February 28th - Dilutions!

Today, we learned all about dilutions! Yay!

When two solutions are mix together, the concentration of the final solution changes. Dilution is the process of decreasing the concentration by adding a solvent (usually water). The amount of solute, which is the chemical that is present in lesser amounts, does not change. As you can see below, the number of moles (n) of solute doesn’t change. Therefore, n₁ = n₂.

Because concentration is in mol/L, we can write concentration as n/V, where n is the number of moles and V is the volume. We can rearrange this to get n = CV, where C is concentration. This equation is equal to C₁V₁ = C₂V₂.

Know that you know that formula, we can dive into some examples. This is literally all you need to know to be able to calculate dilutions!

Ex.) Determine the concentration when 200 mL of 0.20 M HCl is diluted to a final volume of 400 mL.

We can start this problem by writing out everything we know.

C₁ = 0.20 M

V₁ = 200 mL

C₂ = ?

V₂ = 400 mL

Next, we simply solve the equation using C₁V₁ = C₂V₂.

C₁V₁ = C₂V₂

(C₁V₁)/ V₂= C₂

((0.20)(200 mL))/400 mL = C₂

0.10 M = C₂

Ta-da! Easy, right?

Ex.) 300 mL of 0.500 M Sodium nitrate is mixed with 400 mL of 0.100 M Sodium nitrate. Determine the final [NaNO₃].

You may not know what to do here, since you have been told C₁, V_1, C₂ and V_2. Of course, things become easier when we list what we know.

 C₁ = 0.500 M

V₁ = 300 mL

C₂ = 0.100 M

V₂ = 400 mL

Now, we need to find the final concentration of the solution. To do this, we add the volumes together. We also calculate the moles in both concentrations and add those together. With our total number of moles and total volume, we can divide ‘em to get concentration!

V_tot = 300 mL + 400 mL

V_tot = 700 mL

0.300 L   x   (0.500 mol/L)   =   0.15 mol NaNO₃

0.400 L   x   (0.100 mol/L)   =   0.04 mol NaNO₃

mol_tot = 0.15 mol + 0.04 mol

mol_tot = 0.19 mol

0.19 mol   x   0.7L   =   0.133 M NaNO₃

Ex.) When 75.0 mL of water is added to a 2.00 M solution of NaCl the resulting solution has a concentration of 0.850 M. Determine the initial volume of the solution.

C₁ = 2.00 M

V₁ = ?

C₂ = 0.850 M

V₂ = V₁ + 75.0 mL

Oh, no! This question is more complex! But, how do we solve it?

That’s right: math! We know that we added 75.0 mL to V₁, so our V₂ would actually be (V₁ + 75.0). Observe:

C₁V₁ = C₂V₂

(2.00) V₁ = (0.850)V₁ + 75.0

2V₁ = 0.850V₁ + 63.75

(1.15V₁)/1.15 = 63.75/1.15

V₁ = 55.4 mL

Wasn’t that fun, kids? Join us next class when we perform a lab.

As always, the obligatory video:

Posted by Michael.

February 24 - Sour Titration Lab

After learning about titrations, we had to put our knowledge to the test. We did so using by performing a titration ourselves. Our titrant, or the substance with a known concentration was NaOH, which had a molarity of 0.8. Next, we were given acetic acid and asked to find its concentration.

To recall, the steps to a titrations include making a table:

Trial	1	2	3	4
Final Reading
Initial Reading
Volume Used

We did this to find the volume of titrant used. We then take the average to compensate for potential errors. With this number, we can find the number of moles of NaOH that we used.

Remember, we pour enough titrant in so that the indicator, phenolphthalein, in this case, just turns pink.

With this number, we use simple stoichiometry to find the number of moles of acetic acid used. This requires a the balanced equation, which is: NaOH + CH₃COOH à NaCH₃COO + H₂O

With the found number of moles, we divide it by the volume of acetic acid that was originally present in the Erlenmeyer flask, which was 10mL, and we have our concentration.

Now let’s go over the same thing, utilising numbers this time.

