1.5 - Worked Practice Problems
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Mole-Mass-Particle Conversions
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Today we're diving into mole-mass-particle conversions. Can anyone remind me what a mole represents in chemistry?
Isn't it a way to count particles, like a dozen is for eggs?
Exactly! One mole contains 6.022 Γ 10Β²Β³ entitiesβAvogadro's number. Now, let's look at how to find the number of moles in a sample. If I have 25.0 grams of calcium carbonate, what do I do first?
We need to calculate its molar mass.
Right! The molar mass of CaCOβ is about 100.09 g/mol. So, how many moles would that be?
I think it would be 25.0 g divided by 100.09 g/mol.
Great! So, how many moles do we get?
That would be approximately 0.2498 moles.
Excellent work! Now, can someone tell me how to determine the number of formula units in that sample?
We multiply the number of moles by Avogadro's number.
Exactly! Let's do that calculation together: 0.2498 moles times 6.022 Γ 10Β²Β³ formula units/mole.
That's about 1.505 Γ 10Β²Β³ formula units of CaCOβ!
Fantastic! You've beautifully summarized the mole-mass-particle conversions. Always remember, Moles = Mass Γ· Molar Mass. Any questions before we move on?
No questions, this was clear!
Balancing Equations
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Next, letβs talk about balancing equations. Why is it important to have a balanced equation in stoichiometry?
Because the law of conservation of mass tells us that matter canβt be created or destroyed.
Exactly! We have to balance the number of atoms for each element on both sides. Letβs balance the equation for the reaction of aluminum sulfide and water: AlβSβ + HβO β Al(OH)β + HβS. What do we start with?
I think we should count how many aluminum and sulfur we have.
Great approach! How do we balance the sulfur atoms?
We need 3 HβS because there are 3 sulfur atoms in AlβSβ.
Right! So now our equation looks like this: AlβSβ + HβO β Al(OH)β + 3 HβS. But we still need to balance the aluminum. Whatβs next?
We should add a coefficient of 2 in front of Al(OH)β.
Exactly. Now, letβs check our hydrogen and oxygen to balance everything out. We have a balanced reaction now. Can someone summarize the balancing strategy?
Write the correct formulas and count the atoms, adjusting coefficients without changing subscripts!
Spot on! Understanding this will set you up for success in stoichiometry.
Limiting Reagent and Percent Yield
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Now letβs discuss limiting reagents. Why is it essential to identify them?
Because they determine how much product can be formed!
Thatβs right! If we have excess of one reactant, it wonβt affect the yield. Letβs look at an example. Suppose we have 10.0 g of Al and 50.0 g of FeβOβ for the thermite reaction. Whatβs our first step?
We should calculate the moles of each reactant.
Perfect! After that, we compare the mole ratios to see which one runs out first. Once thatβs clear, whatβs the next step?
We can calculate the theoretical yield based on the limiting reagent.
Then we would determine the actual yield from the experiment and use that to find percent yield.
Fantastic recap! Percent yield is a way to measure the efficiency of a reaction, calculated as (Actual Yield / Theoretical Yield) x 100%. What if our actual yield was 12.0 g? If the theoretical yield was 24.72 g, how would we find the percent yield?
We divide 12.0 by 24.72 and multiply by 100.
Well done! Identifying the limiting reagent and calculating yields is fundamental in real-world reactions and industrial applications.
Solutions and Concentrations
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Now, let's explore solutions and concentrations. Who can explain what molarity is?
Molarity is the number of moles of solute per liter of solution.
Correct! And how do we calculate the molarity of a solution if we know the mass of solute and the volume of the solution?
We divide the mass in grams by the molar mass and then divide that by the volume in liters.
Exactly! Letβs say we have 8.00 g of NaOH in 250 mL of solution. How would we find its molarity?
First, we find the moles of NaOH and then convert 250 mL to liters.
Great! If the molar mass of NaOH is about 40 g/mol, what do we calculate next?
0.200 mol of NaOH, and converting 250 mL to liters is 0.250 L.
Fantastic! So, molarity is 0.200 mol divided by 0.250 L, which gives us 0.800 M. Now, whatβs our next challenge with dilutions?
We can use the dilution formula CβVβ = CβVβ!
Exactly! If our stock solution is 6.00 M and we want 1.00 L of 0.150 M, how do we find Vβ?
We can rearrange it: Vβ = CβVβ / Cβ.
Great teamwork! Understanding concentrations is key to performing all kinds of chemical analysis.
Introduction & Overview
Read summaries of the section's main ideas at different levels of detail.
