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Mole-Mass-Particle Conversions
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Today we'll discuss how to convert between moles and mass, starting with Problem 1. Who can tell me what molar mass is?
Isn't it the mass of one mole of a substance?
Exactly! For example, to find the number of moles in 25 grams of calcium carbonate, we first calculate its molar mass. Molar mass of CaCO₃ is 100.09 g/mol. Can anyone calculate the number of moles?
I think it's 0.2498 moles.
That’s right! Now, how do we find the number of formula units in that sample?
We multiply the number of moles by Avogadro's number!
Exactly! So, 0.2498 moles times 6.022 × 10²³ will give us the number of formula units.
So that would be about 1.505 × 10²³ formula units.
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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Let's move on to balancing equations. Can someone remind me why we need to balance equations?
To follow the law of conservation of mass, right?
Well done! Let’s look at Problem 2. Can anyone balance the reaction between aluminum sulfide and water?
I would start with Al₂S₃ + H₂O, but I need to make sure the number of each atom matches on both sides.
Good start! After balancing all atoms, what does the balanced equation look like?
It should be Al₂S₃ + 6 H₂O ⟶ 2 Al(OH)₃ + 3 H₂S!
Perfect! Now, how do we use this balanced equation to find out how much H₂S is produced from 12.0 g of Al₂S₃?
First, we'd calculate the moles of Al₂S₃, using its molar mass, and then apply the mole ratio to find H₂S produced.
Exactly! So the theoretical mass of H₂S can be calculated from the moles? Great conversation here!
Limiting Reagent and Percent Yield
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Now let's talk about limiting reagents. Why is identifying the limiting reagent important?
Because it determines the maximum amount of product we can create!
Exactly! In Problem 3, if we have 1.20 moles of N₂O₅, can someone explain how to find the products formed from this?
We first find the moles of NO₂ and O₂ using the mole ratio from the balanced equation.
Correct! And how do we calculate percent yield if we only collect a certain amount of product?
We compare the actual yield to the theoretical yield.
Right again! It's calculated as percent yield = (actual yield ÷ theoretical yield) × 100%. Remember, yield over 100% indicates some error! Can anyone think of examples of errors?
Maybe impurities in the product or measurement inaccuracies?
Great points! Let's recap: identifying limiting reagents helps us maximize reactions and understanding percent yield is key in evaluating our methods.
Introduction & Overview
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Quick Overview
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.
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Problem 1(a): Molar Mass Calculation
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- Molar mass of CaCO₃ = 40.08 + 12.01 + (3 × 16.00) = 100.09 g/mol.
- Number of moles = 25.0 g ÷ 100.09 g/mol = 0.2498 mol.
Detailed Explanation
To calculate the number of moles in 25.0 g of calcium carbonate (CaCO₃), 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 CaCO₃, we multiply 16.00 by 3. Adding these values together gives us the molar mass of CaCO₃ 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 CaCO₃), 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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- Number of formula units = 0.2498 mol × 6.022 × 10²³ formula units/mol = 1.505 × 10²³ formula units.
Detailed Explanation
To find the number of formula units of CaCO₃ 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 × 10²³). We multiply the number of moles (0.2498) by Avogadro's number, resulting in about 1.505 × 10²³ formula units of CaCO₃.
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 × 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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- Given 1.50 × 10²³ formula units, moles = (1.50 × 10²³) ÷ (6.022 × 10²³) = 0.2492 mol.
- Mass = 0.2492 mol × 100.09 g/mol = 24.93 g.
Detailed Explanation
To determine the mass from a known number of formula units (1.50 × 10²³), we first convert formula units back to moles using Avogadro's number: we divide 1.50 × 10²³ by 6.022 × 10²³, resulting in approximately 0.2492 moles. Next, we find the mass by multiplying the number of moles by the molar mass of CaCO₃ (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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- Unbalanced: Al₂S₃ + H₂O ⟶ Al(OH)₃ + H₂S.