Trial	1 (disregard)	2	3	4
Final Reading mL	14.6	25.5	36.9	46.9
Initial Reading mL	1.6	14.4	25.5	35.9
Volume Used mL	13.0	11.1	11.4	11.0

We determined trial one was done inaccurately, since it was far away from the other three. Therefore, we will disregard that in our calculations. From these results, we obtained 11.3 mL to be our average. Now time for some stoichiometry.

11.2 mL (0.8 mol NaOH/ 1000 mL) (1 / 1) = 0.00904 mol CH₃COOH (1 / 0.01L) = 0.896 [CH₃COOH]

And there you have it! We completed the titration. The actual value for acetic acids concentration is 0.85M. Therefore, our percent error is: 5.4%. Not bad, but we could have done better.

And that’s all! So, next time you’ll be using vinegar, you’ll know it was 0.8 moles of solute per litre of solution!

Posted by Andrew.

February 22nd - Titrations

Today we learned all about titrations. This lesson was mostly in preparation for next class’s lab. However, we still learned the basics of the topic.

First of all, a titration is an experimental technique used to determine the unknown concentration of a solution. The whole processed can be summed up quite nicely in a neat little diagram. Observe:

Here are some of the parts you can see above:

• The buret (also spelled burette) contains the solution with the known concentration (AKA the stock solution or titrant).
• The stopcock (AKA tap) is a valve that is used to control the flow of the titrant from the buret.
• The pipet (also spelled pipette) is used to accurately measure the volume of the solution with the unknown concentration.
• An Erlenmeyer flask contains the solution with the unknown concentration.
• The indicator is used to identify the end point of the titration if no visible reaction occurs.

You’ll be able to tell when the reaction has taken place because the indicator will change colour. However, don’t add too much titrant! Here is what you should aim for:

And that’s it for notes! We’ll do two quick examples.

Ex.) A solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl? The trials are as follows:

Trial	1	2	3	4
Final Reading (mL) Initial Reading (mL)	11.6 0.2	22.4 11.6	34.0 22.4	44.7 34.0
Volume Used	11.4	10.8	11.6	10.7

First, determine the average. Above, it is 11 mL of NaOH.

Next, write the chemical equation.

NaOH + HCl à HOH + NaCl

Finally, use quantitative analysis to determine the molarity of HCl.

0.025 L   x   (0.5 mol/L)   x   1/1   x   (1/0.050L)   =   0.25 M HCl

Simple!

Ex.) A 12.8 mL solution of Ba(OH)₂ (molarity: 0.75) is tested in a titration experiment. It is reacted with 10 mL samples of HCl. What is [HCl]?

The first step is to write a balanced equation.

2HCl + Ba(OH)₂ à BaCl₂ + 2HOH

In this example, we don’t need to draw a chart. This is because the volume of Ba(OH)₂ is provided for us. It’s only quantitative analysis for this round!

0.0128 L   x   (0.75 mol/L)   x   2/1   x   (1/0.010 L)   =   1.9 M HCl

That’s it! We look forward to next class’s lab! As always, here’s a titration video to wrap things up:

Posted by Michael.

February 20th - Stoichution Chemistry

Today’s class dove into examples right away. Therefore, we will do the same!

1. A 0.750 L aqueous solution contains 90.0g of ethanol, C₂H₅OH. Calculate the molar concentration of the solution.

90.0g C₂H₅OH (1 mol C₂H₅OH / 46.0 g C₂H₅OH) ( 1 / 0.750 L) = [2.61]

2. How many liters of 0.100M HCl would be required to react completely with 5.00 grams of Ca(OH)₂ in Ca(OH)₂ + 2HCl à CaCl₂ + 2H₂O?

5.00 grams Ca(OH)_{2 ­}( 1 mol Ca(OH)₂ ­­/ 74.1 g Ca(OH)₂ ) ( 2 mol HCl / 1 ) ( 1 L / 0.100 mol HCl)

            =1.35 L HCl is required

3. Using the same equation, if I added 15.0 grams of Ca(OH)₂ with 75.0ml of 0.500M HCl, how many grams of CaCl­­₂ would I form?

Remember, since two reactants are given you should find the limiting reactant.

