If you’ve ever made ice tea and decided it needs more sugar, you understand the importance of solution concentration.  Add too little sugar and the sugar concentration is too low to make the ice tea pleasantly sweet.  Add too much sugar and the ice tea just tastes like sugar water because the sugar concentration is too high.  Somewhere in there is the ice tea Goldilocks zone – just the right amount of sweet.  If you have access to water, a measuring cup, a glass, some tea, some sugar, and a teaspoon: fill a glass with one cup (8 ounces) of cold water, add one bag of tea, and let it steep while you continue working.

Back to our lesson: we need to understand the concept of molarity which is defined as the concentration of dissolved substances in a solution, expressed in moles of solute per liter of solution.  We use M as the unit of molarity (M = mol/L).  We already know how to calculate moles from grams, so just take moles and divide by volume (in liters) to calculate molarity.  Here’s an easy example question: glucose has a molar mass of 180 g/mol.  If 90 g of glucose is added to 1 L of water, calculate the molarity of the solution.  Answer: 90 g x 1 mol / 180 g = 0.5 mol, 0.5 mol / 1 L = 0.5 M.  The molarity of the solution is 0.5 M.

Here’s a harder example question: A 20 fluid ounce bottle of Dr. Pepper contains 64 grams of sugar (high fructose corn syrup).  Determine the molarity of sugar Dr. Pepper.

To approach this question, we need to know some additional pieces of information.  Ultimately, we need our answer in units of M, or mol/L.  So we need to convert ounces to liters and we need the molar mass of high fructose corn syrup.

Key information:

• 1 L = 33.814 fluid ounces
• High fructose corn syrup consists of glucose and fructose, both of which have a molar mass of 180 g/mol.

Answer: First, let’s convert 64 grams of sugar to moles: 64 g x (1 mol / 180 g) = 0.36 mol of sugar.  Next, let’s convert 20 ounces to liters: 20 ounce x (1 L / 33.814 ounces) = 0.59 L.  Finally, we just need to divide moles by liters to calculate molarity: 0.36 mol / 0.59 L = 0.61 M.

Let’s turn our attention back to the tea we started making at the beginning of this lesson and wrap this up with a more challenging practice problem.  Question: If we add one teaspoon of table sugar to our 8 ounce glass of tea, what is the molarity of the solution?

Key information:

• 1 L = 33.814 fluid ounces
• molar mass of table sugar (sucrose) = 342 g/mol
• 1 teaspoon of table sugar = 4.2 grams

Answer: Let’s begin by calculating the number of moles of sucrose (table sugar) added to the tea.  One teaspoon = 4.2 grams of sucrose.  4.2 grams x (1 mole / 342 g) = 0.0123 mol of sucrose.  Next, let’s convert 8 ounces to liters: 8 ounces x (1 L / 33.814 ounces) = 0.237 L.  Finally, to calculate molarity (M), divide moles by liters: 0.0123 mol / 0.237 L = 0.052 M.

Time to show what you know!  Complete the Week 36 – Molarity Google Form assignment and then return to Week 36 – Solution Concentration and continue working.

For the visual learners among us, the video below will help guide you through the process of conducting the research required to complete Slide 2.

An example layout of Slide 2 is pictured below.

Return to the Weeks 34-35 Toxin Research page.

Last week, you were introduced to the concept of limiting reactants.  In a chemical reaction, the limiting reactant is the chemical that is completely used up in the process of creating products.  When the limiting reactant is gone, the reaction ends.  Let’s imagine the chemical reaction of sodium bicarbonate (baking soda) and acetic acid (vinegar):

NaHCO₃(s) + C₂H₄O₂(aq) → NaC₂H₃O₂(aq) + H₂O(l) + CO₂(g)

Let’s first notice that this equation is already balanced.  It does not require any coefficients to have equal numbers of atoms of each element on both sides of the equation.  One mole of solid sodium bicarbonate NaHCO₃(s) reacts with one mole of aqueous acetic acid C₂H₄O₂(aq) to produce one mole of aqueous sodium acetate NaC₂H₃O₂(aq), one mole of liquid water H₂O(l), and one mole of carbon dioxide gas CO₂(g).

