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1.9 - Practice Problem Solutions (Outline)

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Mole-Mass-Particle Conversions

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0:00
Teacher
Teacher

Today we'll discuss how to convert between moles and mass, starting with Problem 1. Who can tell me what molar mass is?

Student 1
Student 1

Isn't it the mass of one mole of a substance?

Teacher
Teacher

Exactly! For example, to find the number of moles in 25 grams of calcium carbonate, we first calculate its molar mass. Molar mass of CaCOโ‚ƒ is 100.09 g/mol. Can anyone calculate the number of moles?

Student 2
Student 2

I think it's 0.2498 moles.

Teacher
Teacher

Thatโ€™s right! Now, how do we find the number of formula units in that sample?

Student 3
Student 3

We multiply the number of moles by Avogadro's number!

Teacher
Teacher

Exactly! So, 0.2498 moles times 6.022 ร— 10ยฒยณ will give us the number of formula units.

Student 4
Student 4

So that would be about 1.505 ร— 10ยฒยณ formula units.

Teacher
Teacher

Great work everyone! To summarize, we covered how to calculate moles using molar mass and how to convert moles to formula units. Remember, 'Moles to mass' and 'Mass to moles' is key in stoichiometry.

Balancing Equations and Stoichiometry

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Teacher
Teacher

Let's move on to balancing equations. Can someone remind me why we need to balance equations?

Student 1
Student 1

To follow the law of conservation of mass, right?

Teacher
Teacher

Well done! Letโ€™s look at Problem 2. Can anyone balance the reaction between aluminum sulfide and water?

Student 2
Student 2

I would start with Alโ‚‚Sโ‚ƒ + Hโ‚‚O, but I need to make sure the number of each atom matches on both sides.

Teacher
Teacher

Good start! After balancing all atoms, what does the balanced equation look like?

Student 3
Student 3

It should be Alโ‚‚Sโ‚ƒ + 6 Hโ‚‚O โŸถ 2 Al(OH)โ‚ƒ + 3 Hโ‚‚S!

Teacher
Teacher

Perfect! Now, how do we use this balanced equation to find out how much Hโ‚‚S is produced from 12.0 g of Alโ‚‚Sโ‚ƒ?

Student 4
Student 4

First, we'd calculate the moles of Alโ‚‚Sโ‚ƒ, using its molar mass, and then apply the mole ratio to find Hโ‚‚S produced.

Teacher
Teacher

Exactly! So the theoretical mass of Hโ‚‚S can be calculated from the moles? Great conversation here!

Limiting Reagent and Percent Yield

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Teacher
Teacher

Now let's talk about limiting reagents. Why is identifying the limiting reagent important?

Student 1
Student 1

Because it determines the maximum amount of product we can create!

Teacher
Teacher

Exactly! In Problem 3, if we have 1.20 moles of Nโ‚‚Oโ‚…, can someone explain how to find the products formed from this?

Student 2
Student 2

We first find the moles of NOโ‚‚ and Oโ‚‚ using the mole ratio from the balanced equation.

Teacher
Teacher

Correct! And how do we calculate percent yield if we only collect a certain amount of product?

Student 3
Student 3

We compare the actual yield to the theoretical yield.

Teacher
Teacher

Right again! It's calculated as percent yield = (actual yield รท theoretical yield) ร— 100%. Remember, yield over 100% indicates some error! Can anyone think of examples of errors?

Student 4
Student 4

Maybe impurities in the product or measurement inaccuracies?

Teacher
Teacher

Great points! Let's recap: identifying limiting reagents helps us maximize reactions and understanding percent yield is key in evaluating our methods.

Introduction & Overview

Read a summary of the section's main ideas. Choose from Basic, Medium, or Detailed.

Quick Overview

This section provides a concise outline of solutions to practice problems focused on stoichiometric calculations and concepts.

Standard

The outline covers detailed approaches to solving practice problems that address various key concepts in stoichiometry, including mole conversions, balancing equations, identifying limiting reagents, yield calculations, and solution concentrations.

