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in a Mixture. Gravimetric Determination of NaHCO 3. FINAL Final Exercise – 105 points. FINAL. CHE 133 MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM. Eligible Students are ONLY those who have excused absences from

in a Mixture

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## in a Mixture

Gravimetric Determination of NaHCO3

FINAL Final Exercise – 105 points

FINAL

### CHE 133MAKE-UP LABORATORY EXERCISE Mon 11/19 - 2:00 PM or Tue 11/20 – 2:30 PM

Eligible Students are ONLY those who have

excused absences from

either a TEST (105 point) orPRELIMINARY (55 point) Exercise

SHOULD HAVE SIGNED UPON SHEETS POSTED

IN EACH LABORATORY ROOM

Everyone does the same make-up exercise. You can download the exercise at:

http://www.ic.sunysb.edu/Class/che133/susb/SUSB055.pdf

The Calendar

Purpose:

Determine the percent of NaHCO3 in a mixture by Gravimetry

Concepts:

Thermal DecompositionConstant Weight

Techniques:

Weighing

Thermal Decomposition

Apparatus:

Analytical BalanceHot Plate

CrucibleCrucible Tongs

This Part of the Exercise is, again

ConceptuallySimple.

• Weigh Sample,

wSample.

2. Decompose the sample by heating

2 NaHCO3 (s)

2 NaHCO3 (s) 

Na2CO3 (s)

+ H2O (g)

+ CO2 (g)

+ NaCl(s)

+ NaCl(s)

3. Weigh product to get weightof CO2 & H2O lost

weight loss = wCO2 + wH2O

weight loss

= nCO2 * 44.01

+ nH2O * 18.02

Molar Massof CO2

Molar Mass of H2O

weight = mol X Mol Mass

4. Get moles of CO2 , H2O and NaHCO3 lost

2

NaHCO3 (s)  ½Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g)

½

½

½

nCO2 = nH2O= ½ nNaHCO3

(= nCO2 * 44.01

+ nH2O * 18.02 )

so, weight loss

= ½ nNaHCO3 * ( 44.01 + 18.02 )

or, nNaHCO3 =

2 * ( weight loss ) / 62.03

Molar Mass of NaHCO3

5.wNaHCO3 =

nNaHCO3 * 84.01 g / mol

6. Compute Percent Composition of Sample

PctNaHCO3 = 100 * wNaHCO3 / wSample

How is Pct NaHCO3 related to the weight loss?

wNaHCO3 =

nNaHCO3 * 84.01 g / mol

Showed that:

2 * ( weight loss )

62.03

nNaHCO3 =

2 * ( weight loss )

62.03

wNaHCO3 =

X 84.01

= 2.709 * ( weight loss )

wNaHCO3= ( weight loss ) / 0.3691

PctNaHCO3 = 100 * wNaHCO3 / wSample

= 270.9 * ( weight loss) / wSample

wNaHCO3= ( weight loss ) / 0.3691

How much weight would 1.000 g of pure NaHCO3 lose?

wNaHCO3 = 1 g

1 g = ( weight loss ) / 0.3691

weight loss = 0.3691 g

PctNaHCO3 = 100 * wNaHCO3 / wSample

270.9 * ( weight loss) / wSample

= 270.9 * 0.3691/ 1.000

= 100.0 %

### Each mole of NaHCO3 produces one mole of CO2 and one mole of H2O when decomposed by heating.

• True

• False

Each mole of NaHCO3 produces one mole of CO2 and one mole of H2O when decomposed by heating.

NaHCO3 (s)  ½Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g)

BFalse

Unlike gasometric part, this part is NOT complicated by any special adjustments or considerations.

Accuracy and precision of result are solely a function of care with which weighings and decomposition are conducted!

How do the Various Substances behave

at High Temperature?

NaHCO3 decomposes rapidly at T > 200 oC forming Na2CO3

The melting point of NaClis 800 oC.

(Molten NaClcools to produce a solid that adheres to porcelain. )

Na2CO3(product) melts at 850 oC AND begins to decompose to CO2 + Na2O at that temperature.

The peak temperature on a hot plate is well below 800 oC.

But make sure it is at least 200 oC

Would result in additional weight loss

850 oC is bright red heat

If some sample is lost from the crucible during the heating, the resulting percentage of NaHCO3 that we calculate will be ______ the actual value.

• smaller than

• unchanged from

• larger than

OMG ! ! ! ! !

Let’s try again.

This time, with enthusiasm

(and discussion)

If some sample is lost from the crucible during the heating, the resulting percentage of NaHCO3 that we calculate will be ______ the actual value.

• smaller than

• unchanged from

• larger than

If some sample is lost from the crucible during the heating, the resulting percentage of NaHCO3 that we calculate will be ______ the actual value.

