mass of beaker= 49.340g mass of beaker and unknown= 49.540g What is the Moles of

Question

mass of beaker= 49.340g
mass of beaker and unknown= 49.540g
What is the Moles of Unknown assuming it is sodium carbonate?
what is the moles of unknown assuming it is sodium hydrogen carbonate?
What is the Mass of sodium chloride and the moles of sodium chloride?
Calculate the predicted yield (in g) of NaCl assuming your unknown is sodium carbonate.
Calculate the predicted yield of NaCl (in g) assuming your unknown is sodium hydrogen carbonate.
What is the most probable identity of your unknown?

Suppose your unknown had been hydrated form of sodium hydrogen carbonate with the formula NaHCO3 H2O. What would differ in your calculations? Could you still have distinguished it from sodium carbonate? Explain.

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Explanation / Answer

mass of unknown = 49.54 - 49.34 = 0.2 g

moles of Na2CO3 = 0.2/105.9888 = 1.89 x 10^-3 mols

moles of NaHCO3 = 0.2/84.007 = 2.40 x 10^-3 mols

moles of NaCl starting wth Na2CO3 = 1.89 x 10^-3/2 = 9.45 x 10^-4 mols

mass of NaCl = 9.45 x 10^-3 x 58.44 = 0.55

moles of NaCl starting with NaHCO3 = 2.40 x 10^-3 mols

mass of NaCl = 2.40 x 10^-3 x 58.44 = 0.14 g

mass of NaCl If all has reacted = 0.2 g

predicted yield from Na2CO3 = (0.55/0.2) x 100 = 275% (not expected)

predicted yield from NaHCO3 = (0.14/0.2) x 100 = 70%

If hydrated NaHCO3 was used instead, the overall moles of NaHCO3 would had been differe, Which would change the %yield.

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