Detailed

Detailed Summary

In this section, the practice problem solutions are organized into a comprehensive outline that focuses on key stoichiometric concepts essential for mastering chemistry calculations. The solutions showcase various approaches to critical problems, such as converting between moles, mass, and particles, balancing chemical equations, identifying limiting reagents, calculating theoretical and actual yields, and understanding solutions and their concentrations. For each problem, a step-by-step method is presented to illustrate how the underlying principles of stoichiometry can be applied effectively.

Audio Book

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Problem 1(a): Molar Mass Calculation

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  1. Molar mass of CaCOโ‚ƒ = 40.08 + 12.01 + (3 ร— 16.00) = 100.09 g/mol.
  2. Number of moles = 25.0 g รท 100.09 g/mol = 0.2498 mol.

Detailed Explanation

To calculate the number of moles in 25.0 g of calcium carbonate (CaCOโ‚ƒ), we first need to find its molar mass. The molar mass is the sum of the atomic masses of all atoms present in the compound: calcium (Ca) is approximately 40.08 g/mol, carbon (C) is about 12.01 g/mol, and oxygen (O) is roughly 16.00 g/mol. Since there are three oxygen atoms in CaCOโ‚ƒ, we multiply 16.00 by 3. Adding these values together gives us the molar mass of CaCOโ‚ƒ as 100.09 g/mol. Then, to find the number of moles, we divide the mass (25.0 g) by the molar mass (100.09 g/mol), resulting in approximately 0.2498 moles.

Examples & Analogies

Imagine you have a bag of flour (representing the mass of CaCOโ‚ƒ), and you want to know how many cups (moles) of flour you have. Each cup weighs a certain amount (molar mass). By weighing the flour (25.0 g) and knowing how much one cup weighs (100.09 g), you can determine how many cups you have by dividing the total weight by the weight of one cup.

Problem 1(b): Calculating Formula Units

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  1. Number of formula units = 0.2498 mol ร— 6.022 ร— 10ยฒยณ formula units/mol = 1.505 ร— 10ยฒยณ formula units.

Detailed Explanation

To find the number of formula units of CaCOโ‚ƒ corresponding to 0.2498 moles, we use Avogadro's number, which tells us how many entities are in one mole of any substance (approximately 6.022 ร— 10ยฒยณ). We multiply the number of moles (0.2498) by Avogadro's number, resulting in about 1.505 ร— 10ยฒยณ formula units of CaCOโ‚ƒ.

Examples & Analogies

Think of a dozen eggs. If you have 0.25 dozen eggs, you can figure out the total number of eggs by knowing one dozen has 12 eggs. By doing the calculation (0.25 ร— 12), you find you have 3 eggs. Similarly, with moles and formula units, we calculate how many individual 'units' we have in our sample.

Problem 1(c): Mass from Formula Units

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  1. Given 1.50 ร— 10ยฒยณ formula units, moles = (1.50 ร— 10ยฒยณ) รท (6.022 ร— 10ยฒยณ) = 0.2492 mol.
  2. Mass = 0.2492 mol ร— 100.09 g/mol = 24.93 g.

Detailed Explanation

To determine the mass from a known number of formula units (1.50 ร— 10ยฒยณ), we first convert formula units back to moles using Avogadro's number: we divide 1.50 ร— 10ยฒยณ by 6.022 ร— 10ยฒยณ, resulting in approximately 0.2492 moles. Next, we find the mass by multiplying the number of moles by the molar mass of CaCOโ‚ƒ (100.09 g/mol). This calculation yields a mass of about 24.93 g.

Examples & Analogies

Imagine you wanted to know how much sugar you have if you know how many sugar cubes you hold. By counting how many cubes equal one package (like moles), you can backtrack from cubes to know the total weight of grams of sugar. This is similar to moving between formula units to mass in chemistry.

