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Stoichiometry & Solutions
Academic Boot Camp
Curtis P. Martin
July 18, 2016
Reading
Silberberg: §3.4, 3.5.
___Cu2O + ___C → ___Cu + ___CO
___Cu2S + ___O2 → ___Cu2O + ___SO2
Many times, multiple reactions will occur in sequence
Stoichiometry: Multiple Reactions
1 3 22
___Cu2O + ___C → ___Cu + ___CO
___Cu2S + ___O2 → ___Cu2O + ___SO2
Many times, multiple reactions will occur in sequence
Stoichiometry: Multiple Reactions
1 3 22
___Cu2O + ___C → ___Cu + ___CO
___Cu2S + ___O2 → ___Cu2O + ___SO2
Many times, multiple reactions will occur in sequence
Stoichiometry: Multiple Reactions
1 3 22
__________________________________________________________________
We have assumed thus far that we have everything in excess
What if, in real life, we only have 2 mol F2 per 1 mol Cl2?
Limiting Reagents
___Cl2(g) + ___F2(g) → ___ClF3(g)1 3 2
Example: we have 1 mol ~ 70 g Cl2 and 2 mol ~ 76 g F2. How much ClF3?
Limiting Reagents
___Cl2(g) + ___F2(g) → ___ClF3(g)1 3 2
2 mol F2 ∗1mol Cl2
3 mol F2= _____________ actually used in reaction
2 mol F2 ∗2mol ClF3
3mol F2= ______________ actually created in reaction
Example: we have 1 mol ~ 70 g Cl2 and 2 mol ~ 76 g F2. How much ClF3?
Finding the limiting reagent: Find the amount of moles of each reactant
Compare _______________________________________________________________
Limiting Reagents
___Cl2(g) + ___F2(g) → ___ClF3(g)1 3 2
2 mol F2 ∗1mol Cl2
3 mol F2=
2
3mol Cl2 actually used in reaction
2 mol F2 ∗2mol ClF3
3mol F2=
4
3mol ClF3 actually created in reaction
Rarely does a chemist get the amount of product expected. Why?
Theoretical v. Actual Yield: Introduction
Rarely does a chemist get the amount of product expected. Why?
__________________________________________________
__________________________________________________
__________________________________________________
Theoretical v. Actual Yield: Introduction
_____________ yield:
Amount of substance you would get assuming 100% conversion
100% conversion of limiting reactant, if present
_____________ yield:
Amount of substance you get in an actual experiment
_____________ yield: Actual yield
Theoretical yield∗ 100
Percent Yield
1. Balance reaction equation stoichiometry
2. Determine limiting reagent (or if one even exists)
3. Use limiting reagent as basis for calculations learned earlier to find theoretical yield (otherwise just calculate them as normal)
4. Find percent yield from theoretical and actual
Algorithm for Solving Yield Problems
We have 1 mol ~ 70 g Cl2 and 2 mol ~ 76 g F2. Also, we made 100 g ClF3.
Percent Yield: Example
___Cl2(g) + ___F2(g) → ___ClF3(g)1 3 2
2 mol F2 ∗1mol Cl2
3 mol F2=
2
3mol Cl2 theoretically used in reaction
2 mol F2 ∗2mol ClF3
3mol F2=
4
3mol ClF3 theoretically created in reaction
4
3mol ClF3 = ~123 g ClF3 ∴ Percent yield =
100 g
123 g∗ 100 = 81.3%
Solutions are fundamental to quantitative chemistry
_________________: property of a solution indicating its chemical makeup
Amount of solute dissolved into solvent
[=] mass/volume or moles/volume
Unit we will use: Molarity (M)
Moles / liter
Solution Chemistry
Solutions: homogeneous liquid mixture
____________: what dissolves in solvent
Can be solid, liquid, or gas initially
____________: medium by which the solute disperses
Solutions
Usually have a particular concentration in mind
“Strength” of solution
Determine total volume you want
Concentration (molarity) =mol solute
L solvent
Preparing Solutions
Usually have a particular concentration in mind
“Strength” of solution
Determine total volume you want
Concentration (molarity) =mol solute
L solvent
Amount of solute (mol) = conc.∗ vol. Convert moles of solute to mass if using solids or volume if
using liquids
Preparing Solutions
Want: 1L of 4M solution of HCl
Have: 90% (by mass) HCl powder
Preparing Solutions: Example
Want: 1L of 4M solution of HCl
Have: 90% (by mass) HCl powder
1L soln ∗4mol HCl
L soln∗
35.45+1 g HCl
mol HCl∗1gHCl powder
0.9g HCl=
162g HCl powder
Put 162g 90% HCl powder in 1L solvent
Preparing Solutions: Example
M1V1 = M2V2 M = molarity
V = volume
Diluting Solutions
M1V1 = M2V2 M = molarity
V = volume
Want: 1L of 4M solution of HCl
Dilute
Have: 18M HCl solution Concentrated
Diluting Solutions: Example
M1V1 = M2V2 M = molarity
V = volume
Want: 1L of 4M solution of HCl
Dilute
Have: 18M HCl solution Concentrated
V2 =M1V1
M2=
4M∗1L
18M= 0.22 L
Diluting Solutions: Example
___________________
Questions?
Next time:
Organic chemistry?
Homework #2:
Due Wednesday, July 20