Write your answers in pen and include units Post Lab Questions: Identification o
ID: 880346 • Letter: W
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Write your answers in pen and include units Post Lab Questions: Identification of a tion of a Compound by Mass Relationships to identify an unknown compound by the method used in this experiment. She found beaed a sample weighing 0.4862 g. the mass barely changed, dropping to 0.4855 g When g to 055&.When I. A student attempted to identify anu that when she heateda the product was con 2 g, the mass barely c erted to a chloride, the mass went up, to 0.5247 a. Is the sample a carbonate? Yes/ no (Circle onc) Please provide your reasoning below b. What are the two compounds that might be in the unknown? balanced chemical equation for the overall reaction that occurs when each of these two orig to a chloride. If the compound is a hydrogen carbonate, use the sum of equation for a sodium salt c. Write the balanced chemical equation for the overall reaction that occurs when each of thesetwo oris Reactions I and 2. If the sample is a carbonate, use Reaction 2. Write the and then for a potassium salt. d. How many moles of the chloride salt would be produced from one mole of original compound? e. How many grams of the chloride salt would be produced from one molar mass of original compound? Molar masses: NaHCC,. 2NaCl KHCO3 gK,CO, - If a sodium, salt, g original compound g chloride g original compound g chloride If a potassium salt. What is the theoretical value of Q, as found by Equation 3, if she has the Na salt? What was the observed value of Q? f. if she has the K salt? g. h. Which compound did she have as an unknown?Explanation / Answer
a) The compound can be carbonate as the carbonates are barely affected by heating, they may decompose only at very high temperatures.
b) The molar mass ratio for compound should be
0.4862 / 0.5247 = 0.9266 On trying the ratio of carbonates to chloride of following founds to be in close aggrement to ratio of the given sample
SrCO3 / SrCl2 = 147.63 / 158.52 = 0.9313 almost equal to 0.4862 / 0.5247 (or 0.9266)
K2CO3 / 2KCl = 138 / 149 = 0.926
c) SrCO3 + 2HCl --> SrCl2 + H2CO3
K2CO3 + 2HCl --> 2KCl + H2CO3
d) From first compound the moles of chloride produced will be one from one mole
and from second compound two moles of HCl will be produced from one mole of carbonate.
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