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 compound.

Short answer

The empirical formula is \( \text{Cl}_4\text{O}_7 \).

Step-by-step solution

Calculate Moles of HCl and H2O

Determine the moles of each product formed. For \( \text{HCl} \):- Molar mass of \( \text{HCl} = 36.46 \, \text{g/mol} \)- Moles of \( \text{HCl} = \frac{0.233 \, \text{g}}{36.46 \, \text{g/mol}} = 0.00639 \, \text{mol} \)For \( \text{H}_2\text{O} \):- Molar mass of \( \text{H}_2\text{O} = 18.02 \, \text{g/mol} \)- Moles of \( \text{H}_2\text{O} = \frac{0.403 \, \text{g}}{18.02 \, \text{g/mol}} = 0.02237 \, \text{mol} \)

Determine Moles of Elements

Use the reactants to determine moles of \( \text{Cl} \) and \( \text{O} \).From \( \text{HCl} \):- Moles of \( \text{Cl} = 0.00639 \, \text{mol} \)From \( \text{H}_2\text{O} \):- Moles of \( \text{O} = \frac{0.02237 \, \text{mol}}{1 \, \text{mol} \, \text{H}_2\text{O}/ 0.5 \, \text{mol} \, \text{O}} = 0.01118 \, \text{mol} \)

Calculate the Simplest Mole Ratio

Divide the moles of each element by the smallest moles to find the simplest whole number ratio.Smallest moles is for \( \text{Cl} = 0.00639 \, \text{mol} \).\( \text{Cl} : \frac{0.00639}{0.00639} = 1 \)\( \text{O} : \frac{0.01118}{0.00639} \approx 1.75 \)

Adjust to Obtain Whole Numbers

Multiply through the ratios obtained to get whole numbers:Multiply both by 4 to obtain near whole numbers:- \( \text{Cl} = 1 \times 4 = 4 \)- \( \text{O} = 1.75 \times 4 = 7 \)

Write the Empirical Formula

Using the whole number ratio of \( \text{Cl}_4\text{O}_7 \), write the empirical formula. Hence, the empirical formula of the compound is \( \text{Cl}_4\text{O}_7 \).

Key concepts

Molar Mass

Understanding molar mass is crucial because it connects the mass of a substance to the amount in moles. It represents the mass of one mole of a given substance, usually expressed in grams per mole (g/mol). This is a vital concept in chemistry because it helps convert between grams and moles, which is often necessary in problem-solving. For example, in the problem above, the molar mass of HCl is given as 36.46 g/mol. By knowing this, you can calculate the moles of HCl from its mass by using the formula:\[\text{Moles of HCl} = \frac{\text{mass of HCl}}{\text{molar mass of HCl}}\]This calculation allows us to work with a more manageable number in chemical equations, especially since reactions often deal with very small or very large quantities. Being comfortable with molar mass calculations aids in interpreting laboratory data and solving chemical equations.

Mole Ratio

Mole ratio is a key factor when dealing with chemical reactions and formulas. It's derived from the coefficients of substances in balanced chemical equations. Understanding mole ratio allows chemists to predict how much of a substance will be consumed or produced in a reaction. In the example from the exercise, we calculated the moles of HCl and H₂O formed and then identified the moles of chlorine (Cl) and oxygen (O) based on these products. The mole ratio of the elements is determined by dividing the moles of each by the smallest number of moles present. - This calculation helps in determining empirically how the elements combine in a compound, - Ultimately, reflecting the simplest whole number ratio of the elements involved, guiding how we write the empirical formula. Thus, understanding and determining the mole ratio is critical in synthesizing compounds and achieving balanced reactions.

Chemical Reaction

In a chemical reaction, substances undergo chemical changes to form new substances. Reactants interact, and products are the outcome of such interactions. The given problem involves a reaction between a compound of chlorine (Cl) and oxygen (O) with hydrogen (H₂) to produce HCl and H₂O. Chemical reactions can be depicted in balanced equations, showing the proportionate relationships between the reactants and products. These relationships are crucial in computing theoretical yields and actual yields: - Balanced equations ensure that the law of conservation of mass is upheld. - They also establish the basis for calculating the amounts of reactants needed or products formed, by providing the mole ratio used in further calculations. By understanding the chemical reaction involved, you gain insights into the stoichiometry of the process, accurately predicting the inputs needed to achieve the desired results.

Elemental Composition

Elemental composition refers to the determination of the specific elements and their proportions within a compound. Knowing the composition of a compound helps in deducing empirical formulas and deeper chemical analysis. From the reaction products given in the exercise, the elemental composition of the original Cl and O compound was determined. By finding the moles of Cl and O and analyzing their simplest ratio, we discerned the most basic pattern of atom combination in the compound. - This process starts by breaking down the compound into its basic elements and converting the masses into moles. - The elemental composition then reveals the identity and number of atoms in the simplest terms, often leading to the empirical formula of the compound. Such analysis not only uncovers the formula but also provides insight into the behavior and characteristics of the substance, important for applications ranging from industrial synthesis to environmental science.