Problem 2(a): Balancing a Chemical Equation

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  1. Unbalanced: Alโ‚‚Sโ‚ƒ + Hโ‚‚O โŸถ Al(OH)โ‚ƒ + Hโ‚‚S.
  2. Balance sulfur atoms:
    โ–  There are 3 S atoms in Alโ‚‚Sโ‚ƒ and only 1 S atom in Hโ‚‚S. Place coefficient 3 before Hโ‚‚S:
    Alโ‚‚Sโ‚ƒ + Hโ‚‚O โŸถ Al(OH)โ‚ƒ + 3 Hโ‚‚S.
  3. Balance aluminum atoms:
    โ–  2 Al on left; only 1 Al in Al(OH)โ‚ƒ on right. Place coefficient 2 before Al(OH)โ‚ƒ:
    Alโ‚‚Sโ‚ƒ + Hโ‚‚O โŸถ 2 Al(OH)โ‚ƒ + 3 Hโ‚‚S.
  4. Balance hydrogen and oxygen:
    โ–  On the right side, 2 Al(OH)โ‚ƒ has (2 ร— 3) = 6 O atoms and (2 ร— 3) = 6 H atoms.
    โ–  Also on right, 3 Hโ‚‚S has (3 ร— 2) = 6 H atoms, total H atoms on right = 6 (from Al(OH)โ‚ƒ) + 6 (from Hโ‚‚S) = 12.
    โ–  Oxygen atoms on right = 2 Al(OH)โ‚ƒ gives 2 ร— 3 O = 6 O.
    โ–  Therefore, to supply 12 H atoms and 6 O atoms on left, we need 6 Hโ‚‚O (which has 6 ร— 2 = 12 H atoms and 6 ร— 1 = 6 O atoms).
  5. Final balanced equation:
    Alโ‚‚Sโ‚ƒ(s) + 6 Hโ‚‚O(l) โŸถ 2 Al(OH)โ‚ƒ(s) + 3 Hโ‚‚S(g).

Detailed Explanation

Balancing a chemical equation is crucial because it ensures the law of conservation of mass is observed; atoms cannot be lost or gained in a reaction. We start with the unbalanced equation and analyze the number of atoms for each element on both sides. By iteratively adding coefficients and ensuring each element's number matches on both sides, we arrive at the balanced equation: Alโ‚‚Sโ‚ƒ + 6 Hโ‚‚O โŸถ 2 Al(OH)โ‚ƒ + 3 Hโ‚‚S.

Examples & Analogies

Think of balancing a budget. If you have a certain amount of income (reactants) and expenses (products), you want to make sure your income matches your expenses. Just like every dollar must balance, in a chemical reaction, every atom must balance outโ€”what goes in must equal what comes out.

Problem 2(b): Mass of Product from Reactants

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  1. Molar mass of Alโ‚‚Sโ‚ƒ = (2 ร— 26.98) + (3 ร— 32.06) = 53.96 + 96.18 = 150.14 g/mol (rounded to 150.16 g/mol).
  2. Moles of Alโ‚‚Sโ‚ƒ = 12.0 g รท 150.16 g/mol = 0.07991 mol.
  3. Balanced equation shows: 1 mol Alโ‚‚Sโ‚ƒ โŸถ 3 mol Hโ‚‚S. So 0.07991 mol Alโ‚‚Sโ‚ƒ โŸถ 0.2397 mol Hโ‚‚S.
  4. Molar mass of Hโ‚‚S = (2 ร— 1.008) + 32.06 = 2.016 + 32.06 = 34.076 g/mol.
  5. Mass of Hโ‚‚S = 0.2397 mol ร— 34.076 g/mol = 8.17 g.

Detailed Explanation

To calculate the mass of hydrogen sulfide (Hโ‚‚S) produced from the decomposition of aluminum sulfide (Alโ‚‚Sโ‚ƒ), we first need the molar mass of Alโ‚‚Sโ‚ƒ. After calculating, we find that 12.0 g of Alโ‚‚Sโ‚ƒ corresponds to approximately 0.07991 moles. From the balanced equation, it produces 3 moles of Hโ‚‚S for every mole of Alโ‚‚Sโ‚ƒ decomposed. Therefore, for 0.07991 moles of Alโ‚‚Sโ‚ƒ, we can produce 0.2397 moles of Hโ‚‚S. Finally, we convert this into grams by multiplying the number of moles by the molar mass of Hโ‚‚S (34.076 g/mol), resulting in approximately 8.17 g of Hโ‚‚S.