To determine how much of each substance we need to actually conduct this experiment, let’s calculate the molar masses of sodium bicarbonate and acetic acid:

• NaHCO₃ = 22.99 + 1.008 + 12.01 + (16.00 x 3) = 84.01 g/mol
• C₂H₄O₂ = (12.01 x 2) + (1.008 x 4) + (16.00 x 2) = 60.06 g/mol

Therefore, if we wanted to react one mole of sodium bicarbonate with one mole of acetic acid, we would need to combine 84.01 g of sodium bicarbonate with 60.06 g of acetic acid.  Easy!  We also know that we will produce one mole of carbon dioxide gas in this reaction – there will be bubbles!  Thinking back to our work with gas laws in Unit 3, we know that one mole of any gas at standard temperature and pressure will occupy a volume of 22.4 L.  So there will be a lot of gas bubbles!  Here is the reaction:

What did you observe?  Hopefully you noticed the geyser of bubbles – that was hard to miss!  Did you also notice the white solid material at the bottom of the flask?  That shouldn’t be there if the reactants fully reacted as expected.  The science tells us that one mole of sodium bicarbonate will fully react with one mole acetic acid.  Why was there so much unreacted sodium bicarbonate remaining?  Why was acetic acid the limiting reactant in this chemical reaction?

To answer this question, we need to look closely at our reactant labels:

The baking soda is pure sodium bicarbonate.  The distilled white vinegar has some fine print that says “distilled with water to 5% acid strength” – in other words, the distilled white vinegar actually contains 5% acetic acid and 95% water!  In our balanced equation, we assumed we were using 100% acetic acid, not 5% acetic acid.  Because 5% is 1/20 of 100%, we need to combine 20 times more of the 5% acetic acid (60.06 g x 20 = 1201.2 g) with 84.01 g of sodium bicarbonate to fully react both chemicals.  Or, we could keep the amount of acetic acid the same (60.06 g) and use 20 times less of the sodium bicarbonate (84.01 g / 20 = 4.2 g).  Let’s do that instead, to keep the amounts of each reagent reasonable:

What did you observe?  Can you explain it?  As we wind down the school year, this is an opportunity for you to earn credit in the lab report section of your chemistry grade.  This is an optional assignment and will be worth 50 bonus points. If your grade is not where you want it, and your lab report scores have been less than amazing, this is your chance to make up a lot of ground.  Please take advantage of it.

Lab Report Requirements:

• • Purpose: What are we trying to accomplish with this lab?
  • Procedure: Write out the steps, in order and in detail, that were followed in this experiment.
  • Results: What exactly did you observe?  Be clear, be descriptive, and include images if possible.
  • Conclusion: Clearly explain the results.  Include the concepts of chemical reaction, reactants, products, molar mass, and limiting reactant.

Note: If you have access to baking soda and vinegar and a safe space to work, you are welcome to substitute your own experimental data for what you were provided above.

Need some help setting up your lab report?  The links below will give you either a highly structured template specific to this lab, or a more generic template applicable to any lab report.  If you decide to use either one, click the link and then select File > Make a Copy and get to it!

• Highly structured Limiting Reactants lab report template
• Generic lab report template perfect for any lab report

Finally, if you have read all the way down to this point, enjoy!

When finished, return to Weeks 34-35 – How Much Is Too Much? and continue working.

For the final research part of the Toxin Research Project, it’s time to answer the question: how much is too much?  To answer that question, you need to research how much of your toxin is in a “standard dose” of your toxin.  “Standard dose” can be hard to find, and is often provided as a range.  For example, a “standard dose” of caffeine might be 100-200 mg per “dose” with a dose being an 8 ounce cup of coffee or a 12 ounce can of soda.  In that case, split the difference and use 150 mg as the standard dose for caffeine.  Buckle up – this is often the hardest part of this project.  Be persistent!  If “standard dose” doesn’t work as a search term, try “amount per serving” and see if that works.

Instructions for Slide 3:

• Add Slide 3 to your Google Slides deck and title it “Toxin LD₅₀ Calculations”
• On Slide 3, answer the following:
  • Click the PubChem link to your toxin from Slide 1
  • Locate the molecular weight (synonym for molar mass) in PubChem and enter it on Slide 3
  • Using the molar mass and the LD₅₀ (from Slide 1), calculate the lethal dose of the toxin for humans of three different masses and show your calculation work on the slide:
    • 10 kg
    • 30 kg
    • 75 kg
  • How many “standard doses” of the toxin does it take to reach the lethal dose for a person of mass 10 kg?  30 kg?  75 kg?

Here’s a video of my efforts to determine the “standard dose” of capsaicin per jalapeno pepper:

Need help bring these concepts together?  From the video above, we learned that the average jalapeño contains 2.24 mg of capsaicin.  Here is an example calculation for how many jalapeños a 75 kg person would have to eat to reach the lethal dose for capsaicin (identified in Slide 1 as 47.2 mg/kg):

75 kg x (47.2 mg capsaicin / 1 kg) x (1 jalapeños / 2.24 mg capcaisin) = 1580 jalapeños

When finished with Slide 4, return to Weeks 34-35 – How Much Is Too Much? and continue working.