(0.050) (0.1) = 0.0083 moles b. Pour 8.3 mL of the stock solution to get the amount needed. c. Measure out 8.3 mL in a graduated cylinder 8. Exercise 8: a. 41.8 mL are used b. 0.00079 moles EDTA4- c. 0.00079 moles ZnI2 d. 0.0517 grams of ZnI2 are in the sample e. 0.0517/0.237= 21.8% f. Error Is 6.34% Lab Report: Part 1: In this lab we used the following supplies: * Zinc Iodide * Na2H2EDTA(s) * Calmagite indicator solution * pH 10 buffer solution * 6M Acetic Acid * Unknown Zinc Compound The main purpose for this part of the lab was to determine the amount of zinc ion in a sample of ZnI2 by titration.
In the first part, five 100 mL flasks of 5 mL ligand solution, 5 mL 2 M sodium acetate, 4 mL 3 M NH2OH, and 1-5 mL Fe2+ solution are diluted with water. The absorption spectrum for varying concentrations of Fe2+ are measured using a spectrophotometer and the data is graphed in Excel. The slope of the line is ε in the Beer-Lambart equation A = εcl. In the second part of the experiment, eleven flasks containing diluted stock solutions of Fe2+ and ligand are mixed with 5 mL 2 M sodium acetate and 4 mL 3 M NH2OH and diluted with water. The absorption spectrum is measured using a spectrophotometer and the data is graphed in Excel.
0.05mol/6M=8.3*10-3 L=8.3mL stock solution c. 100mL-8.3mL=91.7mLwater Add 91.7 water to 6M stock solution to prepare 0.5M acetic acid. Exercise 8: a. 42.35 - 0.55 = 41.8 mL b. The moles of EDTA4- : 0.0189M*(41.8*10-3)L=7.9*10-4mol c. Zn2+(aq)+EDTA4-(aq)—Zn(EDTA)2-(aq) The ratio Zn2+ and EDTA4- is 1:1 The moles of Zn2+= the moles of EDTA4-=7.9*10-4mol d. 7.9*10-4mol*65.39g/mol=0.0517g Zn e.
From your three trials, calculate the average volume of Na2S2O3 needed for the titration of 25.00mL of diluted bleach. 3. Use the average volume and the molarity of Na2S2O3 to determine the molarity of the diluted bleach. (Find moles of Na2S2O3, convert to moles of NaClO, and divide by volume of dilute bleach that was titrated in each trial to get M). 4.
2) Percent recovery for isolation of benzoic acid % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.43/1.01) x100% = 42.57% That concludes that the percent recovery is 42,57%. 3) Percent recovery for isolation of hydroquinone dimethyl ether % Recovery = mass of recovered material _________________________________ x100% mass of starting material = (0.16/1.01) x100% = 15.84% That concludes that the percent recovery is 15.84%. Table 2: : Experimental IR peaks compared to literature IR peaks for Benzoic acid Functional groups | Experimental peak (cm-1) | Literature peak (cm-1) | O-H | 3407-2563 | 3400-2564 | C=O | 1689 | 1689 | C-H |
NaOH (aq) + KHP (aq) —› Na+ (aq) + K + (aq) + P2- (aq) + H2O (l) NaOH (aq) + CH3COOH (aq) —› Na+ (aq) + CH3COO- (aq) + H2O (l) The titration of NaOH with KHP will identify the concentration of the NaOH provided. KHP will be weighed and its mass will be recorded. The moles of KHP will be calculated: Moles KHP = Mass KHP recorded / Molar mass of KHP = Mass of KHP recorded / 204.22g/mol The volume of NaOH will be determined by reading the
The following mistakes were made when carrying out the experiment. What effect does each have on the calculated molar mass? Be specific. For example, too large because… Only part of the pipet was immersed in the boiling water, so the temperature in part of the pipet was less than that of the water bath. If the temperature was less than the water bath in some places because only part of the pipet was immersed in the boiling water, the molar mass calculated would become lower.
Repeat the titration until there are two titres within 0.1cm3 of each other. Record results in a suitable table. Results: Rough Titre: 7.653 First Run: 6.553 Second Run: 6.453 Third Run: 6.553 Calculations: During the titration, iron(II) ions are oxidised to iron(III) ions and manganate(VII) ions are reduced to manganese(II) ions. The equation is as follows: 5Fe2+(aq) + MnO4-(aq) + 8H+(aq) ? 5Fe3+(aq) + Mn2+(aq) + 4H2O(l) The above equation shows that one mole of manganate(VII) ions reacts with 5 moles of iron(II) ions in acid solution.
- Test tube - Distilled water - Volumetric flask. - LP4 tubes First 25ml of 0.1M Cupric Sulphate was made where the amount required was found in the following way... n=CV n-moles C-concentration V- Volume n= 0.1 x 25 x 10-3 =2.5 x 10-3 moles Then n = w/m n-moles w-mass required m- molecular weight 2.5 x 10-3 = w/249.71 molecular weight of CUSO4.5H2O=249.71 W = 0.624g Then as usual the solution was made by using the measured Cupric sulphate with the addition of water up to the point of 25ml in the volumetric flask. Then from the 0.1M stock solution of Cupric Sulphate the series of dilution was done as in the following table. The below mentioned volume of cupric sulphate was mixed with the corresponding volume of water as the final volume would be 2ml. The volume of cupric sulphate required for each concentration was calculated in the following way.
The solution was pooled, concentrated and redissolved in 1.0 ml methanol. The samples were centrifuged at 10,000 rpm before HPTLC analysis. The referred chemical is Morphine, Codeine, Thebaine, Papaverine and Narcotine were provided from Govt. Opium & Alkaloid Works, Neemuch, India. The purity of each opiate was ~98% by HPLC peak and normalization method.