Problem 92
Question
How many molecules are contained in each compound? a. 1.35 mol of carbon disulfide (CS_{2} ) b. 0.254 mol of diarsenic trioxide \(\left(\mathrm{As}_{2} \mathrm{O}_{3}\right)\) c. 1.25 mol of water d. 150.0 mol of \(\mathrm{HCl}\)
Step-by-Step Solution
Verified Answer
a. CS_{2}: \(8.130 x 10^{23}\) molecules
b. As_{2}O_{3}: \(1.530 x 10^{23}\) molecules
c. H_{2}O: \(7.528 x 10^{23}\) molecules
d. HCl: \(9.033 x 10^{25}\) molecules
1Step 1: Recall Avogadro's number
The number of particles (atoms, molecules, ions) in one mole of any substance is known as Avogadro's number, which is equal to \(6.022 x 10^{23}\) particles/mol.
2. Calculate the number of molecules for each compound.
2Step 2: a. Determine the number of molecules for CS_{2}
Given moles of carbon disulfide (CS_{2}) = 1.35 mol
To calculate the number of molecules, multiply the given moles by Avogadro's number:
Number of molecules = 1.35 mol x \(6.022 x 10^{23}\) molecules/mol = \(8.130 x 10^{23}\) molecules
3Step 3: b. Determine the number of molecules for As_{2}O_{3}
Given moles of diarsenic trioxide (As_{2}O_{3}) = 0.254 mol
To calculate the number of molecules, multiply the given moles by Avogadro's number:
Number of molecules = 0.254 mol x \(6.022 x 10^{23}\) molecules/mol = \(1.530 x 10^{23}\) molecules
4Step 4: c. Determine the number of molecules for H_{2}O
Given moles of water (H_{2}O) = 1.25 mol
To calculate the number of molecules, multiply the given moles by Avogadro's number:
Number of molecules = 1.25 mol x \(6.022 x 10^{23}\) molecules/mol = \(7.528 x 10^{23}\) molecules
5Step 5: d. Determine the number of molecules for HCl
Given moles of hydrochloric acid (HCl) = 150.0 mol
To calculate the number of molecules, multiply the given moles by Avogadro's number:
Number of molecules = 150.0 mol x \(6.022 x 10^{23}\) molecules/mol = \(9.033 x 10^{25}\) molecules
In summary:
a. CS_{2}: \(8.130 x 10^{23}\) molecules
b. As_{2}O_{3}: \(1.530 x 10^{23}\) molecules
c. H_{2}O: \(7.528 x 10^{23}\) molecules
d. HCl: \(9.033 x 10^{25}\) molecules
Key Concepts
Mole to Molecule CalculationStoichiometryChemical Compounds
Mole to Molecule Calculation
Understanding how to convert moles to molecules is a fundamental aspect of chemistry that informs us about the quantity of a substance on the molecular level. The bridge between these two measurements is known as Avogadro's number, which is approximately equal to \(6.022 \times 10^{23}\) particles per mole. Whether you are dealing with molecules of water or carbon disulfide, this conversion factor remains constant.
For example, to find the number of carbon disulfide molecules (\(CS_2\) in the given problem, you would multiply the amount in moles by Avogadro's number:
\[\text{Number of } CS_2 \text{ molecules} = 1.35 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}\]
This calculation yields \(8.130 \times 10^{23}\) molecules of carbon disulfide, providing insight into the actual number of individual CS2 molecules present in a given amount.
For example, to find the number of carbon disulfide molecules (\(CS_2\) in the given problem, you would multiply the amount in moles by Avogadro's number:
\[\text{Number of } CS_2 \text{ molecules} = 1.35 \text{ mol} \times 6.022 \times 10^{23} \text{ molecules/mol}\]
This calculation yields \(8.130 \times 10^{23}\) molecules of carbon disulfide, providing insight into the actual number of individual CS2 molecules present in a given amount.
Stoichiometry
At the heart of chemistry lies stoichiometry, a branch that quantitatively examines the relationships between reactants and products in a chemical reaction. Understanding stoichiometry enables students to predict the outcomes of reactions and determine how much of a chemical is needed to react completely with another. The stoichiometric coefficients from the balanced chemical equation guide us in these calculations.
For exercises involving stoichiometry, always start by identifying the balanced equation for the chemical reaction. From there, use the moles of one substance to find the moles of another based on their stoichiometric ratio. If the exercise had asked for the amount of CO2 produced from burning carbon disulfide in oxygen, stoichiometry would be the tool to use to solve it. This technique shows how interconnected these concepts are in the study of chemistry.
For exercises involving stoichiometry, always start by identifying the balanced equation for the chemical reaction. From there, use the moles of one substance to find the moles of another based on their stoichiometric ratio. If the exercise had asked for the amount of CO2 produced from burning carbon disulfide in oxygen, stoichiometry would be the tool to use to solve it. This technique shows how interconnected these concepts are in the study of chemistry.
Chemical Compounds
Chemical compounds, such as water (\(H_2O\)), hydrochloric acid (\(HCl\)), carbon disulfide (\(CS_2\)), and diarsenic trioxide (\(As_2O_3\)), are substances formed from two or more different types of atoms held together by chemical bonds. The composition of a compound is represented by its chemical formula, which indicates the types and numbers of atoms involved.
Understanding the chemical formula is crucial when performing mole-to-molecule calculations. Each compound has a unique molecular makeup, affecting its physical properties, reactivity, and role in chemical reactions. For instance, the water molecule's polar nature makes it an excellent solvent, while the strong covalent bonds in carbon disulfide give it distinct chemical properties. By recognizing the different compounds and their formulas, students gain a deeper appreciation of the diversity and complexity of material substances.
Understanding the chemical formula is crucial when performing mole-to-molecule calculations. Each compound has a unique molecular makeup, affecting its physical properties, reactivity, and role in chemical reactions. For instance, the water molecule's polar nature makes it an excellent solvent, while the strong covalent bonds in carbon disulfide give it distinct chemical properties. By recognizing the different compounds and their formulas, students gain a deeper appreciation of the diversity and complexity of material substances.
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