By now you are all wizards at balancing equations.  You know the number of atoms of each element must be the same on the reactant (left) side of the arrow as on the product (right) side of the arrow.  Let’s use a reaction you should be familiar with from the conservation of mass lab (Lesson 71): aqueous calcium chloride reacts with aqueous sodium carbonate to produce solid calcium carbonate and aqueous sodium chloride:

Here’s the balanced equation:

CaCl₂(aq) + Na₂CO₃(aq) → CaCO₃(s) + 2NaCl(aq)

Quick vocabulary review and a piece of important new learning:

• (aq) means aqueous (in water)
• (s) means solid
• In a chemical equation, the arrow points from the reactants toward the products.  As written above, the reactants are on the left and the products are on the right.
• The letters represent element symbols from the periodic table
• The numbers represent subscripts and coefficients.
  • Subscripts are the small integers (whole numbers) to the lower right of an element symbol and they cannot be changed.  Subscripts tell how many atoms of each element are present in a given molecule.  In the equation above, the small 2 after Cl tells us there are two chlorine atoms bonded with 1 calcium atom in one molecule of CaCl₂.
  • Coefficients are the regular-sized integers to the left of a molecule.  This is new and important: Coefficients tell us the correct mole ratio in which the reactants combine to form the products.  In the equation above, we now know that one mole of aqueous calcium chloride reacts with one mole of with aqueous sodium carbonate to produce one mole of solid calcium carbonate and two moles of aqueous sodium chloride.

Another example: N₂(g) + 3H₂(g) → 2NH₃(s)  This equation tells us that one mole of nitrogen gas reacts with 3 moles of hydrogen gas to produce one mole of solid ammonia.

Even more examples:

Your turn!  Complete the Mole Ratios Google Form before returning to Week 33 – Stoichiometry to continue working.

So far this week, we’ve learned that the coefficients of balanced equations provide us with mole ratios, which we can then use to predict the amount of products generated by reactants in a chemical reaction.  Once we have a theoretical value for the amount of reactant we will generate (our theoretical yield), we can then run the  experiment and measure how much reactant we actually recover (our actual yield).  By comparing the theoretical and actual yeilds, we can calculate our percent yield: how much reactant we are able to recover given real-world experimental constraints.

In the Going Through the Mole Tunnel section, we investigated the reaction of copper with silver nitrate to produce silver and copper nitrate.  We went on to calculate the number of moles of copper and silver nitrate we would have to combine to produce 30 g of silver.  Now, imagine we actually did that experiment and our recovery of pure silver was 28.5 g.  What was our percent yield of silver?

• Theoretical yield of silver = 30 g
• Actual yield of silver = 28.5 g

Percent yield = (actual yield) / (theoretical yield) x 100%

For our experiment: Percent yield = 28.5 / 30 x 100% = 95%

Not bad!  Our actual yield was only 5% less than our theoretical yield.  Why wasn’t the percent yield 100%?  Experimental error!  Maybe we mis-measured one of our reagents.  Maybe we used a solution that wasn’t prepared as accurately as it should have been.  Maybe our balance needs to be re-calibrated, or the volume measured by our beaker wasn’t perfectly accurate.

Another example:

Your turn!  Complete the Percent Yield Google Form before returning to Week 33 – Stoichiometry to continue working.

You went through the Mole Tunnel and came out the other side. Along the way, you got hungry. You go to the refrigerator and find a 10-pack of hot dogs. You check the cupboard and find an unopened 12-pack of hot dog buns. A hot dog requires 1 hot dog and 1 bun. How many complete hot dogs can you make? Do you have exactly the right amount of hot dogs and buns to use everything up?  Will you be left with too many hot dogs or too many buns? Hmmm….10 hot dogs, 12 buns…looks like you’re going to have 2 leftover hot dog buns. Turns out, when it comes to hot dogs and hot dog buns, hot dogs are the limiting reactant.  You run out of hot dogs first. The number of hot dogs determines how many complete hot dogs you can make.

