Separating a Mixture of Compounds Exp. No. 2 name Chemistry 145 February 2‚ 2014 I. Purpose The purpose of this experiment was to separate this mixture into components‚ determine the mass and percentage of each component. In addition to test the accuracy of an alternative method of subtraction to find the mass of the sodium chloride directly. Start with a mixture of sand (SiO2)‚ sodium chloride (NaCl)‚ and ammonium chloride (NH4Cl)‚ and use these separation processes: sublimation filtration
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weighed them out and obtained a net weight for the two of them as well as a mixture percentage for all four components. Data Table 1: Experiment Data | | Grams | Percent of mixture | Iron filings | 1.9 g | 28.8% | Sand | 1.3 g | 19.7% | Table salt | .6 g | 9.09% | Benzoic acid | 2.8 g | 42.4% | Total | 10.6 g | 99.99% | Data Tables: **NOTE: The initial total mass I got was 6.6 g‚ however after the experiment was complete I had a mass of 10.6 g. I calculated the percent of mixture using
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Homogenous or Heterogenous 1. Examine 5 mixtures and record your observations in a table similar to the one drawn below 2. Classify the mixtures as either homogenous or heterogenous B. Solutions 1. The salt‚ sodium chloride‚ has a solubility of 35.9g/100mL at 25 C. Determine the amount of salt that you would need if only 15mL of distilled water solvent was used. (5.885 g) 2. Using the measuring cylinder‚ measure and pout the 15mL of distilled water into one of the 100mL beakers. Measure and record
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world. It is a beautiful land of Sufi Saints‚ brave‚ hardworking‚ kind-hearted people‚ classical poets‚ writers and rich history. It has one of the oldest great Indus civilization. it holds four of the world’s highest peaks. It has the second largest salt mines in the world and also has the world’s largest deep sea. Pakistan has so many other beautiful places and resides so many great people‚ which the media totally ignores. World’s
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immediately. Lead Nitrate Stock Solution— Dissolve 159.8 mg of lead nitrate in 100 mL of water to which has been added 1 mL of nitric acid‚ then dilute with water to 1000 mL. Prepare and store this solution in glass containers free from soluble lead salts. Standard Lead Solution— On the day of use‚ dilute 10.0 mL of Lead Nitrate Stock Solution with water to 100.0 mL. Each mL of Standard Lead Solution contains the equivalent of 10 µg of lead. A comparison solution prepared on the basis of 100 µL of Standard
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standard curve of conductivity. Because water contains very few ions it does not conduct electricity very well. When NaCl solution is dissolved in water‚ the solution conducts very well‚ because the solution contains ions. The ions come from the table salt‚ whose chemical name is sodium chloride. Sodium chloride contains sodium ions‚ which have a positive charge‚ and chloride ions‚ which have a negative charge. Because sodium chloride is made up of ions‚ it is called an ionic substance. After completing
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NaBr is the bottom layer because NaBr has a density of 3210 kg/m3. For experience 4d: Question 1. Presumably I have an acidic compound and a neutral compound in the mixture. The acid is represent by RCOOH. So with the NaOH added‚ it form the sodium salt (which is soluble in water and the neutral compound is not). RCOOH goes to RCOONa. Then I separate the layers‚ add HCl to the RCOONa which forms the original acid and that is what is precipitating. If I have all of the acid removed from the mixture
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remained on the filter. Data: Item | Mass (grams) | Weighing Dish | 0.6 g | Dish + Mixture | 7.9 g | Net Mass of Mixture | 7.3 g | | Mass (Grams) | Percent of Mixture | Iron Fillings | 1.8g | 26.9% | Sand | 1.5g | 22.4% | Table Salt | 1.9g | 28.4% | Benzoic Acid | 1.5g | 22.4% | Total: | 6.7g | | Observations: Some things I observed during this laboratory include the magnet not picking up some of the smaller iron particles without getting really close to the solids
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Section 4 Experiment#: PROP0375 Separating the Components of a Ternary Mixture Hassan Alhammad 2554122 T.A Katherine Turner 9/11/2014 Introduction The purpose of the experiment is to separate components of mixture of sand (SiO2)‚ salt (NaCl)‚ and chalk (CaCO3). 1. CaCO3+2HCl-----CaCl2+CO2+H2O 2. CaCl2+K2CO3====CaCO3+2KCl The first reaction shows the separation of calcium carbonate from the mixture. In the second reaction we use the product of the first reaction (CaCl2) and change
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mathematical error analysis was needed. Discussion This experiment was designed to acquaint one with the techniques of quantitative transfer‚ volumetric glassware and electric balances. Carefully measure the weight of the 250ml beaker as well as KCl salt without the use of weighing paper or spatulas. Add 25ml of deionized
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