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.