1.5. Worked Practice Problems

1. (a) Calculate the number of moles in 25.0 g of calcium carbonate (CaCO₃). (b) Determine the number of formula units of CaCO₃ in that sample. (c) Find the mass of 1.50 × 10²³ formula units of CaCO₃.

In Problem 1, students will work with the concept of moles and how they relate to mass and particles.

1. (a) To find the number of moles in a substance, we use the formula:

   $\text{Number of moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}$

   First, we need the molar mass of calcium carbonate (CaCO₃):

   • Calcium (Ca): 40.08 g/mol
   • Carbon (C): 12.01 g/mol
   • Oxygen (O): 16.00 g/mol (there are 3 oxygen atoms in CaCO₃)

   Therefore, the molar mass of CaCO₃ = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol. Now we can calculate the number of moles:

   $\text{Number of moles} = \frac{25.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.2498 \, \text{mol}$

2. (b) To find the number of formula units, we use Avogadro's number (6.022 × 10²³). The relationship is:

   $\text{Number of formula units} = \text{number of moles} \times 6.022 \times 10^{23}$

   So:
   $\text{Number of formula units} = 0.2498 \, \text{mol} \times 6.022 \times 10^{23} = 1.505 \times 10^{23} \text{ formula units}$

3. (c) To find the mass of a specific number of formula units, first convert the formula units to moles and then to grams:

   $\text{number of moles} = \frac{1.50 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.2492 \, \text{mol}$
   $\text{mass} = 0.2492 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 24.93 \, \text{g}$

Think of moles like a dozen eggs. Just as 1 dozen means 12 eggs, 1 mole means a specific number of entities (6.022 x 10²³). If you have 25 grams of calcium carbonate, it’s like saying you have a portion of a dozen eggs and you want to find out how many complete dozens (moles) you have. This analogy can help students visualize the conversion process.

2. (a) Balance the following chemical equation for the reaction between aluminum sulfide and water: Al2S3(s) + H2O(l) ⟶ Al(OH)3(s) + H2S(g). (b) If 12.0 g of Al₂S₃ reacts with 50.0 g of water (excess), calculate the mass of H₂S gas produced.

This problem is focused on balancing chemical equations and performing stoichiometric calculations.

1. (a) To balance the equation, we need to ensure that the number of atoms of each element is equal on both sides. Start with the unbalanced equation:
   $\text{Al}_2\text{S}_3 + \text{H}_2\text{O} \longrightarrow \text{Al(OH)}_3 + \text{H}_2 ext{S}$
   Count the atoms on each side:

   • Aluminum (Al): 2 on left, 1 on right.
   • Sulfur (S): 3 on left, 1 on right.
   • Hydrogen (H) and Oxygen (O) also need to be balanced.

   By adjusting coefficients systematically, we can arrive at the balanced equation:
   $\text{Al}_2\text{S}_3 + 6 \text{H}_2\text{O} \longrightarrow 2 \text{Al(OH)}_3 + 3 \text{H}_2 ext{S}$

2. (b) To find the mass of H₂S produced, we start by converting 12.0 g of Al₂S₃ into moles. The molar mass of Al₂S₃ is 150.14 g/mol:
   $\text{Moles of Al}_2\text{S}_3 = \frac{12.0 \, \text{g}}{150.14 \, \text{g/mol}} \approx 0.07991 \, \text{mol}$
   Since the balanced equation shows that 1 mol of Al₂S₃ produces 3 mol of H₂S, we find moles of H₂S:
   $\text{Moles of H}_2 ext{S} = 0.07991 \, \text{mol Al}_2\text{S}_3 \times 3 = 0.2397 \, \text{mol H}_2 ext{S}$

   Finally, converting moles back to grams using the molar mass of H₂S (34.08 g/mol):
   $\text{Mass of H}_2 ext{S} = 0.2397 \, \text{mol} \times 34.08 \, \text{g/mol} \approx 8.17 \, \text{g}$

Balancing chemical equations is like making sure a recipe has the right number of ingredients on both sides of the equation. If a recipe calls for 2 eggs and you use 1, you won't have enough for your dish. The same goes for balancing reactions; if the ingredients don't match up, the reaction won't be complete.

3. Consider the decomposition reaction: 2 N2O5(g) ⟶ 4 NO2(g) + O2(g). (a) If 1.20 mol of N₂O₅ decomposes, what masses of NO₂ and O₂ are formed, assuming 100 % conversion? (b) In a particular experiment, only 120.0 g of NO₂ is collected. Calculate the percent yield for NO₂.

This problem relates to the concepts of limiting reagents and the percent yield of a reaction.