- Balance sulfur atoms:
■ There are 3 S atoms in Al₂S₃ and only 1 S atom in H₂S. Place coefficient 3 before H₂S:
Al₂S₃ + H₂O ⟶ Al(OH)₃ + 3 H₂S. - Balance aluminum atoms:
■ 2 Al on left; only 1 Al in Al(OH)₃ on right. Place coefficient 2 before Al(OH)₃:
Al₂S₃ + H₂O ⟶ 2 Al(OH)₃ + 3 H₂S. - Balance hydrogen and oxygen:
■ On the right side, 2 Al(OH)₃ has (2 × 3) = 6 O atoms and (2 × 3) = 6 H atoms.
■ Also on right, 3 H₂S has (3 × 2) = 6 H atoms, total H atoms on right = 6 (from Al(OH)₃) + 6 (from H₂S) = 12.
■ Oxygen atoms on right = 2 Al(OH)₃ gives 2 × 3 O = 6 O.
■ Therefore, to supply 12 H atoms and 6 O atoms on left, we need 6 H₂O (which has 6 × 2 = 12 H atoms and 6 × 1 = 6 O atoms). - Final balanced equation:
Al₂S₃(s) + 6 H₂O(l) ⟶ 2 Al(OH)₃(s) + 3 H₂S(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: Al₂S₃ + 6 H₂O ⟶ 2 Al(OH)₃ + 3 H₂S.
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 out—what goes in must equal what comes out.
Problem 2(b): Mass of Product from Reactants
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- Molar mass of Al₂S₃ = (2 × 26.98) + (3 × 32.06) = 53.96 + 96.18 = 150.14 g/mol (rounded to 150.16 g/mol).
- Moles of Al₂S₃ = 12.0 g ÷ 150.16 g/mol = 0.07991 mol.
- Balanced equation shows: 1 mol Al₂S₃ ⟶ 3 mol H₂S. So 0.07991 mol Al₂S₃ ⟶ 0.2397 mol H₂S.
- Molar mass of H₂S = (2 × 1.008) + 32.06 = 2.016 + 32.06 = 34.076 g/mol.
- Mass of H₂S = 0.2397 mol × 34.076 g/mol = 8.17 g.
Detailed Explanation
To calculate the mass of hydrogen sulfide (H₂S) produced from the decomposition of aluminum sulfide (Al₂S₃), we first need the molar mass of Al₂S₃. After calculating, we find that 12.0 g of Al₂S₃ corresponds to approximately 0.07991 moles. From the balanced equation, it produces 3 moles of H₂S for every mole of Al₂S₃ decomposed. Therefore, for 0.07991 moles of Al₂S₃, we can produce 0.2397 moles of H₂S. Finally, we convert this into grams by multiplying the number of moles by the molar mass of H₂S (34.076 g/mol), resulting in approximately 8.17 g of H₂S.
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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- Molar mass of N₂O₅ = (2 × 14.01) + (5 × 16.00) = 28.02 + 80.00 = 108.02 g/mol.
- Balanced decomposition: 2 N₂O₅ ⟶ 4 NO₂ + O₂.
■ Divide both sides by 2 to see per mole basis: 1 N₂O₅ ⟶ 2 NO₂ + 0.5 O₂. - For 1.20 mol N₂O₅:
■ Moles of NO₂ = 1.20 mol × (4 mol NO₂ / 2 mol N₂O₅) = 2.40 mol NO₂.
■ Moles of O₂ = 1.20 mol × (1 mol O₂ / 2 mol N₂O₅) = 0.600 mol O₂. - Molar mass of NO₂ = 14.01 + (2 × 16.00) = 14.01 + 32.00 = 46.01 g/mol.
■ Mass of NO₂ = 2.40 mol × 46.01 g/mol = 110.4 g. - Molar mass of O₂ = 32.00 g/mol.
■ Mass of O₂ = 0.600 mol × 32.00 g/mol = 19.2 g.
Detailed Explanation
In this problem, we start by calculating the molar mass of dinitrogen pentoxide (N₂O₅), which is essential to find out how many grams can be produced from its decomposition. The balanced equation shows that when 2 moles of N₂O₅ decompose, they produce 4 moles of nitrogen dioxide (NO₂) and 1 mole of oxygen gas (O₂). By knowing we have 1.20 moles of N₂O₅, we can use stoichiometry to find we will get 2.40 moles of NO₂ and 0.600 moles of O₂. Each of these amounts can then be converted into grams using their respective molar masses (NO₂ and O₂) to find the total mass produced.
