Hey Chemistry students. Only 2 weeks left of home learning. I will upload some solution stoichiometry questions for you this week. It's the same steps as the previous week, you just need to use the solution formulas instead. Also check out Kahn Academy.

I hope you are all still doing well. Teachers are back in school this week. It's lonely without the students!

This will be our last week on the topic of chemical solutions.

2 more terms you need to know:

Solute: it is the chemical that is dissolved inside the water.

Solvent: it is the liquid that does the dissolving. 95% of the time the solvent is water making it aqueous (aq). It can also be alcohol (al). Water is the universal solvent.

Word problems on dilution and concentration of solutions.

The concentration of a solution tells you how much of the chemical is diluted in water. It is typically measured in mol/L or g/L. The formula to calculate the concentration of a solution is

C = n/V if you want mol/L This is often called Molarity (mol/L of a solution). If you are asked to find the molarity of a solution, this is the formula you use. If in a question you are given a mass (g) and need to calculate molarity, you need to do the question in 2 steps

1. convert the mass into moles using molar mass of the chemical
2. then calculate the molarity

C = m/V can also be used if you want your answer in g/L. This unit is not used often but does exist.

Example 1: What would be the molarity of the solution if we took 4.80 g of sodium chloride and added 75.0 mL of water?

Molarity needs to be in mol/L. We don't' have moles. Also, the volume must be in L. 75.0 mL = 0.75 L

step 1: n=m/M,  m is the mass in the question, and M is the molar mass of sodium chloride (NaCl) found on the periodic table.

n = 4.80 g ÷ 58.44 g/mol = 0.0821355126 mol (don't round off until the very last answer)

Step 2: C = n/V = 0.0821355126 mol ÷ 0.75 L = 0.10955140168 mol/L

Answer: The molarity or concentration of the prepared solution is 0.110 mol/L (significant digits)

Dilution means that you are taking a solution and adding water to it to make it less concentrated. Many chemicals in a lab are purchased as concentrated and the lab technician prepares a diluted solution for work use. This is what I do when I prepare a lab for students. You use diluted solutions in a lab.

Another example of dilution is frozen concentrated fruit juice, lemonade for instance. You can purchase a frozen can of fruit juice and then you add a certain amount of water to make a jug of juice. You dilute the concentrated solution.

The formula associated with dilution is:

C₁V₁ = C₂V₂

C is concentration 

V is volume

The 1 and 2 represent initial and final values

Example 1: What would be the final concentration of a hydrochloric acid solution that is prepared from using a 10.0 mL sample of 2.75 mol/L concentrated solution when water is added to reach a volume of 0.50 L?

• First, units should be the same for the same variable. I will change my 0.50 L to 500 mL since the first volume is in mL. You could also do the reverse and change the L to mL. you will get the same answer in the end as long as in your formula the units cancel out.

C₁ = 2.75 mol/L

V₁ = 10.0 mL

C₂ = ?

V₂ = 500 mL

• Rearrange the formula to solve for C₂

C₂= C₁V₁ ÷ V₂

C₂ = 2.75 mol/L  x 10.0 mL ÷ 500 mL = 0.055 mol/L

Answer: The final concentration of the hydrochloric solution would be 0.055 mol/L.

Example 2: How much of a concentrated 3.85 mol/L sodium hydroxide solution is needed to make 1.0 L of 0.5 mol/L solution?

• This question is asking to solve for V₁

V₁ = C₂V₂ ÷ C₁

V₁= 0.5 mol/L x 1.0 L ÷ 3.85 mol/L = 0.1289 L

• In a practical circumstance of a lab setting, you would not have a graduated cylinder that can measure 0.1289 L. You would need to convert this to mL to measure out.

0.1289 L = 128.9 mL

Answer: The initial amount of concentrated sodium hydroxide solution needed is 128.9 mL.

*****See Solution Unit Review sheet in the document section *******

Hi everyone. I placed for the past 2 weeks information on properties of gases. For the next 2 weeks you will learn about solutions. I assigned one video from Khan academy for you to watch this week. I also uploaded some power point slides for you to look at on here. I need to be at the school to scan some data charts (solubility, flame and colour) for you to be able to do any word problems. For now, just take a look at the video and notes. I will post again mid-week with more info for you to work on. This is a section that you may need more explanation from me. I will most likely video a lab demonstration from school if I can. Skype sessions are available if you are interested.

Yes you can take this course as a credit.

This week we will focus on the mole. The mole is a unit of measurement that is probably the most used in chemistry. I always compare the explanation of the mole with a dozen.  Here goes:

One dozen = 12 items

It doesn't matter if you have a dozen eggs, a dozen cars or a dozen feathers. All are equal to 12 items. The masses of each dozen of eggs, cars, feathers are obviously different, the sizes are also different, however, they are all equal to 12 in numbers.

One mole = 6.02 x 10²³ particles

So if you have one mole of hydrogen, one mole of potassium or one mole of iron, they are all equal to 6.02 x 1023 particles. The masses are different, the sizes of the molecules are different, but the number of particles are all equal to 6.02 x 1023 particles.

This number is referred to as Avogadro's number, named after the Italian scientist who developed this concept.

In a balanced chemical reaction, the number placed in front of the chemicals, what we call the coefficient, represents the moles of that chemical. For example:

     Cl₂ + 2 NaBr -->  Br₂ +  2 NaCl 

This is read as 1 mole of chlorine reacts with 2 moles of sodium bromide to produce 1 mole of bromine and 2 moles of sodium chloride.

