# Percentage Yield Experiment

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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
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

• ## Separation Of Copper Sulfate Essay

Title: Separations of a Mixture Notes: - H¬2O, Copper Sulfate, Starch - H2O (water)- liquid, solvent, boiling point: 100 C, freezing point: 0 C. - Copper Sulfate- solid, soluble, forms an aqueous solution, is a salt. - Starch- solid, macromolecule (long chain sugar), not very soluble in water at low temperatures. - When H2O, Copper Sulfate, and starch are mixed together, it forms a heterogeneous mixture. - Aqueous solution of copper sulfate with a suspension of starch. Research Question: What effect does the use of the separation techniques of filtration for the separating of the starch and evaporation for separating the water, have on the mass of the remaining substance, copper sulfate?…

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• ## Experiment 7 Electrophilic Aromatic Substitution Of Salicyamide

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• ## Analysis Of Meta-Methyl Nitrobenzoate

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• ## Questions: Common Phs For Anions And Cations

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• ## The Saponification Process

It also provided better foam and water-softening in the liquid hand soap. The safe range of pH for soap is between 6 and 9 according to Hornsey (2014). When the pH is above 10, it indicates that the soap is harsh or lye-heavy. This means that it is non-reactant to oils and can burn or irritate the human skin. After the saponification process, sodium chloride or table salt was added to precipitate the soap because soap is less soluble in salt water.…

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• ## Lab Analysis Of An Acid-Base Titration

This type of reaction also known as neutralization reaction which means formation of water. It normally occurs at pH 7. Experiment 2: The aim of standardizing a solution of sodium thiosulfate pentahydrate is to find its exact concentration from the approximate given value. On titrating, the color changes from colorless to pale yellow and only then 2 drop of starch is added to the solution to become dark blue. The reason why starch was added close to the end point is to prevent the production of starch-iodine complex over time.…

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• ## Example Of Acid Titration

Title- Experiment #9- Acid Base Titrations II: Potentiometric and Indicator Titrations Introduction- In this experiment, the reactions of three different acids with a strong base are studied by observing how the pH of the solution changes as the titrant is added. A plot of pH versus amount of titrant added will be constructed. Such a curve is called a pH titration curve. The equivalence point of the reaction will be determined from an analysis of the pH titration curve, and by the use of acid-base indicators. The pH measures the solution whether it is acidic and basic.…

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• ## Langmuir Isotherm Analysis

The parameter KC can be calculated using the following equation as described in the reference [29]: Where Ce is the equilibrium EBT dye concentration on the adsorbent (mg/L) and qe is the solid phase concentration (mg/g). The values of ΔH0 and ΔS0 were calculated from the slope of the linear plot of Ln (KC) versus 1/T (see Figure 9) and the free energy change ΔG0 was calculated at different temperatures and are given in Table 3. We consider firstly the adsorption of EBT dye onto P – PAN adsorbent, the negative values of ΔG0 has decreased from - 2.72 to -7.48 kJ/mol with an increase in temperature from 298 K to 328 K. This behavior indicated that the spontaneous nature of adsorption and favorable at higher temperatures. In the other hand, the positive value of ΔH0 (44.514 kJ/mol) suggest that the process was endothermic nature. While the positive value of ΔS0 (158.51 J/mol K) indicating the entropy of the system increased during the…

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• ## Equation Of Oxidation Chemistry: An Analysis Of Sodium Hypochlorate

Calculation of theoretical yield of product. The mole of pure product was calculated by dividing the grams of product by the total gram found in one mole of 3-hydroxymethyl-4-heptanone (Equation 3). (0.0545 g 3-hydroxymethyl-4-heptanone )/1 × (1 mol 3-hydroxymethyl-4-heptanone )/(144.23 g 3-hydroxymethyl-4-heptanone ) =0.00038 mol 3-hydroxymethyl-4-heptanone of pure product (3) Equation 3. Calculation of mole of pure product. Since 0.00326 mole of product was expected, dividing the actual yield of product by the expected yield would give the percent yield when it was multiplied by 100.…

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• ## An Analysis Of The Colligative Properties Of Freezing Point Depression

The normalized freezing point depression for unknown three which was -2.7 °C was divided by the product freezing point depression constant of water (1.86) and the Van’t Hoff factor which was one for both fructose and sucrose. This calculation provided the value for molality, which was 1.45 m/kg. The molality was then converted to moles by multiplying 1.45 times the kg of solvent, which was 0.02 kg. The moles were calculated to be 0.0029. The same calculation was done for unknown four where the normalized freezing point depression was -1.6 °C this was then divided by the product of the freezing point constant of water and the Van’t Hoff Factor.…

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