Problem 102 - Chemistry

Question

A sample of a compound of \(\mathrm{Cl}\) and \(\mathrm{O}\) reacts with an excess of \(\mathrm{H}_{2}\) to give \(0.233 \mathrm{~g}\) of \(\mathrm{HCl}\) and \(0.403 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O} .\) Determine the empirical formula of the \(\mathrm{com}-\) pound.

Step-by-Step Solution

Answer

The empirical formula of the compound is \(\mathrm{ClO}_7\).

Step 1: Identifying Given Information

The information we know is that \(\mathrm{HCl}\) is produced by the amount of \(0.233 \mathrm{g}\) and \(\mathrm{H}_{2} \mathrm{O}\) is produced by the amount of \(0.403 \mathrm{g}\).

Step 2: Calculating Moles of Compounds

The molar mass of \(\mathrm{HCl}\) is \(35.5+1=36.5 \mathrm{g/mol}\) and the molar mass of water is \(18 \mathrm{g/mol}\). So, the number of moles of \(\mathrm{HCl}\) is \(\frac{0.233}{36.5}=0.00639 \mathrm{mol}\) and the number of moles of \(\mathrm{H_2O}\) is \(\frac{0.403}{18}=0.0224 \mathrm{mol}\). This is important to note because the moles of H2O will give us the moles of O in the compound.

Step 3: Getting Moles of Elements

The number of moles of \(\mathrm{Cl}\) would be equal to the number of moles of \(\mathrm{HCl}\), which is \(0.00639 \mathrm{mol}\). The number of moles of \(\mathrm{O}\) would be equal to the number of moles of \(\mathrm{H_2O}\), which is \(0.0224 \mathrm{mol}\).

Step 4: Determining Empirical Formula

The empirical formula is obtained by dividing the moles of each element by the smallest number of moles calculated, which in this case is the number of moles of \(\mathrm{Cl}\): \(\frac{0.00639}{0.00639}=1\) for Cl and \(\frac{0.0224}{0.00639}\)=3.5 for O. But an empirical formula should consist of whole numbers, multiply both numbers by 2 to give Cl1O7. Hence the empirical formula of the compound is \(\mathrm{ClO}_7\).

Key Concepts

Moles Calculation

Understanding how to calculate moles is essential in chemistry, especially when determining the empirical formula of a compound. The mole is a fundamental unit in chemistry that allows scientists to count atoms, molecules, or ions by weighing them. It is defined as the amount of substance that contains as many entities as there are atoms in 12 grams of carbon-12, which is approximately \(6.022 \times 10^{23}\) entities, also known as Avogadro's number.
To calculate moles, you simply divide the mass of a substance by its molar mass, which is the weight of one mole of a compound or element.

• For example, if you've got 0.233 grams of \(\mathrm{HCl}\) and you want to find out how many moles that is, you would use the molar mass of \(\mathrm{HCl}\), which is \(36.5 \ \mathrm{g/mol}\). The calculation would be: \[\text{Moles of HCl} = \frac{0.233}{36.5} \approx 0.00639 \, \text{moles}\]
• Similarly, for \(0.403 \ \mathrm{g}\) of \(\mathrm{H_2O}\), using the molar mass of water \(18 \ \mathrm{g/mol}\), the moles of water is \[\text{Moles of water} = \frac{0.403}{18} \approx 0.0224 \, \text{moles}\]

This step is crucial because converting masses to moles allows us to use a stoichiometric relationship between volumes or weights and countable unit quantities of substances in chemical reactions.

Chemical Reactions

Chemical reactions involve the transformation of one or more substances into different substances through the breaking and forming of chemical bonds. Understanding chemical reactions is essential to finding out why and how compounds form.
In the exercise, the chemical reaction occurs between a compound containing chlorine \((\mathrm{Cl})\) and oxygen \((\mathrm{O})\), and hydrogen \((\mathrm{H})\) is in excess. The objective is to find the product quantities formed, like \(\mathrm{HCl}\) and \(\mathrm{H_2O}\). This reaction formation helps determine the empirical formula; however, the details of the reaction steps were not directly listed in the problem itself.
Chemical reactions follow the law of conservation of mass. This principle states that matter cannot be created or destroyed in an isolated system. Therefore, knowing the masses of the products helps backtrack and find the moles of each element involved, which is a crucial step in the determination of the empirical formula. The reaction of \(\mathrm{HCl}\) and \(\mathrm{H_2O}\) as products indicates that the compound reacting with hydrogen contains chlorine and oxygen. The aim is to map these back to the original compound to understand its structure better.

Stoichiometry

Stoichiometry is the bridge between balancing chemical reactions and determining the quantities of reactants and products involved. It relies heavily on the concept of moles, as every balanced chemical equation provides the ratio of moles of each substance involved in the reaction.
Stoichiometric calculations are essential for determining empirical formulas because they allow us to use mole ratios to find out the quantitative relationship between reactants and products. In the provided exercise, you calculated the number of moles for \(\mathrm{Cl}\) and \(\mathrm{O}\), based on the moles of \(\mathrm{HCl}\) and \(\mathrm{H_2O}\) produced:

• Moles of \(\mathrm{Cl}\) are deduced from moles of \(\mathrm{HCl}\).
• Moles of \(\mathrm{O}\) are deduced from the moles of \(\mathrm{H_2O}\).

Once you have the moles for each element, you find the smallest mole number to use as a division factor. This helps in simplifying to the smallest whole number ratio. If the ratio isn't whole, as found in the solution (\(\mathrm{Cl}_1\mathrm{O}_{3.5}\)), you multiply by a common factor (in this instance, 2) to create a whole number empirical formula. Thus, improving comprehension in stoichiometry deepens our understanding of chemical formulas and reactions in chemistry.