Quick Overview
Standard
In this section, students will tackle various practice problems designed to apply stoichiometric principles, including mole-mass-particle conversions, balancing chemical equations, limiting reagents, percent yield calculations, and solutions and dilutions. These exercises will enhance problem-solving skills and deepen understanding of the material.
Detailed
Worked Practice Problems
This section is dedicated to reinforcing critical concepts in stoichiometry through worked practice problems. Students will engage in exercises that cover a variety of key topics including mole-mass-particle conversions, balancing chemical equations, calculating limiting reagents, determining theoretical and actual yields, and understanding solution preparations and dilutions. Each problem is structured to guide students through the calculation steps clearly, emphasizing the importance of proper conversions, ratios, and chemical knowledge. This practice aims not only to solidify theoretical knowledge but also to develop practical skills essential for success in chemistry.
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Problem 1: MoleβMassβParticle Conversions
Chapter 1 of 5
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Chapter Content
- (a) Calculate the number of moles in 25.0 g of calcium carbonate (CaCOβ).
(b) Determine the number of formula units of CaCOβ in that sample.
(c) Find the mass of 1.50 Γ 10Β²Β³ formula units of CaCOβ.
Detailed Explanation
In Problem 1, students will work with the concept of moles and how they relate to mass and particles.
- (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 (CaCOβ):
- Calcium (Ca): 40.08 g/mol
- Carbon (C): 12.01 g/mol
- Oxygen (O): 16.00 g/mol (there are 3 oxygen atoms in CaCOβ)
Therefore, the molar mass of CaCOβ = 40.08 + 12.01 + (3 x 16.00) = 100.09 g/mol. Now we can calculate the number of moles:
\[ \text{Number of moles} = \frac{25.0 \, \text{g}}{100.09 \, \text{g/mol}} \approx 0.2498 \, \text{mol} \]
- (b) To find the number of formula units, we use Avogadro's number (6.022 Γ 10Β²Β³). The relationship is:
\[ \text{Number of formula units} = \text{number of moles} \times 6.022 \times 10^{23} \]
So:
\[ \text{Number of formula units} = 0.2498 \, \text{mol} \times 6.022 \times 10^{23} = 1.505 \times 10^{23} \text{ formula units} \]
- (c) To find the mass of a specific number of formula units, first convert the formula units to moles and then to grams:
\[ \text{number of moles} = \frac{1.50 \times 10^{23}}{6.022 \times 10^{23}} \approx 0.2492 \, \text{mol} \]
\[ \text{mass} = 0.2492 \, \text{mol} \times 100.09 \, \text{g/mol} \approx 24.93 \, \text{g} \]
Examples & Analogies
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 10Β²Β³). If you have 25 grams of calcium carbonate, itβs 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.
Problem 2: Balancing Equations and Stoichiometry
Chapter 2 of 5
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Chapter Content
- (a) Balance the following chemical equation for the reaction between aluminum sulfide and water:
Al2S3(s) + H2O(l) βΆ Al(OH)3(s) + H2S(g).
(b) If 12.0 g of AlβSβ reacts with 50.0 g of water (excess), calculate the mass of HβS gas produced.
Detailed Explanation
This problem is focused on balancing chemical equations and performing stoichiometric calculations.
- (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:
\[ \text{Al}_2\text{S}_3 + 6 \text{H}_2\text{O} \longrightarrow 2 \text{Al(OH)}_3 + 3 \text{H}_2 ext{S} \]
- (b) To find the mass of HβS produced, we start by converting 12.0 g of AlβSβ into moles. The molar mass of AlβSβ 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 AlβSβ produces 3 mol of HβS, we find moles of HβS:
\[ \text{Moles of H}_2 ext{S} = 0.07991 \, \text{mol Al}_2\text{S}_3 \times 3 = 0.2397 \, \text{mol H}_2 ext{S} \]
Finally, converting moles back to grams using the molar mass of HβS (34.08 g/mol):
\[ \text{Mass of H}_2 ext{S} = 0.2397 \, \text{mol} \times 34.08 \, \text{g/mol} \approx 8.17 \, \text{g} \]
Examples & Analogies
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.
Problem 3: Limiting Reagent and Percent Yield
Chapter 3 of 5
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Chapter Content
- Consider the decomposition reaction:
2 N2O5(g) βΆ 4 NO2(g) + O2(g).
(a) If 1.20 mol of NβOβ decomposes, what masses of NOβ and Oβ are formed, assuming 100 % conversion?
(b) In a particular experiment, only 120.0 g of NOβ is collected. Calculate the percent yield for NOβ.
