• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Determining the Percent Yield of Calcium Carbonate

Extracts from this document...

Introduction

Determining the Percent Yield of Calcium Carbonate Purpose: To compare the theoretical amount to the actual amount of calcium carbonate and calculate its percent yield in the reaction between solutions of sodium carbonate and calcium chloride. Materials: stirring rods fine filter paper 2 small beakers 0.8 - 1.2 g of sodium carbonate Erlenmeyer flask electronic balance graduated cylinder safety glasses 2 g of calcium chloride deionized water funnel retort stand ring clamp Safety Procedures: Sodium carbonate and calcium chloride are both harmful if swallowed or inhaled and may cause irritation to skin, eyes and respiratory tract. Safety goggles, and proper attire are to be worn at all times during this experiment. In particular, no loose clothing should be worn and long hair should be tied back. Procedure: 1. ...read more.

Middle

The filter paper was measured for mass. Results: Table 1: Masses of the Materials Material Mass (g) Filter Paper 1.77 Beaker 1 50.53 Beaker 1 with sodium carbonate 51.71 Beaker 2 101.46 Beaker 2 with calcium chloride 103.44 Filter paper with product 2.93 Table 2: Qualitative observations of the reagents before and after the reaction Course of time in the experiment Observations Before * sodium carbonate : white, powdery, odorless anhydrous salt that had a slower rate of dissolving in water and required extra water to completely dissolve; the solution was clear and transparent * calcium chloride: odouless white granules that dissolved easily in water; the solution was translucent During * as soon as the calcium chloride solution was poured into the sodium carbonate solution, a white, powdery precipitate was formed After * the residue in the filter paper was a white solid, powdery Calculations: Sample Calculation for determining ...read more.

Conclusion

Similarly, there are 1.11 � 10-2 moles of sodium carbonate in the experiment. NNa2CO3 < N CaCl2 Therefore, Na2CO3 is the limiting reactant. Number of moles of CaCO3 produced= Number of moles of Na2CO3 = 1.11 � 10-2 mol Mass of CaCO3= number of moles of CaCO3� molar mass of CaCO3 1 mol = 1.11 � 10-2 mol � 100.09 g 1 mol = 1.11 g Therefore, the theoretical mass of is CaCO3 1.11 g. Percent yield = experimental yield � 100% Theoretical yield = 1.16 � 0.02 g � 100% 1.11 g = 104.50 % Therefore, the percent yield is 104.50 %. Percent Error = | theoretical yield - experimental yield| � 100% theoretical yield = |1.11 g - 1.16 g| � 100% 1.11 g = 4.50% Therefore, the percent error in this experiment is 4.50%. ?? ?? ?? ?? - 1 - ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Chemistry section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Chemistry essays

  1. Determining the mass of calcium carbonate obtained

    To find the theoretical mass of calcium carbonate, firstly we have to find the limiting reagent in the reaction. The mole ratio from the equation is CaCl2 : Na2CO3 1 : 1 The actual mole ratio of reagents present is Mass in g - 4 : 6 Molar mass in

  2. Chemistry Extended Essay - Viscosity of Xanthan Gum solutions

    of the cathode in an electrolytic purification reaction involving a copper anode and a copper cathode in a copper sulphate solution Weights of cathode and anode before and after electrolytic reaction: Anode mass(g) before electrolytic reaction Anode mass(g) after electrolytic reaction Solution 1 (0%)

  1. The aim of the experiment is determining the percentage yield of the product (copper), ...

    HYPOTHESIS The assumed mass of copper is grams (theoretical yield), hence anything other than this explains that an error occurred during the experiment. If the actual mass of copper is within the range of grams, then the percentage yield is as close to 100% as possible and that the level of experimental errors is acceptable.

  2. FInding the percentage purity of CACO3 in egg shell

    4.931g Pipette solution (±0.060cm3) 25.00 cm3 Burette solution (±0.020cm3) 50.00 cm3 Volumetric flask (±0.120cm3) 250.0cm3 Table shows the chemicals that were used and there respective volumes Chemicals used Volume of chemicals used CaCO3 solution (±0.12 cm3)

  1. Aim: To estimate volumetrically the amount of Calcium carbonate present in the eggshell

    10.5/1000(dm3) X 204.23(gm) Uncertainty: 2/ (10.5/1000±0.1) X 204.23 = so total uncertainty is 0.1 The uncertainty is kept as 0.1 for the volumes and Concentration as

  2. Measuring the fatty acid percentage of the reused sunflower oil after numerous times of ...

    as it is going to be Molarity number of moles volume M n V alcoholic. For 0.01M: 10 0.01 number of moles 1L n 0 .01 0 .01 moles ASLAN Özge Cemre D129077 0.01 moles of KOH x 56 .109 g KOH 0.561 0.001 grams of KOH 1 1 mole of KOH must be used for 0.01M solution.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work