The same goes for chemical reactions. When it comes to reactions, if you combine reactants, you can either combine them in the exactly right amount to use both up, or you are left over with something. The chemical that runs out first is the limiting reactant.

Let’s return to the single replacement reaction we examined as we navigated the Mole Tunnel:

Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(s) + 2Ag(s)

Imagine you have a beaker containing 4 mol of AgNO₃.  To the beaker you add 75 g of Cu.  Will the addition of this amount of Cu completely react with the AgNO₃?  Will either reactant by left over?  Let’s find out.

• Step 1: Calculate the number of moles of Cu are in 75 g.  The molar mass of Cu is 63.55 g/mol, so 75 g x (1 mol / 63.55 g) = 1.18 mol Cu
• Step 2: Evaluate the chemical equation to determine the mole ratios.  How many moles of Cu does it take to fully react with AgNO₃?  1 mol Cu : 2 mol AgNO₃
• Step 3: Compare the mole ratio with the amount of AgNO₃?  We are told we have a beaker containing 4 mol of AgNO₃.  According to the mole ratio, we need 2 mol Cu to fully react with 4 mol of AgNO₃.
• Step 4: Compare the mole ratio with the amount of Cu added.  You added 1.18 mol Cu to 4 mole of AgNO₃.  You need 2 mol of Cu to fully react with the AgNO₃, and the 1.18 mole we added is less than 2 mol.  Therefore, Cu is the limiting reactant in this experiment.

What does that mean?  All of the copper will fully react to form Cu(NO₃)₂.  How many moles of Cu(NO₃)₂?  You added 1.18 moles of Cu as the reactant, and the mole ratio of Cu:Cu(NO₃)₂ is 1:1, so you will produce 1.18 mol of Cu(NO₃)₂.

More examples:

Chemistry Challenge!  Try this one!

Your turn!  Complete the Limiting Reactant Google Form before returning to Week 33 – Stoichiometry to continue working.

Welcome to Week 33!  For this week, we will be focusing on stoichiometry: the quantitative relationship between reactants and products in a chemical reaction.  As a question: How do you convert between grams and moles to determine the mass of product in a chemical equation?  Please work through the list of links below.  Each section contains important information and ends with a portion of the weekly assignment.  You can complete it all in one sitting or break it up as needed.  Ready, set, go!

1. Week 33 Attendance Check-In (required)
2. Mole Ratios (Google Form assignment)
3. Going Through the Mole Tunnel (Google Form assignment)
4. Limiting Reactants (Google Form assignment)
5. Percent Yield (Google Form assignment)
6. Summary of Stoichiometry (encouraged)

You did it!  Just to make sure, here’s a checklist of items you must complete this week by Sunday, May 10 at 11:59pm:

• Weekly Attendance Check-In (school district requirement)
• Mole Ratios Google Form (worth 10 assignment points)
• Mole Tunnel Google Form (worth 10 assignment points)
• Limiting Reactants Google Form (worth 10 assignment points)
• Percent Yield Google Form (worth 10 assignment points)

Remember, you can email me any time.  Office hours for Science are Tuesdays from 11am-12pm and Thursdays from 1pm-2pm.  Check your student Gmail for Zoom instructions.

Finally, by popular demand…click here for the Week 33 Bonus Credit Opportunity!

Now that we understand the role of coefficients in chemical equations, we can convert between grams and moles to determine the mass of the products that result from a chemical reaction.  Let’s conduct an experiment where we combine copper and silver nitrate.

Here is the following balanced chemical reaction:

Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(s) + 2Ag(s)

First off, let’s notice this is a single replacement reaction.

