Hi everyone. Last week of home learning! I know for some this hasn't been easy. Learning on your own is a huge challenge. I really hope next year I can have you in the same room as me, to actually do the teaching/learning the way most students need it to be. We can also do some labs!

I'm attaching one last assignment for you to try out. It's actually an end of unit test on stoichiometry that I gave in the past. Have a good week! 

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 is the last unit of Chemistry 112. It focuses on quantitative analysis and typically this unit is used to strengthen your scientific concepts and laboratory skills/techniques. It is unfortunate that you will not be able to explore the lab this year. To be succesful in this unit, you must take everything you learned in the previous units an apply be able to apply it in a laboratory setting. The questions you will be working on are application based. This week will focus on gravimetric stoichiometry.

Excess reagent:  the reactant that is left over after the chemical reaction has taken place.

           *Use molar mass to calculate mass of the substance asked in the question.

See Khan Academy assignment to read an article about stoichiometry and watch 3 videos of exmple questions. I have placed 10 questions in the document section for you to work on.

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 *******