CH
3
COOH + NaHCO
3
NaCH
3
COO + CO
2
+ H
2

O
PPT Init: 1/24/2011 by Daniel R. Barnes
WARNING: As with all my power points, this one contains graphical images and/or other content taken from the world wide web without the permission of the owners of that intellectual property.
Please do not copy or distribute this presentation. Its very existence may be illegal.

SWBAT . . .
. . . predict how many grams of carbon dioxide should be produced from a given number of grams of baking soda.
. . . use the idea of “limiting reactant” to explain the amounts of carbon dioxide produced by various amounts of baking soda reacting with a fixed amount of vinegar.

CH
3
COOH + NaHCO
3 acetic acid sodium hydrogen carbonate
 NaCH
3
COO + CO sodium
2 carbon
+ H
2
O water acetate dioxide

CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
H
H
C H
C
O O
H
Na
O
C
H
O
O
H
H
C H
C
O O
Na
O
C
O
H
O
H

CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
H
H
C H
C
O O
H
Na
O
C
H
O
O

VOCABULARY FRONT-LOAD
“tare mass” = mass of empty container
“net mass” = mass of what’s in the container
“gross mass” = mass of container + contents gross = net + tare gross = contents + container net = gross - tare

Stabby
Flakes
Stabby
Flakes
NET WT 3 LB
Empty box
= 0.25 pounds
Cereal
= 3 pounds
NET WT 3 LB
Box full of cereal
= 3.25 pounds

Stabby
Flakes
Stabby
Flakes
NET WT 3 LB
Empty box
= 0.25 pounds
Cereal
= 3 pounds
NET WT 3 LB
Box full of cereal
= 3.25 pounds

1. Experimentally determine how many grams of carbon dioxide can be produced by reacting baking soda with 100 grams of vinegar.
2. From this, use stoichiometry to determine how many grams of acetic acid is in 100 grams of vinegar.
Your period will get a bonus to everyone’s grade if you can share your data in such a way that you can produce a graph that shows grams of carbon dioxide produced for a variety of different amounts of baking soda.
Each student will be expected to write his/her own freehand lab report organized into the following sections: purpose, materials, procedure, data (including graphs), and conclusions (including math).

50 mL beaker is ONLY for baking soda
100 mL beaker is ONLY for baking soda
250 mL beaker is ONLY for vinegar
600 mL beaker has no special rules – but DO clean it @ day.
PROCEDURE is what you do with the physical objects, materials, and measuring devices. It may include a small amount of math, but the CONCLUSIONS section of the report is where the big math is.
Your group must have figured out its procedure by halfway through the period on Monday. I suggest conferring over the weekend via social media/e-mail/phone calls/whatever.

PURPOSE: your mission statement (? g CO2 producible w/100 g vinegar)
MATERIALS: chemicals, equipment
PROCEDURE: what you did with the chemicals & equipment
+ gross/net/tare math needed to do procedure
DATA: the information you gathered during the procedure
CONCLUSIONS: calculations based on data gathered (stoich for theo yield), graphs, discussion of various things such as difficulties in execution/technique errors/what you’d do differently next time, interpretation of graph (esp re: limiting & theoretical yields of CO2), . . .

NOTE: The scale pictured here is a digital scale, but, as of this writing, the lab worksheet says you’re supposed to use a triple beam balance. Sorry. TBB’s are too hard to draw. Bear with me.
When you have the 600 mL beaker and the 50 mL beaker on the scale and they’re both empty, you’ve got a certain amount of glass on the scale.
Mass of emtpy container(s)
= “tare” mass
251.4
g

“net” mass mass should be exactly
100 g higher than the
“tare” mass.
g

“net” mass
If you were to put exactly five grams of baking soda in the 50 mL beaker, that should bring the new gross mass up to . . .
. . . 356.4 g g

356.4
g
If you dump the baking soda into the vinegar . . .

356.4
g
If you dump the baking soda into the vinegar . . .
The chemicals react to form bubbles of carbon dioxide

CO
2
CO
CO
2
2 g
When the bubbles rose and popped, carbon dioxide escaped from the 600 mL beaker and into the air.
Therefore, the gross mass went . . .
. . . down, in this case by
2.6 grams.

-
356.4 g = gross mass before the reaction
353.8 g = gross mass after the reaction
2.6 g = amount of CO
2 produced
We assume that however much the mass went down
. . .
. . . is the amount of CO2 produced.
353.8
g

-
356.4 g = gross mass before the reaction
353.8 g = gross mass after the reaction
2.6 g = amount of CO
2 produced
353.8
g
The law of conservation of matter says that . . .
. . . matter can’t be destroyed during a chemical reaction.
The drop in mass doesn’t represent matter that was destroyed. It represents matter that went into the air.