15.0 g Ca(OH)₂ ( 1 mol /  74.1 g) = 0.202 mol Ca(OH)₂ present

0.075L (0.500 mol HCl / 1 L HCl) (1 mol Ca(OH)₂ / 2 mol HCl) = 0.01875 mol Ca(OH)₂ needed

Therefore, HCl is our limiting reactant. Now we can proceed to find the product formed.

0.075L (0.500 mol HCl / 1 L HCl) (1 mol CaCl₂ / 2 mol HCl) (111.1 g CaCl₂ / 1 mol CaCl₂)

            = 2.08 g CaCl­­₂

If you are able to solve the question we just did, then you will have no problem in this area. This is the most difficult question that could be asked at the moment. Just remember, find your limiting reactant and then proceed as usual.

Posted by Andrew.

February 14th - Solution Stoichiometry

Today, we started our new unit: solution stoichiometry! In this unit, we’ll learn how to apply our knowledge of stoichiometry to solution chemistry. We’ll do a few examples, but first, we need to review some key points.

Solutions are homogeneous mixtures composed of a solute and a solvent. If you aren’t familiar, solutes are the chemicals in lesser amounts in a reaction. In a solution, it is whatever dissolves. Solvents are the chemicals present in a great amount, or whatever does the dissolving.

Here is a picture of acetic acid, a common solvent.

As we already know, chemicals dissolved in water are aqueous. They have an _(aq) symbol (ex. NaCl_(aq)). If you want to show a reaction for something dissolved in water, you don’t include water on either side of the ‘reacts to form’ (à) symbol . Instead, it is shown by putting HOH above the à.

This equation shows Sodium chloride dissolving in water.

Concentration of a chemical in a reaction can be expressed in many ways. Grams per litre (g/L), percent by volume, percent by mass, and moles per litre (mol/L) are all acceptable. The most common is mol/L, which is called molarity. It is also important to note that mol/L is often replaced by the symbol M.

Molarity is an expression of the number of moles of the solute divided by the volume of the solvent. When you see the chemical symbol of a compound or element in square brackets, such as [HCl], it is referring to the concentration of said compound/element.

With all the notes out of the way, we can do a few examples!

Ex.) What is the concentration of NaCl_(aq) made from 1.2 mol of the chemical dissolved in 0.75 L of water?

Using quantitative analysis, we must convert 1.2 mol in .75 L to M, or mol/L. We do so like this:

1.2 mol   x   (1/.75 L)   =   1.6 mol/L, or 1.6 M

If we know the number of moles and the volume, we could simply divide them. However, it is better to set up the equation like I just did. When we get into trickier questions later in the unit, it will make things easier.

Diluted solutions would have low molarity, while concentrated ones would have high molarity.

Ex.) Determine the mass of AgNO₃ required to create 0.100 M in 250 mL of water.

We use quantitative analysis yet again. Keep in mind that 250 mL is equal to 0.250 L. Set up the equation like so:

0.250 L   x   (0.100 mol/L)   =   0.0250 mol of AgNO₃

We then simply convert moles to mass using the molar mass of AgNO₃.

0.0250 mol   x   (169.9g/mol)   =   4.25 g of AgNO₃

Ex.) How many litres of water are required to make a solution when 0.250 mol is dissolved to create a 1.75 M solution of Na₂S?

This time, we start with moles first and convert to how many litres of water we need. Simple!

0.250 mol   x   (L/1.75 mol)   =   0.143 L of HOH

The question seems daunting, but the procedure is simple. The hardest part of solution stoichiometry is determining the order in which to convert.

As always, here is a superb video on solution stoichiometry. Enjoy!

Posted by Michael.

February 10th - The Big Test

Today, we had the test. Many people did not finish, as we had very little time. Still, we look forward to receiving our marks and starting Unit 5: Solution Chemistry.

On Friday, time was of the essence.

Posted by Michael.

February 8th - Review!

Today, we did nothing but review! Ms. Z was our sub, so we had all class (& AP) to study! Yay!

Get ready for the unit test on Friday, and then:

Unit 5: Solution Chemistry

Here is a picture of an awesome saline water solution! Wow!

Posted by Michael.

