Chapter 6 + 7: READ OVER QUIZZES + LECTURE NOTES Flashcards
How do we calculate the “Formula Mass” (GMM) of a Compound?
To calculate the “formula mass of a compound in grams per mole” we simply take the Atomic Masses of each Element in that compound times the number of times that element appears. Do this for each element in the formula then add up the contributions from each element. (Get the Atomic Masses from the Periodic Table or an Alphabetical list of Elements and their masses)
EX: What is the GMM of table sugar, the compound “Sucrose”, C12H22O11 ?
Solve by making a chart:
Carbon = 12.01 g/ml x12 atoms = 144.12 g /mol
Hydrogen = 1.008 g/mol x 22 atoms = 22.176 g /mol
Oxygen = 16.000 g/mol x 11 atoms = 176.000 g/mol
Total = 342.296 g/ mole = GMM of C12H22O11
How to solve for empirical formula?
We need to do two things- put all the data in table form for ease of analysis and do two simple math tricks along the way.
- The first trick is to convert the percentages directly into mass in grams.
- From there we will convert masses of the elements into their corresponding moles (divide mass in grams by GMM of corresponding element)
- The second trick is to divide all of the resulting mole values by the smallest number of moles.
- Then round off to the nearest integer values for the empirical Formula.
what is the difference between Empirical formula for a compound and Molecular Formula? Can they be the same?
Molecular Formula is the actual formula for the compound. Empirical formula is the simplest ratio of all the elements in that compound.
How to find the relationship between Molecular Formula and Empirical Formula?
Divide the GMM of the molecule from our “other” experiment by the GMM of our Empirical Formula and we should get an integer (or really close). Take that integer and multiply it by our Empirical formula and we get the Molecular Formula.
EX: Acetone has a GMM of approximately 60 g/mole, Empirical formula we found = C3H6O. The formula Mass (GMM) for three carbons, six hydrogens and one oxygen is 58.08 g/mole
- Molecular Formula / Empirical formula = (60 g/mol)/ (58 g/ mol) ≈1
What is molarity?
- The Number of Moles of Solute per Liter of solution (final volume).
- The symbol for Molarity is “M” = moles solute/ Liter.
Molarity “formulas”
1) Concentration (M) = moles/liter
2) Concentration (mol/L) x volume (L) = #moles
3) # grams of solute = moles of solute X GMM
4) Volume needed (L) = (# moles solute needed)/ (“M” as moles/L)
What is dilution?
Dilution means adding more solvent to an already existing solution
What is mixing?
Mixing is a related phenomenon – adding more of both solute and solvent to an existing solution
Dilution problem: A) We start with 250 ml of a .500M salt solution (say KBr) and add 100 ml of solvent (water) to it. What is the Final Concentration of the salt solution?
we already know the concentration and the volume of the solution we start with so we know the number of moles of Solute. (moles/ liter x liters = # moles). And we also know the amount of solvent we add to it so we know the Final Volume of the solution (= 250 ml + 100 ml = 350 ml = .350 Liters).
So moles of solute divided by final volume in Liters = New Concentration
= (.500 mol/L x .250L) divided by .350 L= .125 moles / .350L = .357 Mol/L final conc.
Mixing problem: If we mix 200 ml of .400M NaCl with 350 ml of .100M NaCl what is the Concentration of the resulting solution?
we have .400Mol/L x .200L = .0800 moles (NaCl) Plus .100 mol/L x .350 L = .0350 moles NaCl
Final Concentration = Total moles/ Final volume
= (.0800moles + .0350moles)/ (.200L + .350 L) = (.115 moles NaCl/ .550L) = .209 M Final conc.
Dilution of stock solution problem: We have a large quantity of 5.00M HCl and we want to make 500 ml of .250M HCl. How much of the concentrated stock do we need to take?
The equation we use to solve all of these dilution problems is: C1V1 = C2V2
On the left side concentration times volume = the number of moles of solute we need to take from the Stock solution.
This is the number of moles of solute we will have in our “working” solution represented by concentration x volume on the right side.
So “moles of stock = moles of working”- The volumes and concentrations are different on each side.
What do we need to solve for in this equation? We know the concentrations for the “stock” and “working” solutions (C1 and C2) and we know how much of the “working” solution we need (V2).
So the only thing we need to solve for is V1- the volume of Stock solution we take. So rearrange to solve for V1 = (C2/ C1) x V2.
So here: V1 = (.250M/ 5.00M) x 250 ml = 12.5 ml of V1
How do you balance a chemical reaction?
EX: C4H10 (g) + O2 (g) -> CO2 (g) + H2O (g)
To balance this equation first find and balance the elements that start in one compound and end up in only one compound.
Here these would be Carbon and Hydrogen. When we do this we produce 4 CO2 and 5 H2O. We still don’t know how much Oxygen to use- so we work back words from the Oxygen
we find in the products to the O2 needed =13/2 O2.
- double everything to get rid of fractions
Balanced equation:
2 C4H10 (g) + 13 O2 (g) -> 8 CO2 (g) + 10 H2O (g)
How to find theoretical yield?
Step #1- Find moles that reacted
Step #2 Determine the “Limiting Reagent”: Divide the number of moles of each component by its stoichiometric ratio – The lower resulting number is the “Limiting Reagent”- This runs out first- (nearly) all other calculations are based on the moles of “Limiting Reagent”.
- “reaction ratio”: mole reaction ratio
Step #3: use limiting reagent to find theoretical yield. Convert limiting reagent to grams/mole of substance you’re trying to find the theoretical yield of
- make sure to use moles you found BEFORE finding limiting reagent (look at test #2 if you still don’t understand this mistake dummy)
How to find how much excess reactant is left over after reaction?
1) Find the moles of reactant that actually reacted (convert limiting reagent to moles of excess reactant).
2) Subtract reacted moles from the starting moles. This = the “unreacted moles” left over.
3) If we want we can find how many grams the “unreacted” moles represents by multiplying by molar mass of the element.
What is the stoichiometry “roadmap”?
grams “A” -> moles “A” -> moles “B” -> grams “B”