Examples & Analogies
Imagine a factory assembly line that converts raw materials into final products. Each input (N₂O₅) results in multiple final outputs (NO₂ and O₂) 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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- Theoretical yield of NO₂ = 110.4 g (from part a). Actual yield = 120.0 g.
- Percent yield = (120.0 g ÷ 110.4 g) × 100 % = 108.7 %.
■ 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 NO₂) to the theoretical yield calculated earlier (110.4 g). Using the formula for percent yield: (actual yield / theoretical yield) × 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 were—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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- Molar mass of K₂CrO₄ = (2 × 39.10) + 52.00 + (4 × 16.00) = 78.20 + 52.00 + 64.00 = 194.20 g/mol.
- To prepare 0.200 M in 0.500 L: required moles = M × V = 0.200 mol/L × 0.500 L = 0.100 mol.
- Mass of K₂CrO₄ = 0.100 mol × 194.20 g/mol = 19.42 g.
- Procedure:
■ Weigh 19.42 g of solid K₂CrO₄.
■ Transfer to a 500 mL volumetric flask.
■ Add distilled water until about 400 mL, swirl until dissolved.
■ Fill to the 500 mL mark with distilled water.
■ Invert the flask several times to ensure uniform mixing.
Detailed Explanation
To prepare a solution of potassium chromate (K₂CrO₄) 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 K₂CrO₄, 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 K₂CrO₄), 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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- Initial concentration C1=0.200 MC₁ = 0.200 M, initial volume V1=50.0 mL=0.0500 L.
- Final volume V2=250 mL=0.250 L. Use C1V1=C2V2.
- Solve for C2: C2=C1×V1/V2=0.200 mol/L×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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- Molar mass of NaOH = 22.99 + 16.00 + 1.008 = 39.998 g/mol (≈ 40.00 g/mol).
- Moles of NaOH = 8.00 g ÷ 40.00 g/mol = 0.200 mol.
- Volume of solution = 250.0 mL = 0.250 L.
- Molarity = 0.200 mol ÷ 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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- Density of solution = 1.07 g/mL; total volume = 250.0 mL ⇒ mass of solution = 1.07 g/mL × 250.0 mL = 267.5 g.
- Mass of solute (NaOH) = 8.00 g; therefore, mass of solvent (water) = 267.5 g – 8.00 g = 259.5 g = 0.2595 kg.
- Molality = moles of solute ÷ kilograms of solvent = 0.200 mol ÷ 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
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Mole and Molar Mass: Understand the mole as a unit and how to convert between mass and moles using molar mass.
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Balancing Chemical Equations: Importance of balancing to comply with the law of conservation of mass.
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Identifying Limiting Reagents: How to recognize which reactant limits the production of products and influences yield.
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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
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To calculate moles from mass, divide the mass of the substance by its molar mass. For example, 25 g of CaCO₃ gives you approximately 0.25 moles.
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Balancing the equation Al₂S₃ + 6 H₂O ⟶ 2 Al(OH)₃ + 3 H₂S ensures conservation of mass for the reaction.
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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.
🎵 Rhymes Time
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Moles, moles, count them right, over grams they take their flight!
📖 Fascinating Stories
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Imagine a race where the limiting reagent is the slowest runner, deciding how many can finish the race!
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Picture a chef adjusting recipes to get the perfect cookie; balancing ingredients until the taste is just right. This is like balancing an equation!
🎯 Super Acronyms
Don't forget
- M.B.A. = Moles
- Balance
- Amount – the steps for hands-on stoichiometry!
Flash Cards
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Glossary of Terms
Review the Definitions for terms.
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Term: Mole
Definition:
The amount of substance that contains 6.02214076 × 10²³ entities.
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Term: Molar Mass
Definition:
The mass of one mole of a substance, usually expressed in grams per mole (g/mol).
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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.
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Term: Limiting Reagent
Definition:
The reactant that is completely consumed first, regulating the maximum amount of product formed.
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Term: Theoretical Yield
Definition:
The maximum amount of product that could be formed from the given amounts of reactants.
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Term: Actual Yield
Definition:
The measured amount of product obtained from a chemical reaction.
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Term: Percent Yield
Definition:
A measure of the efficiency of a reaction calculated as (actual yield ÷ theoretical yield) × 100%.