Journal Reflection 5: Stoichiometry

Moles and Molar Mass

The amount of substance is measured amount of the number (n) of any given particle (atoms, ions, molecules) in a solution. This value is also known as the mole (mol) of a substance. The mole is a standard unit of measurement for the amount of a substance. A mole of a substance is equal to Avogadro's Number, which is approximately 6.022 x 10 ^23 particles. Scientists use moles as a way to measure the amount of particles in any substance and measure the molar mass of substances.

Molar Mass: The molar mass is the mass of a mole of any given substance (g/mol). It is calculated by adding all the atomic masses of a substance. For single elements, the molar mass is equal to the atomic mass of the element on the periodic table. Listed below are examples of molar mass calculations:

Covalent Compounds:

• CH4 molar mass = 12.011 + (1.0078 x 4) = 16.0422g/mol

• NH3 molar mass = 14.007 + (1.0078 x 3) = 17.0304g/mol

Ionic Compounds:

• NaCl molar mass = 22.990 + 35.453 = 58.443g/mol

• MgO molar mass = 24.305 + 15.999 = 40.304g/mol

The mass of a substance is the measurement of the amount of matter contained, usually measured in grams or kilograms. The mole of a substance is the measurement of the amount of particles in a substance relative to Avogadro's number (6.022 x 10 ^23 particles). The number of atoms or molecules in a solution is measured by comparing the mole of a substance to Avogadro's number, with one mole being equivalent to 6.022 x 10 ^23 atoms or molecules. 

Calculations for Comparison of Values: 

Initial Mass of a sample of NaCl: 9.07g  (this is the mass of a substance, measure in grams)

Molar Mass of NaCl: 22.990 + 35.453 = 58.443g/mol

Mole Calculation for NaCl sample: 9.07g / 58.443g = 0.155mol (this is the mole of the substance in moles)

# of Particles in 0.155mol: 0.155 x (6.022 x 10 ^23) = 9.3341 x 10 ^22 particles (this is the number of molecules in our sample of NaCl)

Limiting Reactants 

Real-world Situation Application of Limiting Reactant: A tricycle factory gets a shipment with 400 seats and 600 wheels.

Applying my knowledge of limiting reactant, I first had to figure out how many of each reactant I would need to make one product. I would require three wheels for each seat (3:1 ratio). I decided to create a BCA table to illustrate my calculations and conclusion.

BCA Table:

1 St + 3 W > 1 StW3

Before 400 seats 600 wheels 0 StW3

Change -1(200) seats -3(200) wheels +1(200) StW3

After 200 seats 0 wheels 200 tricycles!

Recipe Factors :  400/1 = 400 R.F

  600/3 = 200 R.F (this will be our lowest recipe factor, so it will be used for our BCA table, this makes the wheels the limiting reactant)

Conclusion: After determining the lowest recipe factor of the two reactants, we found out that the wheels would be the limiting reactant of this reaction, being the first to be fully expended. Applying the recipe factor to our BCA table, we were left with 200 seats that weren't utilized in our reaction, and with a final total of 200 tricycles produced. if we had had more wheels in our reactants, we could have had greater efficiency in the production of tricycles.

Application of Limiting Reactants to Chemistry:

2 H2 + 1 O2 > 2H20

The Law of Definite Proportions states that a given chemical compound always contains the same elements in the exact same proportions by mass (Libretexts).

Example: any sample of pure water contains 11.9% hydrogen and 88.81% oxygen by mass. 

It does not matter where the sample of water came from or how it was prepared. Its composition, like that of every other compound, is fixed.

The reason why the molecule simulation does not allow for changes in the ratio of products like the sandwich simulation does is because the given compounds, such as O2 and H2, will always contain the same elements in the same proportion of mass. There will never be another variation of the given product compounds that contains other elements and/or proportions of mass within the compound.

References:

Libretexts. (2025, March 21). 4.4: Law of definite proportions. Chemistry LibreTexts. https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/04%3A_Atomic_Structure/4.04%3A_Law_of_Definite_Proportions

Which reactant amounts would get the most water with the least amount of leftovers?

Prediction: Using my knowledge on limiting reactants and the law of definite proportions, I predict that the reaction with five H2 and three O2 will have the least leftover. Since the ratio for water is 2H:1O, with our 6 oxygen molecules, we could get max of 6 water molecules, but since oxygen is the limiting reactant and runs out first, we can only make 4 water molecules with a leftover 02 and H2 molecules. If there was just one more H2, we could have a reaction with 100% water molecule formation efficiency.

Game Reflection: Playing the game built into the simulation has allowed me to better visualize limiting reactants. Whenever I would do limiting reactant problems, I would usually solve them by using math and ratios to find out which reactant ran out first and what would be leftover in the product. Doing these puzzles showed me that limiting reactant problems can also be solved efficiently with particle diagrams. This was also my first time doing one of the problems in reverse, where instead of finding the leftovers of the reactants, I was finding the reactants of the leftovers.

Chemistry Café:

The Chemistry Café was out of bread. The cook went next door to the bakery and bought a loaf of bread which has 33 slices. Then, she sells 12 sandwiches (two pieces of bread per sandwich). How much bread did she have left?

Answer: If 12 sandwiches are made, that means that 24 slices of bread were used. That would leave use with 9 slices of bread that were not utilized.

The Chemistry Café cook has a loaf which has 33 slices and a package of cheese that has 15 pieces. She is making sandwiches that have 2 pieces of both bread and cheese. How many sandwiches can she make?

Answer: For this question, I decided to utilize a BCA table

2 Bread + 1 Cheese > 1 Sandwich

Before 33 Bread 15 Cheese 0 Sandwich

Change -2(15) -1(15) +1(15)

After 3 Bread 0 Cheese 15 Sandwich

Recipe Factor: 15/1 = 15 cheese R.F (lowest R.F, cheese is limiting reactant)

    33/2 = 16.5 bread R.F

BCA Answer: 15 Sandwiches can be made

Final Reflection: We are close to the end of the semester. Please help me describe to future students the kind of person that will be most likely to succeed in this course. 

I believe that the most succesful students of this course will be students that are willing to learn new topics and concepts and dont shy away from challenging themselves. A student needs to want to actually understand the material and desire the ability to apply newfound knowledge. Most importantly of all, I believe the most successful students will be students with a curiosity for the mysteries of how the universe works.