Question

Chloral hydrate \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\right)\) is a drug formerly used as a sedative and hypnotic. It is the compound used to make "Mickey Finns" in detective stories. a. Calculate the molar mass of chloral hydrate. b. What amount (moles) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3} \mathrm{O}_{2}\) molecules are in \(500.0 \mathrm{~g}\) chloral hydrate? c. What is the mass in grams of \(2.0 \times 10^{-2}\) mol chloral hydrate? d. What number of chlorine atoms are in \(5.0 \mathrm{~g}\) chloral hydrate? e. What mass of chloral hydrate would contain \(1.0 \mathrm{~g} \mathrm{Cl} ?\) \(\mathbf{f}\). What is the mass of exactly 500 molecules of chloral hydrate?

Short answer

a) The molar mass of chloral hydrate is 165.37 g/mol. b) There are 3.02 moles of chloral hydrate in 500.0 g of the compound. c) The mass of \(2.0 \times 10^{-2}\) mol of chloral hydrate is 3.31 g. d) There are \(5.45 \times 10^{22}\) chlorine atoms in 5.0 g of chloral hydrate. e) 1.0 g of Cl is contained in 1.55 g of chloral hydrate. f) The mass of exactly 500 molecules of chloral hydrate is \(1.37 \times 10^{-19}\) g.

Step-by-step solution

Determine the molar mass of each element in the formula

Look up the molar mass of each element present in the formula \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}\mathrm{O}_{2}\): Carbon (C): 12.01 g/mol Hydrogen (H): 1.01 g/mol Chlorine (Cl): 35.45 g/mol Oxygen (O): 16.00 g/mol

Calculate the molar mass of chloral hydrate

Multiply each element's molar mass with its subscript in the formula and sum them up. Molar mass of chloral hydrate = (2 × 12.01) + (3 × 1.01) + (3 × 35.45) + (2 × 16.00) = 165.37 g/mol Answer a: The molar mass of chloral hydrate is 165.37 g/mol. #b. What amount (moles) of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}\mathrm{O}_{2}\) molecules are in \(500.0 \mathrm{~g}\) chloral hydrate?

Use the mass-mole conversion formula

The formula for the conversion is : Moles = mass (g) / molar mass (g/mol) Moles of chloral hydrate = (500.0 g) / (165.37 g/mol) = 3.02 mol Answer b: There are 3.02 moles of chloral hydrate in 500.0 g of the compound. #c. What is the mass in grams of \(2.0 \times 10^{-2}\) mol chloral hydrate?

Convert moles to grams

Apply the following formula: Mass (g) = moles × molar mass (g/mol) Mass of chloral hydrate = \(2.0 \times 10^{-2}\) mol × 165.37 g/mol = 3.31 g Answer c: The mass of \(2.0 \times 10^{-2}\) mol of chloral hydrate is 3.31 g. #d. What number of chlorine atoms are in \(5.0 \mathrm{~g}\) chloral hydrate?

Convert mass of chloral hydrate to moles

Moles = mass (g) / molar mass (g/mol) Moles of chloral hydrate = 5.0 g / 165.37 g/mol = 0.0302 mol

Calculate the number of moles of chlorine atoms

Since chloral hydrate has three chlorine atoms in its formula \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}\mathrm{O}_{2}\), we multiply the number of moles of chloral hydrate by 3. Moles of Cl = 0.0302 mol × 3 = 0.0906 mol

Convert moles of chlorine atoms to the number of atoms

Use Avogadro's number (\(6.022 \times 10^{23}\) atoms/mol) to convert moles of chlorine atoms to the number of atoms. Number of Cl atoms = 0.0906 mol × \(6.022 \times 10^{23}\) atoms/mol = \(5.45 \times 10^{22}\) atoms Answer d: There are \(5.45 \times 10^{22}\) chlorine atoms in 5.0 g of chloral hydrate. #e. What mass of chloral hydrate would contain \(1.0 \mathrm{~g} \mathrm{Cl} ?\)

Calculate the moles of Cl from the given mass

Moles = mass (g) / molar mass (g/mol) Moles of Cl = 1.0 g / 35.45 g/mol = 0.0282 mol

