chapter 11 stoichiometry 11.2 percent yield and concentration

CHAPTER 11

Stoichiometry

11.2 Percent Yield and Concentration

2 11.2 Percent Yield and Concentration

In theory, all 100 kernels should have popped.

Did you do something wrong?

+

100 kernels 82 popped 18 unpopped


+


In real life (and in the lab) things are often not perfect

In theory, all 100 kernels should have popped.

Did you do something wrong?No


+


82100 82%

100

100amount of corn popped

percent yieldamount of kern

percent yie

els in the bag

ld

What you get to eat!

Percent yield


+


100

82100

10082%

amount of corn poppedpercent yield

amount of kernels in the bag

percent yield

What you get to eat!

Percent yield


100actual yield

percent yieldtheoretical yield

100actual yield

theoreticalpercent yi

yieldeld

actual yield: the amount obtained in the lab in an actual experiment.

theoretical yield: the expected amount produced if everything reacted completely.


Percent yield in the lab

Decomposition of baking soda:

2NaHCO3(s) → Na2CO3(s) + H2O(l) + CO2(g)

Heating





10.00 g 4.87 g

measured experimentally

Can you think of reasons why the final mass of Na2CO3 may not be accurate? (What could be sources of error?)





10.00 g 4.87 g


- There is usually some human error, like not measuring exact amounts carefully






10.00 g 4.87 g



- Maybe the heating time was not long enough; not all the Na2HCO3 reacted






10.00 g 4.87 g




- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too






10.00 g 4.87 g




- Maybe Na2CO3 was not completely dry; some H2O(l) was measured too

- CO2 is a gas and does not get measured






10.00 g 4.87 g


Let’s calculate the percent yield

100actual yield


obtained in experiment

calculated





10.00 g 4.87 g



4.810

70percent yield

theoretical y eld

g

i

calculated





10.00 g 4.87 g



4.87100

theoretica

gpercen

l yieldt yield

calculated



10.00 g

Use the mass of reactant NaHCO3(s) to calculate the mass of the product Na2CO3(s).

This is a gram-to-gram conversion:



10.00 g

10.00 g



10.00 g

10.00 g

33 3

3

3

22.99 1.0079 12.011

110.00 0.1

84.01

19084.0

3(15.999

1

/)molar mass of NaHCO g mo

mole NaHCOg NaHCO moles NaHCO

g NaHCO

le



10.00 g

33

33

3 22.99 1.0079 12.011

0.1

3(15.999) 84.01 /

110.00 1 0

4.019

8

molar mass of NaHCO g mo

mole NaHCOg NaHCO

g NaHCOmole

l

s

e

NaHCO

10.00 g 0.1190 moles



10.00 g

0.1190 moles

2 3

33 30

1

2.1190 0.05950

mole Na CO

mmoles N

oles NaaHCO moles Na

HCOHCO



10.00 g

2 333

3

0.0591

0.11902

50mole Na CO

moles NaHCOm

mooles NaHC

les N OO

aHC

0.1190 moles 0.05950 moles



10.00 g

0.05950 moles

2 3

2 32 3 2 3

2 3

105.990.

22.99 2 12.011 15.999 3

05950 6.3061

105.99 /

g Na COmoles Na CO g Na CO

mole Na CO

molar mass of Na CO g mole



10.00 g

2 3

2 32 33

32

2

105.990.

22.99 2 12.011 15.999 3

059501

6.3

105.9

0

9 /

6g Na CO

moles Na COmole Na

molar mass of N

g Na CO

a CO g mo

O

e

C

l

0.05950 moles 6.306 g



10.00 g

0.05950 moles 6.306 g10.00 g 0.1190 moles

4.87 g

For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g.

The actual yield (measured) is 4.87 g.



10.00 g

For 10.00 g of starting material (NaHCO3), the theoretical yield for Na2CO3 is 6.306 g.

The actual yield (measured) is 4.87 g.

4.87 g

100

4.87100 77.2%

6.306

actual yieldpercent yield

theoretical yield

gpercent yield

g




10.00 g

Reaction of solid zinc with hydrochloric acid:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)

50.0 mL of a 3.0 M solution

Convert to moles

Convert to moles

Reactions in solution

Stoichiometry with solutions


A sample of zinc metal (Zn) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl) according to: Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq). How many grams of hydrogen gas (H2) will be produced? Assume zinc metal is present in excess.



Asked: grams of H2 produced

Given: 50.0 mL of 3.0 M HCl reactingwith excess zinc

Relationships: M = mole/LMole ratio: 2 moles HCl ~ 1 mole H2

Molar mass of H2 = 1.0079 x 2= 2.02 g/mole



Solve:





22

22 2

2

3.0 0.0

10.150 0.075

500 0.150

02

2.020.0750 0.15

1

mol

molemoles of HCl HCl m

e Hmoles HCl moles H

moles HCl

g Hmoles H g H

mole

L moles HCl

H

L



Solve:





2 2

22

2

2

3.0 0.0

10.150 0.075

500 0.150

2.020.0750 0 5

0

1

2

.1

mol

molemoles of HCl HCl m

g Hmoles H

e H

g

moles HCl moles Hmoles HCl

L moles

Hmol

H l

e H

CL



Solve:





22

22

22

3.0 0.0500 0.150

10.150 0.0750

0.2.02

0.07501

2

15

molemoles of HCl HCl mL moles HCl

Lmol

g Hmole

e

g H

Hmoles HCl moles H

mol

s Hmo

HCl

le H

es



Solve:





22

22 2

2

3.0 0.0500 0.150

10.150 0.0750

2

2.020.0750 0.15

1

molemoles of HCl HCl mL moles HCl

Lmole H

moles HCl moles Hmoles HCl

g Hmoles H g H

mole H

Answer: 0.15 grams of H2 are produced


Reaction of solid zinc with hydrochloric acid:

Zn(s) + 2HCl(aq) → H2(g) + ZnCl2(aq)

50.0 mL of a 3.0 M solution

Convert molarity to moles

Vinegar is 5% acetic acid by mass

Sometimes the concentration is written in mass percent

% 100mass of compound

mass of compoundtotal mass of solution


Commercial vinegar is reported to be 5% acetic acid (C2H4O2) by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (Assume the density of vinegar is the same as pure water, 1.0 g/mL.)



Asked: grams of acetic acid in 120 mL of vinegar

Given: 120 mL of vinegar and 5% acetic acid by mass

Relationships: 120 mL = 120 g, given a density of 1.0 g/mL

% 100mass of acetic acid

massmass of solution






Solve:



%

100

5 %

100 120

mass mass of acetic acid

mass of solution

mass of acetic acid

g






Solve:



%

100

5 %

100 120


mass of solution

mass of acetic acid

g

0.05120

0.05 120

6.0

mass of acetic acid

g

mass of acetic

mass of acetic acid

acid

g

g






Solve:



%

100

5 %

100 120


mass of solution

mass of acetic acid

g

Answer: 6.0 g of acetic acid.

0

0.05

.05

6

12

2

.

0

1

0

0

mass of acetic acid g

mass o

mass of acetic ac

f acetic acid

id

g

g


100actual yield


100actual yield

theoreticalpercent yi

yieldeld

% 100mass of compound

mass of compoundtotal mass of solution

Calculate using molar masses and mole ratios

Obtained from the experiment

chapter 11 stoichiometry 11.2 percent yield and concentration

Documents

concentrationpercent

theoretical yield

concentration actual

sources of error

human error

final mass of na2co3

heating time

gmeasured experimentallylets