Examples & Analogies

Think of a recipe where you know how much of the main ingredient (like flour) you have and how much final dish (like cookies) can be made from it. By knowing how much flour (reactant) you have and the recipe (balanced equation), you can determine the total number of cookies produced (product). In chemistry, we follow a similar process to predict how much product we can make from a specified amount of reactant.

Problem 3(a): Masses from Decomposition

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  1. Molar mass of Nโ‚‚Oโ‚… = (2 ร— 14.01) + (5 ร— 16.00) = 28.02 + 80.00 = 108.02 g/mol.
  2. Balanced decomposition: 2 Nโ‚‚Oโ‚… โŸถ 4 NOโ‚‚ + Oโ‚‚.
    โ–  Divide both sides by 2 to see per mole basis: 1 Nโ‚‚Oโ‚… โŸถ 2 NOโ‚‚ + 0.5 Oโ‚‚.
  3. For 1.20 mol Nโ‚‚Oโ‚…:
    โ–  Moles of NOโ‚‚ = 1.20 mol ร— (4 mol NOโ‚‚ / 2 mol Nโ‚‚Oโ‚…) = 2.40 mol NOโ‚‚.
    โ–  Moles of Oโ‚‚ = 1.20 mol ร— (1 mol Oโ‚‚ / 2 mol Nโ‚‚Oโ‚…) = 0.600 mol Oโ‚‚.
  4. Molar mass of NOโ‚‚ = 14.01 + (2 ร— 16.00) = 14.01 + 32.00 = 46.01 g/mol.
    โ–  Mass of NOโ‚‚ = 2.40 mol ร— 46.01 g/mol = 110.4 g.
  5. Molar mass of Oโ‚‚ = 32.00 g/mol.
    โ–  Mass of Oโ‚‚ = 0.600 mol ร— 32.00 g/mol = 19.2 g.

Detailed Explanation

In this problem, we start by calculating the molar mass of dinitrogen pentoxide (Nโ‚‚Oโ‚…), which is essential to find out how many grams can be produced from its decomposition. The balanced equation shows that when 2 moles of Nโ‚‚Oโ‚… decompose, they produce 4 moles of nitrogen dioxide (NOโ‚‚) and 1 mole of oxygen gas (Oโ‚‚). By knowing we have 1.20 moles of Nโ‚‚Oโ‚…, we can use stoichiometry to find we will get 2.40 moles of NOโ‚‚ and 0.600 moles of Oโ‚‚. Each of these amounts can then be converted into grams using their respective molar masses (NOโ‚‚ and Oโ‚‚) to find the total mass produced.

Examples & Analogies

Imagine a factory assembly line that converts raw materials into final products. Each input (Nโ‚‚Oโ‚…) results in multiple final outputs (NOโ‚‚ and Oโ‚‚) based on the production ratio defined (balanced equation). By knowing how much raw material you start with, you can calculate how much end product you'll receive.

Problem 3(b): Calculating Percent Yield

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  1. Theoretical yield of NOโ‚‚ = 110.4 g (from part a). Actual yield = 120.0 g.
  2. Percent yield = (120.0 g รท 110.4 g) ร— 100 % = 108.7 %.
    โ–  Yield > 100 % indicates experimental error (e.g., impure product, measurement inaccuracies).

Detailed Explanation

To calculate the percent yield, we compare the actual yield of the product obtained (120.0 g of NOโ‚‚) to the theoretical yield calculated earlier (110.4 g). Using the formula for percent yield: (actual yield / theoretical yield) ร— 100 %, we find that the percent yield is 108.7%. A yield greater than 100% suggests there may have been an error in the experiment, possibly from impurities in the product or an inaccurate measurement.

Examples & Analogies

Consider a garden where you estimate you will harvest 10 tomatoes (theoretical yield), but when you collect, you find 12 (actual yield). The percent yield here signifies how productive your gardening efforts wereโ€”anything over 100% hints at a possible overestimation in the initial forecast or strange occurrences, like extra tomatoes growing unexpectedly.