Detailed Explanation
This problem relates to the concepts of limiting reagents and the percent yield of a reaction.
- (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 NOβ and Oβ is generated from 1.20 mol of NβOβ . Using mole ratios from the balanced equation: - For NOβ: \( 1.20 \, \text{mol N}_2 ext{O}_5 \times \frac{4 \, \text{mol NO}_2}{2 \, \text{mol N}_2 ext{O}_5} = 2.40 \, \text{mol NO}_2 \)
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For Oβ: \( 1.20 \, \text{mol N}_2 ext{O}_5 \times \frac{1 \, \text{mol O}_2}{2 \, \text{mol N}_2 ext{O}_5} = 0.600 \, \text{mol O}_2 \)
Now to find the mass of each gas, we can use their molar masses (NOβ: 46.01 g/mol; Oβ: 32.00 g/mol):
\[ \text{Mass of NO}_2 = 2.40 \, \text{mol} \times 46.01 \, \text{g/mol} = 110.4 \, \text{g} \]
\[ \text{Mass of O}_2 = 0.600 \, \text{mol} \times 32.00 \, \text{g/mol} = 19.2 \, \text{g} \] -
(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.
Examples & Analogies
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.
Problem 4: Solutions and Dilutions
Chapter 4 of 5
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Chapter Content
- (a) Describe precisely how to prepare 500 mL of a 0.200 M KβCrOβ solution starting from solid KβCrOβ. Show all massβbased 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?
Detailed Explanation
This problem focuses on preparing solutions and understanding dilutions.
- (a) To prepare a specific molarity:
First, calculate the molar mass of potassium chromate (KβCrOβ): - 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.
- (b) For dilutions, use the dilution equation:
\[ Cβ Vβ = Cβ Vβ \]
Where: - Cβ = initial concentration = 0.200 M
- Vβ = initial volume = 50.0 mL = 0.0500 L
- Vβ = final volume = 250 mL = 0.250 L
Rearranging gives:
\[ Cβ = \frac{Cβ Vβ}{Vβ} = \frac{0.200 \, \text{mol/L} \times 0.0500 \, ext{L}}{0.250 \, ext{L}} = 0.0400 \, ext{mol/L} \]
Examples & Analogies
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.
Problem 5: Concentration Units
Chapter 5 of 5
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Chapter Content
- 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.
Detailed Explanation
This problem involves calculating both molarity and molality of a solution.
- (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:
\[ M = \frac{0.200 \, \text{mol}}{0.250 \, \text{L}} = 0.800 \, ext{M} \]
- (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:
\[ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.200 \, \text{mol}}{0.2595 \, \text{kg}} \approx 0.771 \, m \]
Examples & Analogies
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.
Key Concepts
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Conversions between moles, mass, and particles using Avogadro's Number.
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Importance of balancing equations to adhere to the conservation of mass.
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Identifying limiting reagents to calculate theoretical yields for reactions.
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Understanding how to calculate percent yields to determine reaction efficiency.
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Preparation and dilution of solutions defined by molarity and concentration.
Examples & Applications
Calculating the moles in 25.0 g of CaCOβ by using its molar mass.
Balancing the equation: AlβSβ + HβO β Al(OH)β + 3 HβS and determining the mass of product formed.
Identifying limiting reagents in a reaction with 10.0 g of Al and 50.0 g of FeβOβ.
Calculating the percent yield given theoretical and actual yields of a product.
Preparing a solution of NaOH and determining its molarity.
Memory Aids
Interactive tools to help you remember key concepts
Rhymes
For moles to mass, use grams as your guide; divide by molar mass, let precision be your pride.
Stories
Imagine a factory where for every cake (product), there are ingredients (reactants). The one ingredient that runs out first determines how many cakes can be madeβthis is your limiting reagent!
Memory Tools
Remember: L for Limiting and S for SufficientβLimiting Reagents are the ones that limit your yield.
Acronyms
M.A.P.S
Molar mass
Avogadro's number
Percent yield
Stoichiometry are essentials to master!
Flash Cards
Glossary
- Molar Mass
The mass of one mole of a substance, usually expressed in grams per mole (g/mol).
- Avogadro's Number
The number of entities (atoms, molecules, etc.) in one mole of a substance, 6.022 Γ 10Β²Β³.
- Limiting Reagent
The reactant that is completely consumed first in a chemical reaction, determining the maximum amount of product that can form.
- Percent Yield
A measure of the efficiency of a reaction calculated as (Actual Yield / Theoretical Yield) Γ 100%.
- Molarity
The concentration of a solution expressed as the number of moles of solute divided by the volume of solution in liters.
Reference links
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