Lab 3 Stoichiometry Precipitation
The calculations completed for this experiment include determining the amount of Na2CO3 needed to do a full reaction. This was calculated through stoichiometry calculations:
Molar mass was first calculated for CaCl2*2H2O
Ca = 40.078g Cl2 = 35.453g*2 = 70.906g 2H2 = 1.00794g*4 = 4.03176g 2O = 15.9994g*2 = 31.9988g
40.078g + 70.906g + 4.03176g + 31.9988g = 147.01456g or 147.0 g CaCl2
1g CaCl2 * 2H2O x (1 mol CaCl2 *2H2O/147g CaCl2 *2H2O) = 0.0068 mol of CaCl2*2H2O
Molar mass was then calculated for Na2CO3:
Na2 = 22.9898g*2 = 45.9796g C = 12.0107g O3 = 15.9994g*3 = 47.9982g
45.9796g + 12.0107g + 47.9982g = 105.9885g or 105.99g Na2CO3
Then using the molar mass of Na2CO3 and the 0.0068 mol of CaCl2*2H2O to calculate the grams required of Na2CO3 for the experiment:
105.99g/mol Na2CO3 * 0.0068 mol of CaCl2 = 0.72g of Na2CO3
Final calculation completed was to remove the mass of the filter paper, 1 g from the mass of the filter paper and precipitate:
1.6g (filter paper and precipitate) – 1g (filter paper) = 0.6g precipitate
Questions:
A. From your balanced equation what is the theoretical yield of your product?
Because we have a balanced 1:1 ratio and have 0.0068 mol of CaCl2 then we will get 0.0068 mol of CaCO3 which turns out to be 0.68g of CaCO3.
0.0068 mol CaCO3*100.09g/mol CaCO3 = 0.68 g CaCO3
B. According to your data table, what is the actual yield of the product?
0.6 grams CaCO3 (However the scale actually tipped 1.7g and 1.6g and finally settled on 1.6g)
C. What is the percent yield?
0.6g CaCO3/0.68g CaCO3 = 0.88*100% = 88% yield
D. A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error.
The degree of accuracy is only a tenth of a gram, the scale could not measure in 100ths of a gram and therefore it is likely the yield could have been higher. This would be a source of error, as well as not retrieving the entire product from
Molar mass was first calculated for CaCl2*2H2O
Ca = 40.078g Cl2 = 35.453g*2 = 70.906g 2H2 = 1.00794g*4 = 4.03176g 2O = 15.9994g*2 = 31.9988g
40.078g + 70.906g + 4.03176g + 31.9988g = 147.01456g or 147.0 g CaCl2
1g CaCl2 * 2H2O x (1 mol CaCl2 *2H2O/147g CaCl2 *2H2O) = 0.0068 mol of CaCl2*2H2O
Molar mass was then calculated for Na2CO3:
Na2 = 22.9898g*2 = 45.9796g C = 12.0107g O3 = 15.9994g*3 = 47.9982g
45.9796g + 12.0107g + 47.9982g = 105.9885g or 105.99g Na2CO3
Then using the molar mass of Na2CO3 and the 0.0068 mol of CaCl2*2H2O to calculate the grams required of Na2CO3 for the experiment:
105.99g/mol Na2CO3 * 0.0068 mol of CaCl2 = 0.72g of Na2CO3
Final calculation completed was to remove the mass of the filter paper, 1 g from the mass of the filter paper and precipitate:
1.6g (filter paper and precipitate) – 1g (filter paper) = 0.6g precipitate
Questions:
A. From your balanced equation what is the theoretical yield of your product?
Because we have a balanced 1:1 ratio and have 0.0068 mol of CaCl2 then we will get 0.0068 mol of CaCO3 which turns out to be 0.68g of CaCO3.
0.0068 mol CaCO3*100.09g/mol CaCO3 = 0.68 g CaCO3
B. According to your data table, what is the actual yield of the product?
0.6 grams CaCO3 (However the scale actually tipped 1.7g and 1.6g and finally settled on 1.6g)
C. What is the percent yield?
0.6g CaCO3/0.68g CaCO3 = 0.88*100% = 88% yield
D. A perfect percent yield would be 100%. Based on your results, comment on your degree of accuracy and suggest possible sources of error.
The degree of accuracy is only a tenth of a gram, the scale could not measure in 100ths of a gram and therefore it is likely the yield could have been higher. This would be a source of error, as well as not retrieving the entire product from