Problem 4(a): Preparing a Solution

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  1. Molar mass of Kโ‚‚CrOโ‚„ = (2 ร— 39.10) + 52.00 + (4 ร— 16.00) = 78.20 + 52.00 + 64.00 = 194.20 g/mol.
  2. To prepare 0.200 M in 0.500 L: required moles = M ร— V = 0.200 mol/L ร— 0.500 L = 0.100 mol.
  3. Mass of Kโ‚‚CrOโ‚„ = 0.100 mol ร— 194.20 g/mol = 19.42 g.
  4. Procedure:
    โ–  Weigh 19.42 g of solid Kโ‚‚CrOโ‚„.
    โ–  Transfer to a 500 mL volumetric flask.
    โ–  Add distilled water until about 400 mL, swirl until dissolved.
    โ–  Fill to the 500 mL mark with distilled water.
    โ–  Invert the flask several times to ensure uniform mixing.

Detailed Explanation

To prepare a solution of potassium chromate (Kโ‚‚CrOโ‚„) with a certain molarity, we start by determining the molar mass. After calculating, we find the molar mass to be 194.20 g/mol. Next, we need 0.200 moles per liter in 0.500 liters, which amounts to 0.100 moles. Converting moles to grams gives us 19.42 g of Kโ‚‚CrOโ‚„, which we then measure out. The procedure involves adding the solid to a volumetric flask, dissolving it in some distilled water, and then filling it to the final mark with water to create a precise solution.

Examples & Analogies

Think about making a special drink mixture where you want a specific concentration of flavors. You first measure the right amount of each ingredient (like Kโ‚‚CrOโ‚„), mix them well, and make sure to follow the instructions to achieve the perfect taste in your drink (homogeneous solution).

Problem 4(b): Dilution Calculation

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  1. Initial concentration C1=0.200 MCโ‚ = 0.200 M, initial volume V1=50.0 mL=0.0500 L.
  2. Final volume V2=250 mL=0.250 L. Use C1V1=C2V2.
  3. Solve for C2: C2=C1ร—V1/V2=0.200 mol/Lร—0.0500 L/0.250 L=0.0400 mol/L.

Detailed Explanation

In this problem, we are diluting an initial solution of potassium chromate. When diluting, we use the dilution formula C1V1=C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution used, C2 is the final concentration after dilution, and V2 is the final volume. Here, we start with an initial concentration of 0.200 M and a volume of 50.0 mL, and we want to know the concentration after diluting it to 250 mL, resulting in a final concentration of 0.0400 M.

Examples & Analogies

Diluting a juice concentrate to make a refreshing drink is similar. If your recipe calls for a strong concentrate (C1), and you add water to increase the total volume (V2), your drink becomes less concentrated (C2). Each dilution makes a tastier drink, just like in chemistry!

Problem 5(a): Molarity of a Sodium Hydroxide Solution

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  1. Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 g/mol (โ‰ˆ 40.00 g/mol).
  2. Moles of NaOH = 8.00 g รท 40.00 g/mol = 0.200 mol.
  3. Volume of solution = 250.0 mL = 0.250 L.
  4. Molarity = 0.200 mol รท 0.250 L = 0.800 M.

Detailed Explanation

In this problem, we determine the molarity of a sodium hydroxide (NaOH) solution. We start by calculating the molar mass of NaOH to be about 40.00 g/mol. Given 8.00 g of NaOH, we find the number of moles by dividing the mass by the molar mass, resulting in 0.200 moles. Since the total volume of the solution is 250.0 mL (equivalent to 0.250 L), we calculate the molarity with the formula M = moles/volume, yielding a solution molarity of 0.800 M.

Examples & Analogies

This is like measuring how concentrated your sugar solution is for tea. If you know how much sugar you have and how much total liquid you mixed it into, you can determine how sweet your tea will taste. In chemistry, we measure how concentrated our chemical solutions are the same way!

