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

Preparation and analysis of a complex metal salt

Extracts from this document...


Experiment 7 Preparation and analysis of a complex metal salt (15-12-2006) Objective 1. To prepare a sample of copper-ammonia complex salt. 2. To determine the ammonia content in the sample of salt. 3. To determine the metal content of the sample of salt. Results A. Determination of ammonium content HCl was added in excess into the complex salt. The NH3 ligands were protonated to NH4+ and were detached from the metal cation. The volume of HCl in excess was determined by titration using NaOH. The ammonium content of the salt was determined. * Weight of the complex salt used: (6.321 - 5.896) = 0.425g * Volume of HCl (0.6M) added onto the flask: 25cm3 No. of moles of HCl added: 25/1000 x 0.6 = 0.015 moles * The reaction of the titration is given by the following chemical equation: HCl + NaOH --> NaCl + H2O (1) Volume of NaOH (0.5M) used in the titration: 20.8 - 4.3 = 16.5 cm3 (2) No. of moles of NaOH used: 16.5 / 1000 x 0.5 = 8.25 x 10-3 moles (3) No. of moles of HCl in excess = No. of moles of NaOH used in titration = 8.25 x 10-3 moles * No. ...read more.


So, the no. of moles of SO42- in the sample of salt is 1.74 x 10-3 mole * Molar mass of SO42- : 32.1 + 16 x 4 = 96.1 g/mole * Mass of SO42- in the sample = 96.1 x 1.74 x 10-3 = 0.167 g. * % by mass of SO42- in the sample = 0.167 / 0.434 x 100% = 38.5% * The metal cation, anion and ligand account for (38.5 + 25.5 + 27.0) = 91% of the weight of the salt. So, the salt consists of 9% of water. C. Determination of the chemical formula of the complex salt * % by mass of NH3 in the sample was 27.0%. * % by mass of Cu2+ in the sample was 25.5% * Assuming 100g of the salt, the no. of moles of NH3 and Cu2+ are (27 / 17) = 1.59 mole and (25.5 /63.5) = 0.40 mole respectively. * The molar ratio of NH3 to Cu2+ is (1.59 / 0.40) = 3.97, which is similar to the theoretical ratio of 4:1. * % by mass of water was 9%. Assuming 100g of the salt, there are 9 g of water in the salt sample. ...read more.


But if there is an excess amount of HCl present, it would be ensured that all the ligands are detached from the metal ion and have reacted with HCl. 2. In the preparation of the complex salt, 95% ethanol was added into the solution. Since the complex salt is less soluble in ethanol then in water, adding ethanol into the flask would make the salt precipitate. 3. After some calculations, it was found that there was 5% of water in the sample of salt. The water comes from 2 sources. First, the water molecules may form the ligands in the complex, as suggested by the chemical formula [Cu(NH3)4(H2O)1.25]SO4. Second, the solution, while being precipitated, may be crystallized and forms water of crystallization. 4. In part B of the experiment, the procedures in part A was repeated but no indicators were added. This is because the indicators would affect the color of the solution, which would in turn affect the color change during the EDTA titration. Thus, no indicator was added. 5. It was found that the ratio of water to Cu in the complex is not an integer. This may be due to the assumption that 'all the weight of the salt, apart from Cu2+, NH3 and SO42-, would be water' being wrong. There may be other substances in the complex salt. The other explanation is that the arrangement of the complex is irregular. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our AS and A Level Inorganic 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 AS and A Level Inorganic Chemistry essays

  1. Peer reviewed

    Determining the concentration of acid in a given solution

    5 star(s)

    alkali and I would think that the acid was of a higher concentration than it actually was so the titres would not be accurate. If I read the meniscus slightly below the calibration line in the volumetric flask then the concentration of sodium carbonate used would be slightly higher that I would think.

  2. effects Concentration and Temperature on the Rate of Reaction

    line, thus enabling me to work out an accurate value for the activation enthalpy. I could not easily draw the triangle on the line of best fit, which allows me to calculate the gradient, using a computer. Graph 7 shows that there is a negative correlation between the two sets

  1. Determination of Total Hardness in Water by EDTA Titration

    = 40 + (16x3) + 12 = 100 The hardness of tap water sample = (0.000102 x 100 x 1000)/ 0.025 = 40.8mg/L Discussion We have to determine the molarity of EDTA solution indirectly by titrating it against calcium chloride solution, but not dissolving a known amount of EDTA into water instead.

  2. Lab report Determination of Enthalpy Change of Neutralization

    2781.679552 / 0.05 = 55.6 KJ mol-1 The enthalpy change of neutralization is -55.6 KJ mol-1 Reaction 5: Mass of the solution = (25 + 25) /1000 = 0.05 Kg Heat given out = (0.05 * 4200 + (3/29 * 0.02073 + 3/26 * 0.02234)

  1. Energy and Rates Analysis of Chemical Reactions

    After carrying out the lab, this was proven wrong. The carboxylic acid was actually the slowest of the three. As previously discussed, this could be due to the fact that it is more polar than Hydrochloric and Sulfuric acid. This could also be due to the more simplistic structures of

  2. Determination of Chemical Oxygen Demand (COD) of a Given Sample of Waste Water

    Hence a blank sample is required in the determination of COD. A blank sample is created by adding all reagents to a volume of distilled water. COD is measured for both the water and blank samples, and the two are compared.

  1. The preparation, analysis, and reactions of an ethanedioate complex of iron

    The temperature must be kept below 50?. Then the mixture was heat to the boiling point and 2.503g of oxalic acid dihydrate was added. After stirring for 2-3 minutes, a clear green solution was obtained. Then, it was filtered using a filter paper and a long-stemmed funnel.

  2. Preparation and Standardization of 0.010 M EDTA

    Calculations Amount EDTA added plus error propagation: Amt EDTA = End volume - Start Volume = 30.39-15.19mL = 15.20 � 0.014mL Error: M Standard Zinc Solution: M = (wt. Zinc metal / atomic wt Zinc)(0.500mL)-1 = (0.5012g / 65.39g*mol-1) (0.500mL)

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