Question

In: Chemistry

Calculate the percent of calcium oxide in the original sample provided by first determining the number of moles of the precipitate

 

Calculate the percent of calcium oxide in the original sample provided by first determining the number of moles of the precipitate, calcium oxalate monohydrate, and converting that to moles of calcium oxide; or use the gravimetric factor and do it in one step. Next calculate the mass of calcium oxide and the % CaO in the unknown. Find the mean % CaO of the three samples and determine the standard deviation and 95 % confidence interval. Record the mean, standard deviation, and confidence interval in your notebook.

 

Solutions

Expert Solution

First consider all results from the crucibles. The resulting masses are those that correspond to your precipitate, which is stated to be calcium oxalate monohydrate (CaC2O4 · H2O).

Now the molecular weight for this molecule is : 146.11228g/mol

So, the moles can be calculated with moles= mass/molecular weight ---> moles= 0.5941g/146.11228g/mol = 0.0041moles of CaC2O4·H2O

From this amount, consider that you only need what corresponds to the calcium oxalate, so lose the percentage that corresponds to water.

CaC2O4 = 128.097g/mol, so 128.097/146.11228 = 0.8767(0.0041moles of CaC2O4·H2O) = 0.0036moles of CaC2O4.

Now consider the reaction to form CaC2O4 from CaO and ammonium oxalate

First, CaO must react with the HCl you added, so as to form ions:

CaO + 2HCl ---> Ca2+ + 2Cl- + H2O

The second reaction that takes place is:

(NH4)2C2O4 + Ca2+ + 2Cl- -----> 2NH4Cl + CaC2O4

Hence, you need the same amount of moles of Calcium to form the calcium oxalate, and also it is the same amount of moles of Calcium oxide from the beginning. This is because your reaction always has a coefficient 1 for all your calcium species. Therefore, the moles of CaO are also 0.0036 moles. Now this amount of moles, in grams, can be obtained as follows:

m= moles (molecular weight) ----> m= 0.0036moles(56.0774g/mol) = 0.1999 g of CaO.

If the mass of the unknown is 0.4670g, then : 0.1999g/0.4670g*100 = 42.81% of CaO in the unknown.


Now follow the same process to obtain it for each of the crucibles:
For crucible 2, you get: 42.27%

For crucible 3, you get: 39.28%


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