AB + C → A + CB, where Cu = A, NO₃ = B, and Ag = C

• Question: Find the mole ratio of AgNO₃ to Cu(NO₃)₂  Answer: 2:1
• Question: Find the mole ration of AgNO₃ to Ag  Answer: 1:1
• Suppose you have 6.0 mol of Cu.
  • Question: How many moles of AgNO₃ are needed to react completely with 6.0 mol of Cu? Answer: The mole ratio is 2 moles of AgNO₃ to 1 mole of Cu, so to completely react 6.0 mol of Cu, you need twice as many moles of AgNO₃ (12 mol).
  • Question: How many moles of Cu(NO₃)₂ are produced? Answer: The mole ratio of Cu(NO₃)₂ to Cu is 1:1, so if you have 6.0 mol of Cu, you produce 6.0 mol of Cu(NO₃)₂.
• Suppose you want to make 30.0g of Ag.
  • Question: How many moles of Ag is that? Answer: The molar mass of Ag is 107.9 g/mol.  30g x (1mol / 107.9g) = 0.278 mol Ag
  • Question: How many moles of Cu do you need? Answer: The mole ration of Ag to Cu is 2:1, so with 0.278 mol Ag, you would need 0.139 mol of Cu.

Imagine we actually want to do this single replacement experiment where we produce 30.0 g of silver.  You go to the chemical supply shelf and locate the bottle of copper and silver nitrate.  Both are in solid form.  Now what?  The Mole Tunnel!  In the real world, we work in mass quantities like grams.  To move through a chemical reaction from products to reactants (or vice versa), you go from grams to moles to grams.

We know we need 0.139 mol of Cu.  The copper:silver nitrate ratio is 1:2, so we need twice as many moles of silver nitrate, or 0.278 mol AgNO₃.  To actually do this experiment, we need to use counting by weighing, which means converting moles to grams.  To convert moles to grams, we need molar mass (g/mol).  For Cu, the molar mass is 63.55 g/mol.  For AgNO₃, the molar mass is 169.9 g/mol.

• 0.139 mol Cu x 63.55 g/mol = 8.83 g of Cu
• 0.278 mol AgNO₃ x 169.9 g/mol = 47.2 g of AgNO₃

Done!  You successfully navigated the Mole Tunnel!  You started with grams (30 g), converted to moles, compared mole ratios, then converted back to grams.  By combining 8.83 g of Cu with 47.2 g of AgNO₃, you will produce 30 g of Ag.  Math!

Your turn!  Complete the Mole Tunnel Google Form before returning to Week 33 – Stoichiometry to continue working.

This section of the lesson will serve as a review of working with moles.  Our textbook defines a mole as a counting unit used to keep track of large numbers of particles.  Just like one dozen represents 12 items, one mole represents 6.02 x 10²³ items. Here are some examples:

• One mole of gold atoms is equal to 6.02 x 10²³ gold atoms.
• One mole of glucose molecules is equal to 6.02 x 10²³ glucose molecules.  The molecular formula of a molecule of glucose is C₆H₁₂O₆.  

How many carbon atoms are in one mole of glucose?  Since each glucose molecule includes 6 carbons, and we have 6.02 x 10²³ glucose molecules (equal to 1 mole), then we have a total of 6 x 6.02 x 10²³ carbon atoms, which is equal to 36.12 x 10²³ carbon atoms.  For proper scientific notation, we need to move the decimal one position to the left which then increases the exponent by 1.  Therefore, we have 3.612 x 10²⁴ carbon atoms per mole of glucose.  Because we also have 6 oxygen atoms per molecule of glucose, the same math applies, so we have 3.612 x 10²⁴ oxygen atoms per mole of glucose.  Finally, since we have 12 hydrogen atoms per molecule of glucose, and 12 is 2 x 6, we can double the number of carbon (or oxygen) atoms per mole of glucose to determine the number of hydrogen atoms per mole of glucose: 2 x 3.612 x 10²⁴ = 7.224 x 10²⁴ hydrogen atoms.

How many moles of atoms are in one mole of glucose?  Well, for each glucose molecule, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.  That adds up to 24 atoms (6+6+12) per molecule of glucose.  That means there are 24 moles of atoms per mole of glucose molecules!  Of those 24 moles of atoms, there are 6 moles of carbon, 12 moles of hydrogen, and 6 moles of oxygen atoms per mole of glucose molecules.