1. (a) Given the balanced reaction
   $2 ext{N}_2 ext{O}_5(g) \longrightarrow 4 ext{NO}_2(g) + ext{O}_2(g)$
   We need to determine how much NO₂ and O₂ is generated from 1.20 mol of N₂O₅. Using mole ratios from the balanced equation:

   • For NO₂: $1.20 \, \text{mol N}_2 ext{O}_5 \times \frac{4 \, \text{mol NO}_2}{2 \, \text{mol N}_2 ext{O}_5} = 2.40 \, \text{mol NO}_2$
   • For O₂: $1.20 \, \text{mol N}_2 ext{O}_5 \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol N}_2 ext{O}_5} = 0.600 \, \text{mol O}_2$
     Now to find the mass of each gas, we can use their molar masses (NO₂: 46.01 g/mol; O₂: 32.00 g/mol):
     $\text{Mass of NO}_2 = 2.40 \, \text{mol} \times 46.01 \, \text{g/mol} = 110.4 \, \text{g}$
     $\text{Mass of O}_2 = 0.600 \, \text{mol} \times 32.00 \, \text{g/mol} = 19.2 \, \text{g}$

2. (b) To calculate the percent yield, we compare the actual yield to the theoretical yield. Here, the actual yield is given as 120.0 g:
   $\text{Percent yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\% = \left( \frac{120.0 \, \text{g}}{110.4 \, \text{g}} \right) \times 100\% = 108.7\%$
   Since the percent yield is greater than 100%, it indicates a possible error in measurement or product purity.

Imagine baking a cake. The recipe says you need 2 cups of flour (limiting ingredient) to yield a certain number of slices. If you think you've actually made more cake than the recipe says, but it's actually because you overmeasured the flour, it's similar to how experimental yields can exceed theoretical yields in chemistry.

4. (a) Describe precisely how to prepare 500 mL of a 0.200 M K₂CrO₄ solution starting from solid K₂CrO₄. Show all mass‐based calculations. (b) If you take 50.0 mL of that solution and dilute it to a total volume of 250 mL, what is the resulting concentration?

This problem focuses on preparing solutions and understanding dilutions.

1. (a) To prepare a specific molarity:
   First, calculate the molar mass of potassium chromate (K₂CrO₄):

   • Potassium (K): 39.10 g/mol (there are 2 potassium atoms)
   • Chromium (Cr): 52.00 g/mol
   • Oxygen (O): 16.00 g/mol (4 oxygen atoms)

   Molar mass = 2(39.10) + 52.00 + 4(16.00) = 194.20 g/mol.

   To find the mass for a 0.200 M solution in 0.500 L:
   $\text{Required moles} = 0.200 \, \text{mol/L} \times 0.500 \, \text{L} = 0.100 \, \text{mol}$
   $\text{Mass} = 0.100 \, \text{mol} \times 194.20 \, \text{g/mol} = 19.42 \, g$

   The procedure involves weighing this mass, dissolving in water in a volumetric flask, and bringing the total volume up to 500 mL.

2. (b) For dilutions, use the dilution equation:
   $C₁ V₁ = C₂ V₂$
   Where:

   • C₁ = initial concentration = 0.200 M
   • V₁ = initial volume = 50.0 mL = 0.0500 L
   • V₂ = final volume = 250 mL = 0.250 L

   Rearranging gives:
   $C₂ = \frac{C₁ V₁}{V₂} = \frac{0.200 \, \text{mol/L} \times 0.0500 \, ext{L}}{0.250 \, ext{L}} = 0.0400 \, ext{mol/L}$

Preparing a solution can be compared to making a drink. If you want a 0.200 M solution like you would want a specific level of sweetness. Instead of adding too much syrup too quickly, which would create a concentrated drink, you take a proper amount and dilute it to reach just the right level of sweetness, ensuring you mix it well to achieve uniformity.

5. A laboratory sample labeled 'sodium hydroxide solution' contains 8.00 g of NaOH dissolved in enough water to make 250.0 mL of solution. (a) Calculate the molarity of the solution. (b) Calculate the molality of the solution, assuming the density of the final solution is 1.07 g/mL. Show all steps, including how you find the mass of the solvent.

This problem involves calculating both molarity and molality of a solution.

1. (a) To find molarity (M), we use:
   $M = \frac{\text{moles of solute}}{\text{liters of solution}}$
   First, calculate the molar mass of NaOH:

   • Sodium (Na): 22.99 g/mol
   • Oxygen (O): 16.00 g/mol
   • Hydrogen (H): 1.008 g/mol

   So, the molar mass = 22.99 + 16.00 + 1.008 = 40.00 g/mol.

   Now, find the number of moles:
   $\text{Moles of NaOH} = \frac{8.00 \, \text{g}}{40.00 \, \text{g/mol}} = 0.200 \, \text{mol}$

   Finally, convert volume from mL to L:
   250.0 mL = 0.250 L.

   Molarity:
   $M = \frac{0.200 \, \text{mol}}{0.250 \, \text{L}} = 0.800 \, ext{M}$

2. (b) To calculate molality (m), we need the mass of the solvent. First, find the total mass of the solution using density:
   $\text{Density} = \frac{\text{mass}}{\text{volume}} \implies \text{Mass} = \text{Density} \times \text{Volume} = 1.07 \,\text{g/mL} \times 250.0 \, \text{mL} = 267.5 \, \text{g}$

   The mass of the solvent (water):
   $\text{Mass of solvent} = 267.5 \, \text{g} - 8.00 \, \text{g} = 259.5 \, \text{g} \approx 0.2595 \, \text{kg}$

   Then calculate molality:
   $m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.200 \, \text{mol}}{0.2595 \, \text{kg}} \approx 0.771 \, m$

Calculating concentrations can feel like mixing paint. When you know how much pigment you have (like knowing the grams of NaOH), you can determine how vibrant your final color will be based on how much water (the solvent) you add. Too much water and you dilute your paint, akin to a lower molarity. Balancing these elements ensures you get the desired color intensity.

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