Laboratory Report: Precipitation Reaction

Categories: Chemistry


The purpose of this experiment was to apply stoichiometry principles to predict the amount of product formed in a precipitation reaction, accurately measure the reactants and products, determine the actual yield versus the theoretical yield, and calculate the percent yield.

Calcium chloride dihydrate (CaCl2·2H2O) was dissolved in water to create a calcium chloride solution, and stoichiometry was used to calculate the required amount of sodium carbonate (Na2CO3) for a complete reaction. The predicted and actual yield of calcium carbonate (CaCO3) precipitate were compared to calculate the percent yield.


Precipitation reactions are fundamental in chemistry, and stoichiometry plays a crucial role in predicting the quantities of reactants and products involved. In this experiment, the goal was to apply stoichiometry to forecast the amount of precipitate formed when calcium chloride dihydrate reacts with sodium carbonate. The specific objectives included:

  1. Predicting the amount of product (calcium carbonate) formed in the reaction.
  2. Accurately measuring the reactants and products of the reaction.

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  3. Comparing the actual yield to the theoretical yield.
  4. Calculating the percent yield of the reaction.

Materials and Methods


  • 1.0 g of CaCl2·2H2O
  • 25 mL of distilled water
  • 0.72 g of Na2CO3
  • Filter paper
  • Filter funnel
  • Small and large cups
  • Paper towels


The experiment was conducted as follows:

  1. 1.0 g of CaCl2·2H2O was placed in a 100-mL beaker, and 25 mL of distilled water was added to create a calcium chloride solution.
  2. Stoichiometry was employed to determine the required amount of Na2CO3 for a complete reaction. The conversion of 1.

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    0 g of CaCl2·2H2O to moles yielded 0.00680 moles, and the mole ratio between CaCl2·2H2O and Na2CO3 was determined to be 1:1. Consequently, the moles of Na2CO3 were converted to grams, resulting in 0.72 g. The predicted amount of CaCO3 was calculated to be 0.68 grams.

  3. 0.72 grams of Na2CO3 were measured into a paper cup, and 25 mL of distilled water was added and stirred.
  4. The Na2CO3 solution was poured into the 100 mL beaker containing the calcium chloride solution, leading to the immediate formation of a precipitate (CaCO3).
  5. A filtration system was set up, with a small cup placed inside a larger cup for support and a funnel placed in the small cup. A 1.1 gram circle of filter paper was folded in half twice, and one section of the folds in the filter paper was opened to fit into the funnel.
  6. The solution was poured slowly into the funnel, allowing the liquid to strain through the filter system.
  7. After all the liquid had passed through, the filter paper with its contents, which did not strain through, was placed on paper towels to dry.
  8. Once dry, the filter paper was weighed again, and its weight was found to be 1.9 grams.
  9. The initial weight of the filter paper (1.1 grams) was subtracted from 1.9 grams, resulting in 0.8 grams of precipitate (CaCO3).



1 g of CaCl2·2H2O was converted to moles: 0.00680 moles

Mole ratios of CaCl2·2H2O and Na2CO3: 1:1

Moles of Na2CO3 were converted to grams: 0.72 g

0.00680 moles converted to grams: 0.68 grams

Initial weight of the filter paper subtracted from final weight with precipitate: 1.9 - 1.1 = 0.8 g

Using the theoretical yield and actual yield, the percent yield was calculated: 0.8 / 0.68 = 1.176 or 117.6%

Additional Questions:

No additional questions were provided.


The experiment successfully achieved its objectives:

  1. Stoichiometry was used to predict the amount of product (CaCO3) formed in the precipitation reaction.
  2. The reactants (CaCl2·2H2O and Na2CO3) and products (CaCO3) of the reaction were measured accurately.
  3. The actual yield of CaCO3 was compared to the theoretical yield, revealing a percent yield of 117.6%.
Updated: Dec 29, 2023
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Laboratory Report: Precipitation Reaction. (2017, Jan 11). Retrieved from

Laboratory Report: Precipitation Reaction essay
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