Why does this matter?  Glad you asked!  The mole connects the microscale world (number of atoms or molecules) with the macroscale world (number of grams – something we can actually measure in the lab).  The number of grams per mole of a substance is the same as the number of atomic mass units (amu) of a substance.  For example:

• Gold, Au (element 79), has an average atomic mass of 197.0 amu.  Therefore, 1 mole of gold has a mass of 197.0 grams.  The molar mass of gold is 197.0 g/mol.
• Glucose, C₆H₁₂O₆ has a molecular mass of 180.2 amu.  The molecular mass is calculated by multiplying the average atomic mass of each element by the number of atoms of that element and then adding them all up:
  • carbon = 12.01 amu x 6 atoms = 72.06 amu
  • hydrogen = 1.008 amu x 12 atoms = 12.096 amu
  • oxygen = 16.00 amu x 6 atoms = 96.00 amu
  • 72.06 amu + 12.096 amu + 96.00 amu = 180.156 amu
  • 72.06 and 96.00 both have 4 significant digits, so we round 180.156 amu to 180.2 amu (which has 4 significant digits).
  • Note: One formula unit of glucose has a mass of 180.2 amu.  One formula unit of glucose is the same thing as one glucose molecule.
  • Therefore, 1 mole of glucose has a mass of 180.2 grams.  The molar mass of glucose is 180.2 g/mol.

Finally, imagine you have two samples: 100 g of C₆H₁₂O₆ and 100 g of CO₂.  Which sample has more carbon atoms?  Which sample has more oxygen atoms?  One way to solve this problem is to determine how many moles of each substance are in 100 g:

• From our work above, we know that C₆H₁₂O₆ has a molar mass of 180.2 g/mol.  Therefore, 100 g x 1 mol / 180.2 g = 0.55 mol of C₆H₁₂O₆
• For CO₂, the molar mass works out to 44.01 g/mol.  We’ve calculated that in class many, many times, but you should also try the math yourself to be sure you get it.  Therefore, 100 g x 1 mol / 44.01 g = 2.27 mol of CO₂

Next, we can determine how many moles of carbon are in each substance:

• 0.55 mol of C₆H₁₂O₆ x 6 moles of carbon / mole of C₆H₁₂O₆ = 3.3 moles of carbon in 100 g of C₆H₁₂O₆
• 2.27 mol of CO₂ x 1 mole of carbon / mole of CO₂ = 2.27 moles of carbon in 100 g of CO₂
• Therefore, C₆H₁₂O₆ has more carbon atoms than CO₂ in 100 g of each sample

Finally, we can repeat the process to determine how many moles of oxygen are in each substance.  Give it a try.  Then click here to check your work.

Time to move into the textbook:

1. Create a Google Doc titled Week 32 – Name (example: Week 32 – Pickles Swart)
2. Share the Doc with david.swart@g.highlineschools.org
3. Read Lesson 76 in the textbook and complete Exercise questions 4-7 in your Google Doc in a section labeled Lesson 76
4. Read Lesson 77 in the textbook and complete Exercise questions 3-8 in your Google Doc in a section labeled Lesson 77
5. Read Lesson 78 in the textbook and complete Exercise questions 1-8 in your Google Doc in a section labeled Lesson 78

Return to Week 32 – Mass-Mole Conversions and continue working.

Welcome to Week 32!  For this week, we will be focusing on mass-mole conversions.  In other words, how are particle mass and particle number connected?  Please work through the list of links below.  Each section contains important information and ends with a portion of the weekly assignment.  You can complete it all in one sitting or break it up as needed.  Ready, set, go!

1. Week 32 Attendance Check-In (required)
2. The Return of the Mole (Entry Task)
3. Counting by Weighing (Google Form assignment)
4. Moles and Molar Mass (Textbook Exercises in Google Doc)
5. Comparing Amounts (Exit Task)
6. Chemistry Refresher…review now before it’s too late!
7. Unit 4 Honors Project…the wait is over!

