10.2 Mole–Mass and Mole–Volume Relationships Read the lesson title aloud to students.

Converting Moles to Mass Determine the mass, in grams, of 9.45 moles of aluminum oxide (Al2O3). Tell students: We can use dimensional analysis to convert between moles and mass. Have students identify the knowns and unknowns for the problem. Click to reveal the knowns and unknowns. Ask: How is the unknown, the mass of aluminum oxide, related to the known, the number of moles of aluminum oxide? Answer: The equation “mass = number of moles × molar mass” relates the two quantities. Point out that we will first need to calculate the molar mass of aluminum oxide, and then we will use the molar mass as a conversion factor.

Converting Moles to Mass, cont. Determine the mass, in grams, of 9.45 moles of aluminum oxide (Al2O3). Ask: How can the molar mass of aluminum oxide be determined? Answer: Add together the atomic masses of two aluminum atoms and three oxygen atoms. Click to reveal the calculation of molar mass for aluminum oxide. Ask: Which form of the molar mass should we use as a conversion factor? Why? Answer: the form that has grams in the numerator, because we need to calculate the number of grams Click to reveal the conversion equation. Point out where the conversion factor shows up in the equation. Emphasize how units in the numerator and denominator cancel during calculation. Click to show the cancellation. Ask a volunteer to write the answer to the conversion on the board. Click to reveal the answer. = 964 g Al2O3

Converting Mass to Moles Calculate how many moles of iron(III) oxide are contained in 92.2 grams of pure Fe2O3. Tell students: We can use dimensional analysis to convert between moles and mass. Have students identify the knowns and unknowns for the problem. Click to reveal the knowns and unknowns. Point out that we will first need to calculate the molar mass of iron(III) oxide, and then we will use the molar mass as a conversion factor.

Converting Mass to Moles, cont. Calculate how many moles of iron(III) oxide are contained in 92.2 grams of pure Fe2O3. Ask: How can the molar mass of iron(III) oxide be determined? Answer: Add together the atomic masses of two iron atoms and three oxygen atoms. Click to reveal the calculation of molar mass for iron(III) oxide. Ask: Which form of the molar mass should we use as a conversion factor? Why? Answer: the form that has moles in the numerator, because we need to calculate the number of moles Click to reveal the conversion equation. Point out where the conversion factor shows up in the equation. Emphasize how units in the numerator and denominator cancel during calculation. Click to show the cancellation. Ask a volunteer to write the answer to the conversion on the board. Click to reveal the answer. = 0.578 mol Fe2O3

Avogadro’s Law and STP According to Avogadro’s law, equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Ask: What are the units that are used for the mass of a mole of gas and the volume of a mole of gas? Answer: grams per mole, g/mol; liters per mole, L/mol Point out that, unlike solids and liquids, the molar volume of gases can be greatly affected by temperature and pressure. Therefore, when comparing the molar volume of gases, it is necessary to have the gases at the same temperature and pressure.

Using Avogadro’s Law Determine the volume, in liters, of 0.60 moles SO2 gas at STP. Tell students: The molar volume of a gas can be used to convert between moles and volume in the same way molar mass can be used to convert between moles and mass. Have students identify the knowns and unknowns for the problem. Click to reveal the knowns and unknowns. Ask: Why must the problem specify that the gas is at STP? Answer: The conversion factor 1 mol/22.4 L is valid only at STP.

Using Avogadro’s Law, cont. Determine the volume, in liters, of 0.60 moles SO2 gas at STP. Ask: How can you convert between moles and volume for a gas at STP? Answer: Use Avogadro’s law and the molar volume of a gas. Click to reveal the conversion between volume and moles. Click to reveal the calculation for volume of SO2. Emphasize the importance of writing a unit for every quantity in a calculation and carrying the units through the calculation. Click to show how the units cancel. Ask a volunteer to write the answer on the board. Click to reveal the correct answer. = 13 L SO2

Calculating Molar Mass The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass of the compound? Have students identify the knowns and unknowns for the problem. Click to reveal the knowns and unknowns. Ask: What two units are related through the density of a substance? Answer: mass and volume Ask: What two units are related through the molar mass of a substance? Answer: mass and number of particles Click to reveal the strategy for calculating molar mass.

Calculating Molar Mass, cont. The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass of the compound? Ask: What are the units that must be included in the conversion factor for this problem? Answer: liters and moles Ask: How do you determine which part of the conversion factor is the numerator and which part is the denominator? Answer: The conversion factor is set up so that the volume factors cancel each other. Ask: What information does the molar mass convey? Answer: the mass of one mole of a substance Click to reveal the conversion equation. Click to show how the units cancel. Ask a volunteer to write the answer on the board. Click to reveal the correct answer. = 44.0 g/mol

Mole Relationships Guide students through examples of the various mole conversions. For example, start with 50.0 grams of a compound or element and convert it to moles and then to particles. Then, start with a given volume of a gas and convert it to mass or particles. Have students re-create this road map in a visual way that helps them clearly see the various relationships. Encourage students to use their visual as a reference when doing homework problems.