In the scientific world, there is a significant difference between calculated values and values measured in investigations. Percentage Yield is the ratio, between masses, of the actual yield to the theoretical yield. The actual yield is the amount or mass of product actually collected during an experiment or industrial process, while the theoretical yield is the amount or mass of product predicted based on the stoichiometry of the chemical equation. In order to calculate the theoretical yield, the number of moles (the amount of substance containing 6.02x1023 entities) of each compound should be calculated according to the formula: n = m/M
Where n is the number of moles, m is the given mass of a substance in grams, and M is the molar mass of …show more content…
When both solutions are mixed a white precipitate (Calcium Carbonate) is formed alongside Sodium Chloride. A clear colourless liquid is observed as the white precipitate is filtered through the filter paper. …show more content…
Amount of CaCl2 = m/M = (0.5g )/( 110.98 g/mol ) = 0.004505316 mols of CaCl2.
Amount of CaCO3 = 0.004505316 x 1/1 = 0.004505316 mols of CaCO3.
Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol.
Amount of Na2CO3 = m/M = (0.7g )/( 105.99 g/mol ) = 0.006604396 mols of Na2CO3.
Amount of CaCO3 = 0.006604396 x 1/1 = 0.006604396 mols of CaCO3.
Thus the Calcium Chloride is the limiting reagent as it produces less Calcium Carbonate.
Molar Mass of CaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol of CaCO3.
Theoretical yield of CaCO3 = M x n = 100.09 g/mol x 0.004505316 mols = 0.450937078g of CaCO3.
Actual yield of CaCO3 = Mass of product and filter paper - Mass of dry filter paper = 1.02g – 0.59g = 0.43g of CaCO3.
Percentage Yield = (0.43g )/( 0.450937078g ) x 100% = 95.35698459 % 95%.
Amount of used Na2CO3 = amount of limiting reagent x mole ratio = 0.004505316 x 1/1 = 0.004505316 mols of Na2CO3.
Amount of excess Na2CO3 = 0.006604396 - 0.004505316 = 0.00209908 mols of
Where n is the number of moles, m is the given mass of a substance in grams, and M is the molar mass of …show more content…
When both solutions are mixed a white precipitate (Calcium Carbonate) is formed alongside Sodium Chloride. A clear colourless liquid is observed as the white precipitate is filtered through the filter paper. …show more content…
Amount of CaCl2 = m/M = (0.5g )/( 110.98 g/mol ) = 0.004505316 mols of CaCl2.
Amount of CaCO3 = 0.004505316 x 1/1 = 0.004505316 mols of CaCO3.
Molar mass of Na2CO3 = 2(22.99 g/mol) + 12.01 g/mol + 3(16.00 g/mol) = 105.99 g/mol.
Amount of Na2CO3 = m/M = (0.7g )/( 105.99 g/mol ) = 0.006604396 mols of Na2CO3.
Amount of CaCO3 = 0.006604396 x 1/1 = 0.006604396 mols of CaCO3.
Thus the Calcium Chloride is the limiting reagent as it produces less Calcium Carbonate.
Molar Mass of CaCO3 = 40.08 + 12.01 + 3(16.00) = 100.09 g/mol of CaCO3.
Theoretical yield of CaCO3 = M x n = 100.09 g/mol x 0.004505316 mols = 0.450937078g of CaCO3.
Actual yield of CaCO3 = Mass of product and filter paper - Mass of dry filter paper = 1.02g – 0.59g = 0.43g of CaCO3.
Percentage Yield = (0.43g )/( 0.450937078g ) x 100% = 95.35698459 % 95%.
Amount of used Na2CO3 = amount of limiting reagent x mole ratio = 0.004505316 x 1/1 = 0.004505316 mols of Na2CO3.
Amount of excess Na2CO3 = 0.006604396 - 0.004505316 = 0.00209908 mols of