PctNaHCO3 = 270.9 * ( weight loss) / wSample

If sample is lost, the apparent weight loss will be too large and, therefore, the

% NaHCO3 will be too large.

CLarger

Last Year

Right to Right 48

Right to Wrong 31

Wrong to Right 46

Wrong to Wrong 114

What precision & accuracy should you expect?

• The only sources of error are:

• loss of sample during run

• don’t spill any sample

• incomplete heating

• reheat until two successive

• weighingsdiffer by Δ mg

• overheating

• not likely on hotplate

• weighing errors

• record numbers carefully

To be specified

To find weight loss, you need the weight of the dry, empty crucible

A word from Obviousman

This exercise depends heavily on the proper use of the analytical balance!

• try to use same balance throughout a run

DO NOT TINKER WITH BALANCE SETTINGS!

?? Standard Penny ??

Weigh a coin before anything else

Check the weight of the “standard” coin before each subsequent weighing.

How do we know when the reaction is done?

When all the H2O and CO2 have been lost, the sample will cease to lose weight.

I.e., two successive weighings before and after heating should be the same

How should we define “the same” (Δ)?

Our operational definition is that they differ by less than ± 5 mg. I.e., |w2 – w1| < 0.0050 g

NOT

0.5 mg!

e.g.

w1 16.5386

w216.5294

- 0.0092

w3 16.5242

w4 16.5214

- 0.0028

w2 16.5294

w3 16.5242

- 0.0052

w3 16.5242

w4 16.5276

+ 0.0034

X

X

- 9.2 mg

- 5.2 mg

- 2.8 mg

+ 3.4 mg

### Bonus Question (5 extra points)

What are the last 3 digits of your student ID?

Reproducibility

Uncertainty in Weighing:

wsample analytical balance (0.0004 / 1.0000) < 0.1%

wresidue analytical balance (0.0004 / 0.7000) < 0.1%

But

Our criterion for final weight is one that differs from the previous weight by less than 5mg, so our weight loss should differ from the “true” weight loss by less than 5mg .

For a 1 g sample, the minimumweight loss is

~369.1/2 mg, and

5mg constitutes less than a

100 X 5 X 2 / 369.1 = 2.7 % error

in the weight loss

50%

NaHCO3

Do the analysis twice,

and

Handle the crucible ONLY with crucible tongs!

DATA SHEET

Wt of crucible + sample16.0755 g

Wt of crucible14.9842 g

174.2 mg

Wt of sample

1.0913 g

Wt of crucible + residue –aft heat115.9013 g

19.7 mg

Wt of crucible + residue –aft heat215.8816 g

9.3 mg

Wt of crucible + residue –aft heat315.8723 g

3.4 mg

Wt of crucible + residue –aft heat415.8689 g

Wt of residue [ 15.8689 – 14.9842 ]

0.8847 g

Wt loss per gram of NaHCO3

Wt loss [ 1.0913 – 0.8847 ]

0.2066 g

Wt NaHCO3 [ 0.2066 / 0.3691 ]

0.5597 g

% NaHCO3 [100 X 0.5597 / 1.0913 ]

51.29 %

### If the NaHCO3 is not completely decomposed when we record our final weight, the % NaHCO3 we compute will be ________ the actual value.

• Smaller than

• Equal to

• Larger than

If the NaHCO3 is not completely decomposed when we record our final weight, the % NaHCO3 we compute will be ________ the actual value.

PctNaHCO3 = 270.9 * ( weight loss) / wSample

If NaHCO3 is not completely decomposed, the apparent weight loss will be too small and, therefore, the

% NaHCO3 will be (too) small.

ASmaller than

Review - Avg, Avg Deviation, Pct Deviation

Suppose our second result is 52.41%

51.29 + 52.41

Average = ------------------ = 51.85 %

2

0.56 + 0.56

Avg Dev = ----------------- = 0.56 %

2

100 X 0.56

Pct Dev = ------------------- = 1.1 %

51.85

### At the end of the exercise, the crucible should contain only NaCl.

• True

• False

At the end of the exercise, the crucible should contain only NaCl.

NaHCO3 (s)  ½Na2CO3 (s) + ½ H2O (g) + ½ CO2 (g)

The crucible contains NaCl and Na2CO3

BFalse

Next Week – No Lecture

Nov 19 & 20 - Make-up Labs

Fri, Nov 30, Mon, Dec 3 – REVIEW

Attendance Optional

When you prepare, and eat your holiday

turkey,

tofu,

tuna,

tomatoes,

turnips,

pita,

pasta,

whatever

peas,

potatoes,

WEAR YOUR

It has been fun learning with you

Have a good rest of the semester