Calculating the number of particles based on the amount of moles:

To calculate the number of particles of a substance, we simply multiply Avogadro's number by the number of moles.

        # particles = moles x Avogadro

Example: How many particles in 3 moles of potassium chloride?

        # particles = 3 moles x 6.02 x 10²³ particles/mole = 1.806 x 10²⁴ particles

**** Every calculator brand is different for exponents. Learn how to use yours properly. If you can't get the same answer, then please ask.****

The formula can be rearranged to find moles instead. For example: How many moles does 1.65 x 10²⁴ particles of helium represent?

        moles = # particles / Avogadro = 1.65 x 1024 particles / 6.02 x 10²³ = 2.74 moles of helium

We use the mole in most chemical quantities. If you look at your periodic table, the units for the mass indicated for each element (molar mass) is in g/mol (grams per mole). This is just one simple example of how the mole is used. As you progress in chemistry, you will see it almost everywhere.

Calculating molar mass of a compound:  Molar mass is the mass of one mole of a substance. You find the value of individual elements on the periodic table. For example molar mass of hydrogen is 1.01 g/mol; aluminum is 26.98 g/mol; radon is 226.03 g/mol. TO calculate the molar mass of a compound, you just have to add up all the individual masses of the elements inside the compound. For example:

    H₂O  contains 2 hydrogen and 1 oxygen.  The molar mass is (2 x mass of hydrogen) + (1 x mass of oxygen)= (2 x 1.01 g/mol) + (1 x 16.00 g/mol) = 18.02 g/mol.  This means that one mole of water has a mass of 18.02 grams.

    H₂SO₄ contains 2 hydrogen, 1 sulfur, 4 oxygen. The molar mass is (2 x 1.01 g/mol) + (1 X 32.06 g/mol) + (4 x 16.00 g/mol) = 98.08 g/mol. This means that one mole of hygrogen sulphate has a mess of 98.08 grams.

Molar mass in a mathematical formula is represented with a capital M and is always measured in g/mol.

Mass in a mathematical formula is represented with a lower case m and is measured in g, Kg, mg, etc...

Moles in a mathematical formula is represented with a lower case n and is measured in mol, Kmol, mmol, etc...

Using a formula to calculate mass or moles based on molar mass:

                                     n=m/M  

moles = mass divided by molar mass

This formula can be rearranged to find mass is moles is given in the question. m=nM (mass= moles x molar mass)

Molar mass (M) is found on the periodic table. Just add up the elements within the compound as explained earlier.

Example 1:  How many moles is 78.23 g of sodium chloride?

First, calculate the molar mass (M) of NaCl from the periodic table by adding Na + Cl = 58.44 g/mol

78.23 g is the mass (m)

n= m/M = 78.23g ÷ 58.44 g/mol = 1.339 mol

Example 2: What is the mass of 1.5 mol of H₂0?

from the periodic table, M of H₂O = 18.02 g/mol

m= nM = 1.5 mol x 18.02 g/mol = 27.03 g

Example 3: What is the mass of 3.8 mol of oxygen?

** Here you must remember that an oxygen molecule never has only one oxygen. It is O₂.

m = nM = 3.8 mol x 32.00 g/mol = 121.6 g

Using STP and SATP values of gases to calculate moles: We have seen before that STP stands for Standard Temperature & Pressure of a gas; SATP stands for Standard Ambient Temperature & Pressure.

At STP (0°C and 101.325 kPa) the volume of a gas is 22.4 L/mol.

At SATP (25°C and 100 kPa) the volume of a gas is 24.8 L/mol.

The unit L/mol (liters per mole) represents how much space a gas takes up for each mole of that gas. Imagine the size of a 2 L bottle of pop as a reference when trying to visualize the size of these gas containers. The values mentioned above are also found on the back of your periodic table (bottom righthand side). They can be used to find either the volume or the amount of moles of a gas. For the sake of this section, we are going to refer the STP and SATP values as concentrations (how much gas is in a certain contained area).

   c= v/n  (concentration = volume ÷ mole)

The formula is rearranged as follows:

  v = cn (volume = concentration x moles)

  n = v/c (moles = volume ÷ concentraion)

For this type of question, C is measured in L/mol; v is measured in L, kL, mL, etc...; n is measured in mol, kmol, mmol, etc...

Example 1: What is the volume 0.68 mol of neon gas at STP?

Look for the STP value of a gas for c

    v = cn = 22.4 L/mol x 0.68 mol = 15.2 L of neon gas

Example 2: How many moles does 5.63 L of hydrogen gas represent at SATP?

Look for the SATP value of a gas fo c

    n= v/c = 5.63 L ÷ 24.8 L/mol = 139.6 mol of hydrogen gas

Multistep calculations:  You may often need to use more than one formula to get to your final answer. You need to be able to know which formula to use for which problem.

Example 1: What is the mass of 3.49 L of argon gas at SATP?

I didn't give you a formula with mass and volume in it. However, you can first find moles and then convert that to a mass. 

step 1: n= v/c = 3.49 L ÷ 24.8 L/mol = 0.140725806 mol (never round off until the very end of your calculation. This is EXTREMELY important)

step 2: m= nM = 0.140725806 mol x 39.95 g/mol = 5.62 g of argon gas.

Hint:  Placing your units in your work will be SUPER useful in the future when questions become more complicated. It helps you figure out what the final unit will be and also helps you figure out if you are using the right formula. Units should cancel out in your calculations to leave you with just one in the end.