You did it!  Just to make sure, here’s a checklist of items you must complete this week by Sunday, May 3 at 11:59pm:

• • Return of the Mole Entry Task (worth 10 assignment points)
  • Counting by Weighing (worth 10 assignment points)
  • Textbook Exercises (worth 30 assignment points)
  • Comparing Amounts Exit Task (worth 10 quiz points)

Remember, you can email me any time.  Office hours for Science are Tuesdays from 11am-12pm and Thursdays from 1pm-2pm.  Check your student Gmail for Zoom instructions.

Finally, by popular demand…click here for the Week 32 Bonus Credit Opportunity!

For this section of our weekly lesson, we will be working between the macroscale world (the one we inhabit) and the microscale world (the world of atoms and molecules).

To begin, imagine you have a 10-pound (4.54 kg) bag of potatoes.  You are tasked with counting the number of potatoes in the bag.  For potatoes, that’s pretty straight-forward.  You literally count each potato and the number you count is the number of potatoes in a 10-pound bag of potatoes.  Counting large objects is pretty easy.

Next, you are tasked with counting the number of grains of rice in a 20-pound (9.07 kg) bag of rice.  Suddenly, this job is a lot less easy.  Could you count all those grains of rice by hand?  Sure.  Do you want to?  No.  So what can do you do to estimate the number of grains of rice in the bag?  One solution is to use the technique of counting by weighing.

• You begin by trying to measure the mass of a single grain of rice. It turns out, the mass is to small to read out on your balance.  Also, what if the grain of rice you selected is not representative of a typical grain of rice?  Maybe you selected a grain of rice that is bigger or smaller than normal.
• You decide to be more scientific.  You count out a random sample of 100 grains of rice and then measure the mass using your balance.  You find that 100 grains of rice have a mass of 2.9 grams.  To calculate the average mass of one grain of rice, you divide 2.9 grams by 100 grains of rice, for an average of 0.029 grams per grain of rice.
• Finally, you have 9.07 kg of rice and you know that each grain of rice has an average mass of 0.029 g.
  • 9.07 kg x 1000 g/kg = 9070 g of rice in the bag
  • 9070 g x 1 grain of rice / 0.029 g = 312,759 grains of rice!
  • Actually, since 0.029 grams only has 2 significant digits, we would actually estimate a total of 310,000 grains of rice.

Finally, it’s time to enter the microscale world of counting by weighing in chemistry.  Imagine you are tasked with counting how many atoms of gold are in a gold ring.  The ring has a mass of 10 grams.  Are you going to count each atom by hand?  Nope.  Instead, you turn to your trusty periodic table and find that gold, atomic symbol Au, has an average atomic mass of 197.0 amu.  You also remember Avogadro’s Number which tells you that 1 mole of gold has 6.02 x 10²³ atoms.  That’s all you need!

• First, 197.0 amu = 197.0 g/mole
• Next, to figure out how many moles of gold are in 10 grams of gold: 10 g x 1 mole / 197.0 g = 0.051 moles of gold
• Finally, to figure out how many atoms of gold are in the ring: 0.051 moles x 6.02 x 10²³ atoms / mole = 0.31 x 10²³ atoms of gold (or 3.1 x 10²² atoms using correct scientific notation)

The video below will help reinforce your learning. Also, Lesson 75 in the textbook is titled “Counting by Weighing” and is an excellent resource for more review.  While not required, you are encouraged to read the lesson and practice the exercises at the end of the lesson.  These will not be entered into Synergy but I am happy to answer any questions you have and review your work if you need help.  When ready, complete the Week 32 – Counting by Weighing Google Form assignment.  Note: For this and all Google Form assignments, you can re-assess and your highest score will be entered into Synergy.  Obviously, do your best the first time through!

Return to Week 32 – Mass-Mole Conversions and continue working.