Problem 5(b): Calculating Molality

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  1. Density of solution = 1.07 g/mL; total volume = 250.0 mL โ‡’ mass of solution = 1.07 g/mL ร— 250.0 mL = 267.5 g.
  2. Mass of solute (NaOH) = 8.00 g; therefore, mass of solvent (water) = 267.5 g โ€“ 8.00 g = 259.5 g = 0.2595 kg.
  3. Molality = moles of solute รท kilograms of solvent = 0.200 mol รท 0.2595 kg = 0.771 m.

Detailed Explanation

To find the molality of the sodium hydroxide (NaOH) solution, we first need to determine the total mass of the solution, using its density (1.07 g/mL) and the volume (250.0 mL), which gives us a mass of 267.5 g. We then subtract the mass of the solute (8.00 g of NaOH) from the total mass to find the mass of the solvent (water), which is approximately 259.5 g or 0.2595 kg. Finally, we divide the moles of solute (0.200 mol) by the mass of the solvent in kilograms to calculate the molality, resulting in around 0.771 m.

Examples & Analogies

This process is like figuring out how rich your oatmeal is in nutrients. If you know how much oatmeal (solute) you have and how much total mixture you've added water (solvent) into, you can gauge how concentrated your nutritional intake is. Similarly, in chemistry, we ascertain how concentrated solutions are based on solute and solvent masses.

Definitions & Key Concepts

Learn essential terms and foundational ideas that form the basis of the topic.

Key Concepts

  • Mole and Molar Mass: Understand the mole as a unit and how to convert between mass and moles using molar mass.

  • Balancing Chemical Equations: Importance of balancing to comply with the law of conservation of mass.

  • Identifying Limiting Reagents: How to recognize which reactant limits the production of products and influences yield.

  • Calculating Theoretical vs. Actual Yield: Understand how to derive yields and their comparative calculation for efficiency.

Examples & Real-Life Applications

See how the concepts apply in real-world scenarios to understand their practical implications.

Examples

  • To calculate moles from mass, divide the mass of the substance by its molar mass. For example, 25 g of CaCOโ‚ƒ gives you approximately 0.25 moles.

  • Balancing the equation Alโ‚‚Sโ‚ƒ + 6 Hโ‚‚O โŸถ 2 Al(OH)โ‚ƒ + 3 Hโ‚‚S ensures conservation of mass for the reaction.

  • To identify the limiting reagent, compare the calculated moles of the reactants against their stoichiometric ratios.

Memory Aids

Use mnemonics, acronyms, or visual cues to help remember key information more easily.

๐ŸŽต Rhymes Time

  • Moles, moles, count them right, over grams they take their flight!

๐Ÿ“– Fascinating Stories

  • Imagine a race where the limiting reagent is the slowest runner, deciding how many can finish the race!

  • Picture a chef adjusting recipes to get the perfect cookie; balancing ingredients until the taste is just right. This is like balancing an equation!

๐ŸŽฏ Super Acronyms

Don't forget

  • M.B.A. = Moles
  • Balance
  • Amount โ€“ the steps for hands-on stoichiometry!

Flash Cards

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Glossary of Terms

Review the Definitions for terms.

  • Term: Mole

    Definition:

    The amount of substance that contains 6.02214076 ร— 10ยฒยณ entities.

  • Term: Molar Mass

    Definition:

    The mass of one mole of a substance, usually expressed in grams per mole (g/mol).

  • Term: Balancing Equation

    Definition:

    Adjusting the coefficients in a chemical equation to ensure the number of atoms for each element is equal on both sides.

  • Term: Limiting Reagent

    Definition:

    The reactant that is completely consumed first, regulating the maximum amount of product formed.

  • Term: Theoretical Yield

    Definition:

    The maximum amount of product that could be formed from the given amounts of reactants.

  • Term: Actual Yield

    Definition:

    The measured amount of product obtained from a chemical reaction.

  • Term: Percent Yield

    Definition:

    A measure of the efficiency of a reaction calculated as (actual yield รท theoretical yield) ร— 100%.