Lab Alkali Metal Carbonate PDF

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Lab Alkali Metal Carbonate...

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Which Alkali Metal Carbonate? Worksheet

1. Report the number of your

Unknown #1

unknown:

Results: For all calculations, make sure to include your hand written and annotated sample calculations to your report. Part A 1. Mass of barium carbonate collected: - The precipitate is BaCO3. - In order to find the mass of the precipitate, the mass of the precipitate, watch glass, and filter papers is subtracted by the mass of the watch glass and filter papers. These are both given. - 116 – 115.562 = 0.937g 2. Moles of M2CO3 reacted: Balance chemical equation for reaction (given): M2CO3 (aq) + BaCl2( aq) → BaCO3 (s) + 2MCl (aq) - The stoichiometric coefficients for the BaCO3 and the unknown reactant M2CO3 are both 1. Therefore, one mole of M2CO3 is consumed for every mole of BaCO3 that is produced. In order to figure out the moles of M2CO3 consumed, solving for the moles of BaCO3 produced is needed. -

Molar Mass of BaCO3 : 1 Ba = 137.327 g/mol 1 C = 12.0107 g/mol 3 O = + 42.9982 g/mol 197.3359 g/mol = mass of BaCO3

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Moles = Mass / Molar mass Moles = 0.937 / 197.3359 = 0.00474249 Because the ratio of moles for BaCO3 and M2CO3 is 1:1, 0.00475 mol of M2CO3 were consumed in the reaction.

3. Molar mass of the unknown: - Molar Mass = Mass / Moles Mass of unknown compound M2CO3 (given) = 0.487g Moles of unknown compound M2CO3 (from step 2) = 0.004748249 Molar Mass = 0.487 / 0.004748249 = 102.5641242 or 102 g/mol - Metal carbonate: The molar mass found (102 g/mol) is nearest to the molar mass of Sodium Carbonate (105.99 g/mol).

Part B 4. Mass of CO2 produced: - Mass of CO2 = mass of unknown M2CO3 + actual mass of HCl added – mass of products in beaker. -

Mass of unknown M2CO3 = mass of unknown M2CO3 and beaker (given) – mass of empty beaker (given). Mass of unknown M2CO3 = 128.655 – 128.152 = O.503g

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Actual mass of HCl added = mass of graduated cylinder (given) and 20.0mL of HCl (given) – mass of graduated cylinder after delivering HCl (given). Actual mass of HCl added = 153.224 – 133.092 = 20.132g

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Mass of products in beaker = final mass of beaker after reaction (given) – mass of empty beaker (given). Mass of products in beaker = 148.5781404 – 128.152 = 20.4261404g

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Mass of CO2 produced = 0.503 + 20.132 – 20.4261404 = 0.2088596 or 0.209g

5. Moles of CO2 produced: - Molar Mass of CO2: 1 C = 12.017 g/mol 2 O = + 31.988 g/mol 44.0158 g/mol -

Moles = Mass / Molar Mass 0.209g (step 4) / 44.0158 = 0.004748295 or 0.00475 mol

6. Moles of M2CO3 reacted: Balanced chemical equation for reaction (given): M2CO3 + 2HCl → CO2 + 2MCl + H2O Stoichiometric coefficients for M2CO3 and CO2 are both 1. For every mole of CO2 that is produced, one mole of M2CO3 is consumed. -

Moles of CO2 (step 5) = 0.004748295 mol 0.004748295 (1/1) = 0.004748295 mol

7. Molar mass of the unknown metal carbonate: - Molar Mass = Mass / Moles Molar Mass = 0.503g / 0.00475 mol = 105.8947368 or 105.895 g/mol

8. Based upon methods 1 and 2, what is the most likely identity of your metal carbonate? Mass of Method 1 = 102.5641242 g/mol Mass of Method 2 = 105.8947368 g/mol Identity of Metal Carbonate = Sodium Carbonate

9. Calculate the percent error for mass based upon the average values from parts A and B: - Part A Molar Mass = 102.5641242 g/mol - Part B Molar Mass = 105.8947368 g/mol

% error = -0.166%

the molar molar mass

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Average Molar Mass of Parts A + B = 102.5641242 + 105.8947368 = 208.458861 / 2 = 104.2294305 g/mol

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% error = ((measured-real) / real) * 100 Measured Mass = 104.2294305 g/mol Real Mass of Sodium Carbonate (given) = 105.99 g/mol % error = ((104.2294305 – 105.99) / 105.99) * 100 = -1.661071327

Part C 10.

Ion Sodium

Flame Color

Sodium (cobalt glass) Potassium

Purple Pink + White

Potassium (cobalt glass)

Pink

Calcium Barium

Red + Orange

Strontium Copper

Orange

Wha t is the

Orange + Green Red, Orange, + Pink Green

purpose of using the cobalt glass in the identification of sodium and potassium? Cobalt glass absorbs and blocks out colors that are overwhelmingly bright. By doing so, underlying colors are easier observed. For instance, cobalt glass absorbed the bright light that is produced by sodium, leaving only the true color of potassium to be visible.

11. Does the image of the flame test back up the unknown identification from methods 1 and 2? Briefly explain. The unknown metal carbonate was identified to be Sodium Carbonate from methods 1 and 2.

From the picture, it is seen as an orange flame with a bit of yellow. In Part C, it is described as being orange. Therefore, it can be said that the flame test backs up the unknown identification from methods 1 and 2.

Which Alkali Metal Carbonate? Lab Report:

There is not a formal lab report for this lab. Complete the above pages using the Microsoft version of this file that is available for download on the lab D2L page. Once the worksheet is complete save as a .pdf and submit the worksheet on time and to your TA in the dropbox on D2L....