“that portion of chemistry dealing with numerical relationships in chemical reactions; the calculation of quantities of substances invovled in chemical reactions.”
~Prentice Hall Chemistry, glossary, pg R117
That’s what your textbook says, but in this class,
“stoichiometry” pretty much means . . .
“the process of calculating how many grams of one chemical are consumed/produced during a reaction when the number of grams of any other reactant/product is given.”
That’s what you’re going to have to be able to do on the
CST, so that’s what you’re going to have to be able to do on my tests, too. Anything else is extra credit.

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g
8.9 g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3 x
1
1 mol NaHCO
3
84 g NaHCO
3 x
1 mol CO
2
1 mol NaHCO
3 x
44 g CO
2
1 mol CO
2
NaHCO
3
:
Na: 1
H: 1
C: 1 x 23 x 1 x 12
O: 3 x 16
= 23
= 1
= 12
= 48
84 g/mol
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
44 g/mol
84 )
8.9

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g
8.9 g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3 x
1
1 mol NaHCO
3
84 g NaHCO
3 x
1 mol CO
2
1 mol NaHCO
3 x
44 g CO
2
1 mol CO
2
NaHCO
3
:
Na: 1
H: 1
C: 1
O: 3
The mass of the known substance goes on the top of the first fraction.
x 23 = 23 x 1 x 12
= 1
= 12
CO
2
:
C: 1 x 12 = 12 x 16 = 48
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g
8.9 g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
1 x
1 mol NaHCO
3
84 g NaHCO
3 x
1 mol CO
2
1 mol NaHCO
3 x
1
44 g CO mol CO
2
2
The molar mass of the known substance goes on the bottom of the second fraction.
NaHCO
3
:
Na: 1
H: 1
C:
O:
1
3 x 23 x 1 x 12 x 16
= 23
= 1
= 12
= 48
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g
8.9 g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
NaHCO
Na:
1
1
3
: x 23 x
1 mol NaHCO
= 23
3
84 g NaHCO
3 x
1 mol CO
2
1 mol NaHCO
3 x
1
44 g CO mol CO
2
2
The coefficients go in the third fraction, known as the “mole ratio” fraction.
H:
C:
O:
1
1
3 x 1 x 12 x 16
= 1
= 12
= 48
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g
8.9 g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
NaHCO
Na:
1
1
3
: x 23
1 mol NaHCO
= 23
3
1 mol CO
2
44 g CO x
84 g NaHCO
3 x
1 mol NaHCO
3 x
1 mol CO
Yeah. I know. There are no coeffcients in this equation. That’s why I put 1’s in the fraction.
2
2
H:
C:
O:
1
1
3 x 1 x 12 x 16
= 1
= 12
= 48
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g
8.9 g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
1 x
1 mol NaHCO
3
84 g NaHCO
3 x
1 mol CO
2
1 mol NaHCO
3 x
1
44 g CO mol CO
2
2
The molar mass of the unknown substance goes on the top of the last fraction.
NaHCO
3
:
Na: 1
H: 1
C:
O:
1
3 x 23 x 1 x 12 x 16
= 23
= 1
= 12
= 48
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
1
1 mol NaHCO
3
1 mol CO
2
44 g CO x
84 g NaHCO
3 x
1 mol NaHCO
3 x
1 mol CO
Okay. Now you do the same calculation, but for
2
2 the 2 masses of baking soda your group used.
NaHCO
3
:
Na: 1
H: 1
C:
O:
1
3 x 23 x 1 x 12 x 16
= 23
= 1
= 12
= 48
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
1
1 mol NaHCO
3
1 mol CO
2
44 g CO x
84 g NaHCO
3 x
1 mol NaHCO
3 x
1 mol CO
I need to see the four-fraction chain for each
2
2 calculation . . . NaHCO
3
:
Na: 1
H: 1
C:
O:
1
3 x 23 x 1 x 12 x 16
= 23
= 1
= 12
= 48
CO
2
:
C: 1 x 12 = 12
O: 2 x 16 = 32
17 x 44
748
84 )
8.9
84 g/mol
44 g/mol

Imagine you pour 17 grams of baking soda into a large bucket full of vinegar. Assuming that that there is so much vinegar that it can easily neutralize all the baking soda, how much carbon dioxide should be produced? Assume that the baking soda and vinegar react according to the equation below:
17g g
CH
3
COOH + NaHCO
3
 NaCH
3
COO + CO
2
+ H
2
O
17g NaHCO
3
1 x
1 mol NaHCO
3
84 g NaHCO
3 x
1 mol CO
2
1 mol NaHCO
3 x
1
44 g CO mol CO
2
2
. . . and, somewhere on your paper, I need to see
NaHCO
3
:
Na: 1
H: 1
C:
O:
1
3 the molar mass calculations for NaHCO
3 x 23 = 23 x 1 x 12
= 1
= 12
CO
2
:
C: 1 x 12 = 12
17 x 44 x 16 = 48
O: 2 x 16 = 32 748
& CO
2
.
84 )
8.9
84 g/mol
44 g/mol

6
5
4
3
2
1
0
0 30 60 90 120 mass of Zn provided
150
Black line = ?
Grey line = ?
Limiting reactant = ?
Excess reactant = ?