February 6th - Limiting Reagent

Today we learned about limiting reagents or limiting reactants. We learned, that usually one chemical gets used up before the other. The one used up first will “limit” the reaction and the reaction will stop. By knowing the LR, we can determine the quantity of products formed. We already posses all the tools, it is just a matter of applying them!

Ex. A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. A) Which is the limiting reactant B) How much (in grams) nitrogen monoxide is formed and C) how much excess reactant remains after the reaction has stopped?

A) As we learned in class, the first step is to draw a free body diagram, as shown below:

4NH₃(g) + 5O₂(g) => 4NO(g) + 6H₂O(g)

Now we pick a reactant. I like ammonia, so we’ll use it. Figure out how much of ammonia we have and how much is required. By comparing these two values, it will be evident which reactant limits the reaction.

2.00g NH₃ x (1 mol NH₃ / 17.0 g) = 0.118 mol NH_{3 (present)}

4.00g O₂ x (1 mol O₂ / 32.0 g) x (4 mol NH₃ / 5 mol O₂) = 0.100 mol NH_{3 (needed)}

Since we have more ammonia than we need, oxygen is our limiting reactant.

B) Having determined the LR, we can now proceed with further calculations. To find the number of NO produced:

4.00g O₂ (1 mol O₂ / 32.0g) (4mol NO / 5 mol O₂) (30.0 g/ 1 mol NO) = 3.00 grams NO

C) We simply take the amount of ammonia present minus the amount needed.

0.118 mol NH₃ – 0.100 mol NH₃ = 0.018 mol NH₃ (17.0 g / 1 mol) = 0.306 g NH₃

And that’s it! So remember, when two masses of reactants are provided in the question, one of them is the limiting reactant. You’ll have to find it, unless one reactant is in excess.

To learn some more, watch the video below:

Posted by Andrew.

February 2nd - Energy & Percent Yield

Today, we were introduced to enthalpy. Enthalpy is the energy stored in chemical bonds. The symbol for enthalpy is H and the units for enthalpy are Joules (J). Change in enthalpy is represented by ∆H. The symbol delta, as you know, represents a change in something.

In exothermic reactions, enthalpy decreases. This is because energy is released, so less energy is stored in the chemical bonds. In endothermic reactions, enthalpy increases. Energy is absorbed, increasing the amount of energy in the chemical bonds.

Just look at this wicked exothermic reaction!

Calorimetry is used to experimentally determine the heat released in a reaction. To perform calorimetric calculations, we need to know three things:

• Temperature change (∆T)
• Mass (m)
• Specific Heat Capacity (C)

These are all related by the equation ∆H = mC∆T.

Here are a few examples we can do:

Ex.) Calculate the heat required to warm a cup of 400g of water (C = 4.18 J/g ^oC) from 20.0 ^oC to 50.0 ^oC.

All we have to do for this one is plug in the numbers. We start with the formula ∆H = mC∆T.

∆H = mC∆T

∆H = (400)(4.18)(50.0-20.0)

∆H = 50160 = 50.2 kJ

Easy!

We also learned about percent yield. Percent yield is the theoretical yield of a reaction. In other words, it is the amount of product that should be formed. However, the actual amount depends on the experiment. It can be calculated using:

actual (what we found) 

theoretical (what we should’ve found)

Then, multiply everything by 100 to change the decimal into a percentage of the theoretical amount.

Ex.) Determine the percent yield for the reaction between 3.74g of Na and excess O₂ if 5.34g of Na₂O₂ is recovered.

The first step, like always, is to draw a free body diagram write a balanced equation:

2Na   +   O₂   à   Na₂O₂

Then, using what we know about molar mass conversions, we change the amount of sodium we have into our theoretical amount of Na₂O₂:

3.74g   x   mol/23.0g   x   (1/2)   x   78.0g/mol   =   6.34g of Na₂O₂

We’ve just found out the theoretical yield. The actually amount was the amount given in the question. Now, to calculate the percent yield, we use our formula:

(5.34/6.34)   x   100   =   84.2%

We have 82.4% of what we should have had. This means we have quite a bit of error somewhere in our experiment!

Getting perfect percent yield is quite difficult!

So fun, right? To learn more about percent error and enthalpy, visit your local library. You can also watch these videos:

Posted by Michael.