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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

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

Step by step solution

01

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 \( \text{H}_2\text{O} = \frac{0.403 \, \text{g}}{18.02 \, \text{g/mol}} = 0.02237 \, \text{mol} \)
02

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 \( \text{O} = \frac{0.02237 \, \text{mol}}{1 \, \text{mol} \, \text{H}_2\text{O}/ 0.5 \, \text{mol} \, \text{O}} = 0.01118 \, \text{mol} \)
03

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 \( \text{Cl} = 0.00639 \, \text{mol} \).\( \text{Cl} : \frac{0.00639}{0.00639} = 1 \)\( \text{O} : \frac{0.01118}{0.00639} \approx 1.75 \)
04

Adjust to Obtain Whole Numbers

Multiply through the ratios obtained to get whole numbers:Multiply both by 4 to obtain near whole numbers:- \( \text{Cl} = 1 \times 4 = 4 \)- \( \text{O} = 1.75 \times 4 = 7 \)
05

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 \( \text{Cl}_4\text{O}_7 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 Hâ‚‚O 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 (Hâ‚‚) to produce HCl and Hâ‚‚O. 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.

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Most popular questions from this chapter

Carbon dioxide \(\left(\mathrm{CO}_{2}\right)\) is the gas that is mainly responsible for global warming (the greenhouse effect). The burning of fossil fuels is a major cause of the increased concentration of \(\mathrm{CO}_{2}\) in the atmosphere. Carbon dioxide is also the end product of metabolism (see Sample Problem 3.4). Using glucose as an example of food, calculate the annual human production of \(\mathrm{CO}_{2}\) in grams, assuming that each person consumes \(5.0 \times 10^{2} \mathrm{~g}\) of glucose per day, that the world's population is 6.5 billion, and that there are 365 days in a year.

The atomic mass of element \(\mathrm{X}\) is 33.42 amu. A \(27.22-\mathrm{g}\) sample of \(\mathrm{X}\) combines with \(84.10 \mathrm{~g}\) of another element \(\mathrm{Y}\) to form a compound XY. Calculate the atomic mass of Y.

Calculate the percent composition by mass of all the elements in calcium phosphate \(\left[\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right]\), a major component of bone.

A sample containing \(\mathrm{NaCl}, \mathrm{Na}_{2} \mathrm{SO}_{4},\) and \(\mathrm{NaNO}_{3}\) gives the following elemental analysis: 32.08 percent \(\mathrm{Na}\), 36.01 percent \(\mathrm{O}, 19.51\) percent \(\mathrm{Cl} .\) Calculate the mass percent of each compound in the sample.

The annual production of sulfur dioxide from burning coal and fossil fuels, auto exhaust, and other sources is about 26 million tons. The equation for the reaction is $$ \mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) $$ How much sulfur (in tons), present in the original materials. would result in that quantity of \(\mathrm{SO}_{2}\) ?

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