Chem I Lab Report Project 11

Chem I Lab Report Project 11 - Identification, Properties...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Identification, Properties and Synthesis of an unknown Ionic Compound Abstract This paper describes the identification and investigation of the properties of an unknown compound. The compound was identified as ammonium chloride through a series of physical and chemical tests. Physical properties such as appearance, smell, solubility, and pH were tested and eliminated 6 of the possible 15 unknown ionic compounds. Chemical tests for the analysis of the chloride, sulfate, iodide, bromide, nitrate, carbonate, and acetate anions were proven negative and eliminated 7 of the remaining 9 possible ionic compounds. Through the ammonium cation test, the final possible unknown ionic compound was eliminated and the unknown was identified as ammonium chloride. This finding was later verified through the synthesis of ammonium chloride. Introduction The goal of this laboratory project was to identify and synthesize a sample of an unknown ionic compound that was discovered in a local landfill. It is important that the physical and chemical properties of the compound are analyzed in order to make predictions on how it might behave in the landfill as well as protect individuals who live in the areas in close proximity of the landfill.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Results Table 1: Physical property tests. Appearance of compound: White, crystalline powder Smell: None pH test: Neutral ~ 7 Water solubility: soluble Table 2: Chemical Test for the analysis of Anions. chloride Positive sulfate Negative iodide or bromide Negative nitrate Negative carbonate Negative acetate Negative Table 3: Chemical Test for the analysis of Cations ammonium Positive Calculation 1: Sample calculation of the volume of the stock solution to dilute M 1 V 1 = M 2 V 2 V 1 = (M 2 V 2 ) / M 1 (6 M)(x) = (1M)(10 mL) V 1 = 5 mL H 2 O Calculation 2: Sample calculation of limiting reactant 1000 mL x 1 mole NH 4 OH x 1 mole NH 4 Cl x 2.0g NH 4 Cl 6M NH 4 OH 1 mole NH 4 Cl 53.5 g = 6.231 mL NH 4 OH Calculation 3: Sample calculation of percent yield Percent Yield = (actual amount / theoretical) x 100% = [(1.05 grams) / (2.00 grams)] x 100% = 52.5% yield Discussion
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 5

Chem I Lab Report Project 11 - Identification, Properties...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online