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

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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?

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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

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Questions?

Next time:

Organic chemistry?

Homework #2:

Due Wednesday, July 20