Calculate the moles of chloral hydrate containing 0.0282 mol Cl

Since there are 3 Cl atoms in the formula, we divide the moles of Cl by 3. Moles of chloral hydrate = 0.0282 mol / 3 = 0.0094 mol

Calculate the mass of chloral hydrate containing 0.0094 mol

Mass (g) = moles × molar mass (g/mol) Mass of chloral hydrate = 0.0094 mol × 165.37 g/mol = 1.55 g Answer e: 1.0 g of Cl is contained in 1.55 g of chloral hydrate. #f. What is the mass of exactly 500 molecules of chloral hydrate?

Convert 500 molecules to moles

Moles = number of molecules / Avogadro's number Moles = 500 / \(6.022 \times 10^{23}\) = \(8.30 \times 10^{-22}\) mol

Calculate the mass of 500 molecules of chloral hydrate

Mass (g) = moles × molar mass (g/mol) Mass of chloral hydrate = \(8.30 \times 10^{-22}\) mol × 165.37 g/mol = \(1.37 \times 10^{-19}\) g Answer f: The mass of exactly 500 molecules of chloral hydrate is \(1.37 \times 10^{-19}\) g.

Key concepts

Understanding Stoichiometry

Stoichiometry is a section of chemistry that involves the calculation of the quantities of reactants and products involved in a chemical reaction. It's essential for predicting how much product can be produced from given amounts of reactants or determining the quantity of reactants needed to create a specific amount of product.

When dealing with stoichiometry, the first step is usually to write a balanced chemical equation. This represents the conservation of mass, showing that atoms aren't lost or gained in a chemical reaction. Once the equation is balanced, you utilize the coefficients to determine the mole ratio between the substances. These ratios then become the basis for your calculations, whether you're converting between moles and grams, determining the volume of a gas at STP (Standard Temperature and Pressure), or identifying how many particles, like atoms or molecules, are involved.

Improving student understanding in this area involves emphasizing the importance of a balanced equation and proportional reasoning. By mastering stoichiometry, students will be able to confidently translate from the macroscopic scale (grams, liters, and pieces) to the microscopic scale (molecules, atoms, ions) using the mole as the central unit of translation.

The Role of Avogadro's Number

Avogadro's number, denoted by the symbol \(6.022 \times 10^{23}\), is an incredibly significant figure in chemistry because it connects the micro world of atoms and molecules to the macro world we can measure. When dealing with substances at the atomic or molecular scale, we use this constant to convert between the amount of substance in moles and the number of particles.

One mole of any substance, be it an element or a compound, contains \(6.022 \times 10^{23}\) of its fundamental particles. This number is akin to a chemist's dozen. Just like a dozen always refers to twelve items, a mole always includes Avogadro's number of particles. Whether counting molecules of chloral hydrate or atoms of chlorine, Avogadro's number is integral.

An important tip for students is remembering that Avogadro's number is used in the conversion process—typically from moles to particles or the reverse. Highlighting its application in practical problems, like calculating the number of atoms in a given mass, can help to cement understanding of Avogadro’s constant and its role in mole-to-particle conversions.

Executing Mole-to-Mass Conversion

Mole-to-mass conversion is a fundamental process in stoichiometry, allowing chemists to translate between the mole (the base unit of quantity in chemistry) and mass, which is often more practical to measure. To perform these conversions, you need two key pieces of information: the substance's molar mass and the number of moles you're dealing with.

The molar mass, typically expressed in grams per mole (g/mol), is the mass of one mole of a substance. It's calculated by summing the atomic masses of all atoms in the compound's formula, as seen in the calculation of the molar mass of chloral hydrate. Once the molar mass is known, mole-to-mass conversions become straightforward:

• To find the mass from moles, multiply the number of moles by the molar mass.
• To find moles from a given mass, divide the mass by the molar mass.

Students should practice this two-way conversion until it feels intuitive, ensuring they understand that moles relate to the number of particles, while mass connects to measurable, tangible quantities of a substance. By focusing on the relationship between moles, molar mass, and mass, learners can navigate through a variety of stoichiometric calculations with confidence.