Prelab Assignment matca tuloy@ I) A 3.0ob g sample of a MgCO/CaCOs mixture is he
ID: 693005 • Letter: P
Question
Prelab Assignment matca tuloy@ I) A 3.0ob g sample of a MgCO/CaCOs mixture is heated in a crucible. After na(tyl CaCo3 cooling its mass is found to be 1.804 g. After a second heating the mass is found to be 1.780 g, and after a third heating the mass is found to be 1.779 g. a. What is the mass of magnesium carbonate in the sample? 9.009 1-221gC02 b. What is the percent magnesium carbonate in the sample? Why was the sample heated three times? to yemove the Coz from thu miAture c. 2) If some of the calcium carbonate also decomposed in this experiment, w resulting error cause you to get a %MgCO3 that is higher or lower than percentage? ) If you did not heat the mixture sufficiently to decompose all of the M would the resulting error cause you to get a %MgCO3 that is higher the actual percentage?Explanation / Answer
Ans. Step 1: Let the mass of CaCO3 be X grams and that of MgCO3 be Y grams in the given sample.
So,
X gram + Y gram = 3.000 gram
Or, X + Y = 3.000 - equation 1
Now,
Moles of CaCO3 = Mass / molar mass
= (X g / 100.0872 g mol-1)
= 0.0099913X mol
Moles of MgCO3 = (Y / 84.3142) mol = 0.0118604Y mol
# Step 2: Balanced reactions:
I. CaCO3(s) -----heat---> CaO(s) + CO2(g)
II. MgCO3(s) ----heat----> MgO(s) + CO2(g)
According to the stoichiometry of balanced reactions, 1 mol of CaCO3 and 1 mol MgCO3 produces 1 mol of CO2 each.
So,
Moles of CO2 produced by CaCO3 = 1 x moles of CaCO3 = 0.0099913X mol
Moles of CO2 produced by MgHCO3 = 1 x moles of MgCO3 = 0.0118604Y mol
Total moles of CO2 produced = 0.0099913X mol + 0.0118604Y mol
# Step 3: Total loss in mass due to heating the sample =
Initial mass- Mass after 3rd heating
= 3.000 g – 1.779 g
= 1.221 g
# Moles of CO2 produced = Mass / Molar mass
= 1.221 g / (44.0098 g/mol)
= 0.0277438 mol
Therefore, total moles of CO2 produced during the reaction = 0.0277438 mol
Now, comparing #Step 2 and # Step 3, the total moles of CO2 produced during reaction is given by -
0.0099913X mol + 0.0118604Y mol = 0.0277438 mol
Or, 0.0099913 X + 0.011860Y = 0.0277438 - equation 2
# Step 4: Comparing (equation 1 x 0.00999913) – equation 2-
0.0099913 X + 0.0099913 Y = 0.0299739
(-) 0.0099913X + 0.0118604 Y = 0.0277438
Or, -0.0018691Y = 0.0022301
Or, Y = 0.0022301 / (-0.0018691)
Hence, Y = -1.193
Therefore, mass of MgCO3 in sample = Y g = (-)1.193 g
Note: Since the mass can’t be NEGATIVE, there seems to be some erroneous values in the question. That is, mass of CO2 lost seems to be greater than the actual value. Also, the exact molar masses used in the solution may also be an issue. Please use the molar mass of reactants mentioned in the question. Use correct values as mentioned in your test and follow to steps to calculate mass and % mass of MgCO3 in sample.
Also look for any other point that may interfere with the result. That is, there is decomposition of only CaCO3 (as asked in #2 below) or only MgCO3.
#2. If CaCO3 decomposes, the apparent loss if mass of sample and in turn, the apparent mass (and) moles of CO2 formed would be greater than the actual value. Since moles of CO2 is proportional to % mass of MgCO3, apparent increase in moles of CO2 would lead to apparent increase in % MgCO3 in sample.
#3. Insufficient heating would lead to lower than actual mass of CO2 lost.
Lower is the moles of CO2 lost, lower would be the % MgCO3 in sample.
So, insufficient heating would lead to apparently low % MgCO3 in sample.