For our final section of this week’s lesson, we need to connect back with the theme of toxicity.  Simply stated, too much of anything can be toxic.  For example, caffeine is toxic (LD₅₀ = 150 mg/kg).  One cup of strong coffee has about 150 mg of caffeine.  Two cups of strong coffee would have about 300 mg of caffeine.  Therefore, two cups of coffee (300 mg of caffeine) is more toxic than 1 cup of coffee (1 cup of caffeine).  Thankfully, for most people, it takes a whole lot more than a couple of cups of coffee to be lethal.  For this example, since the LD₅₀ for caffeine is the same as the amount of caffeine per cup of coffee, you can figure out how many cups of coffee it would take to reach the LD₅₀ for you by converting your weight in pounds to mass in kg (weight in pounds divided by 2.2 = mass in kg).  A 180 pound person has a mass of about 82 kg, so drinking 82 cups of coffee would result in a 50% chance of death.

Fun fact: A cup of coffee is nearly 8 ounces of water, and there are about 30 mL per ounce.  Therefore, drinking a cup of coffee means drinking about 240 mL of water.  The LD₅₀ for water is 90 mL/kg.  For that 82 kg person in the coffee example, drinking 7380 mL of water (90 mL/kg x 82 kg) would result in a 50% chance of death.  How many cups of water are in 7380 mL?  Divide 7380 mL by 240 mL / cup and that comes out to 30.75 cups of water.  So for all you coffee haters out there…go easy on your delicious water!  Turns out water is more toxic than coffee.  Sorry.

When comparing substances to determine which is more toxic, the key point is you must convert LD₅₀ values to mol/kg.  When you convert from mass to moles, you remove the impact of molecule size and instead normalize the data on a per molecule basis.  When we ask whether one substance is more toxic than another, we want to be able to compare the toxicity of the molecules, irrespective of the mass of those molecules.  In Lesson 79 of the textbook, there is a table comparing the sweeteners fructose (fruit sugar) and aspartame (artificial sweetener).  Fructose is commonly found in regular soda, while aspartame is found in some diet sodas.  The molecules have different molecular formulas, which means they have different molar masses.  Converting the LD₅₀ values to mol/kg shows that aspartame is more toxic than fructose.  That’s an important finding, but doesn’t address the real-world question: how much of each sweetener is added to regular (fructose) or diet (aspartame) soda?  How many cans of regular or diet soda does it take to reach the LD₅₀ for fructose and aspartame?  Read through Lesson 79 in the textbook and find out – the answer might surprise you!

For your final piece of chemistry work this week, complete the Week 32 Exit Task.

Return to Week 32 – Mass-Mole Conversions and continue working.

On the macro scale, if you have 10 small glass marbles and 10 large clay bricks, what is similar and what is different?  Well, clearly you have 10 of each (similar) but the materials have different masses (difference), with large clay bricks having more mass than small glass marbles.  Thinking about everyday items and quantities isn’t too bad.  However, in chemistry, we have to think about atoms and molecules (things that are incredibly small).  When we work with visible amounts of substances composed of atoms or molecules, we have to work with enormous numbers of those substances.  This brings us back to everyone’s favorite topic: The Mole.

In Unit 3, you learned that 1 mole is equal to 6.02 x 10²³ “things” – such as atoms, molecules, or literally anything you want to count.  If you had one mole of glass marbles, you would have 6.02 x 10²³ glass marbles.  If you had one mole of clay bricks, you would have 6.02 x 10²³ clay bricks.

Let’s figure out the mass of one mole of marbles. One marble has a mass of 1.80 g.

• 1 marble x 1.80 g/marble = 1.80 g/marble
• 1.80 g/marble x 6.02 x 10²³ marbles/mole = 10.8 x 10²³ g/mole
• 10.8 x 10²³ g/mole = 1.08 x 10²⁴ g/mole

Your turn!  The Entry Task will guide you through another macroscale calculation (the mass of clay bricks) and then we will transition to the microscale (atoms and molecules).

Need a refresher on the Mole?  Watch the video below, then head back to the mole lesson in Unit 3 and review as needed.

Return to Week 32 – Mass-Mole Conversions and continue working.