be H_zO_w:- Mass of H = 1 g/mol; Mass of O = 16 g/mol.- The molar mass of H_zO_w is given by: \[ z \times 1 + w \times 16 \]- Percent by mass of O in the compound is: \[ \frac{16w}{z + 16w} \times 100 = 94.12 \% \approx z=2, w=1 \]- The formula is \(H_2O\) as calculated by the percentage composition.

Balanced Chemical Reaction

Now, write the chemical reaction between cesium oxide (Cs2O) and water (H2O):- Cs2O reacts with water to form cesium hydroxide. - Balanced reaction: \[ \text{Cs}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{CsOH} \]

Key concepts

Element Identification

C. Scheele and J. Priestley are known for isolating a unique gas in the year 1774. This gas was none other than the element oxygen. Oxygen is pivotal due to its high reactivity and essential role in life on Earth. Understanding the identification of elements like oxygen through experiments helps to streamline the study of chemical reactions.
Oxygen was identified by reacting pyrolusite (\(\mathrm{MnO}_2\)) with sulfuric acid, leading to its isolation by Scheele. On the other hand, Priestley achieved this by heating mercuric oxide. Steps like these underscore the meticulous processes in science that help discover elemental substances.
Element identification often involves recognizing unique properties—such as reactivity—with other substances, which characterizes them. Oxygen, for example, readily combines with many elements, making it highly significant in chemical reactions.

Compound Formulas

Compounds are formed when elements chemically combine. To understand a compound's formula, percentages of elements within structures play a significant role. Let's explore the compounds formed by oxygen with cesium and hydrogen.
For the cesium compound, oxygen constitutes 19.39% of the compound’s mass. The molar mass of cesium is 132.91 g/mol and for oxygen, it’s 16 g/mol. To find the formula, the molecular weight can be approximated by solving for the unknowns in \[ 132.91x + 16y \]. Using percentage calculations, we determine the formula to be \(\mathrm{Cs}_2\mathrm{O}\).
In a similar way, for the hydrogen compound where oxygen is 94.12% by mass, hydrogen's molar mass is 1 g/mol. The calculation gives the well-known formula, \(\mathrm{H}_2\mathrm{O}\). This highlights the importance of mass percentages and how they aid in discovering compound formulas.

Balancing Equations

Balancing chemical equations is a fundamental exercise in chemistry. It requires that the number of atoms of each element is the same on both sides of the equation.
In this scenario, we balance the chemical reaction between cesium oxide (\(\mathrm{Cs}_2\mathrm{O}\)) and water (\(\mathrm{H}_2\mathrm{O}\)). The reaction results in the formation of cesium hydroxide (\(\mathrm{CsOH}\)).
The balanced equation is: \[ \mathrm{Cs}_2\mathrm{O} + \mathrm{H}_2\mathrm{O} \rightarrow 2\mathrm{CsOH} \]
This equation demonstrates that each element maintains its atomic count from reactants to products, reflecting the law of conservation of mass. Balancing ensures that no atoms are lost or gained, only rearranged, during chemical reactions, which is essential for accurate chemical analysis and synthesis.