Search for an essay or resource:

Essay: Calcium chloride

Essay details:

  • Subject area(s): Science essays
  • Reading time: 8 minutes
  • Price: Free download
  • Published: July 26, 2019*
  • File format: Text
  • Words: 2,128 (approx)
  • Number of pages: 9 (approx)
  • Calcium chloride
    0.0 rating based on 12,345 ratings
    Overall rating: 0 out of 5 based on 0 reviews.

Text preview of this essay:

This page of the essay has 2,128 words. Download the full version above.

An unknown compound was discovered in a local landfill. Investigation of the identity of the unknown compound served the purpose to determine the reactivity of the compound and its hazards to the environment to determine if it was safe or toxic for the environment and needed to be removed from the landfill. The goals included determining the identity of the compound, determining its chemical and physical properties by observing the compound and analyzing its chemical behavior, and devising two syntheses of the compound. This exploration began with preliminary observations. Preliminary observations of the sample confirmed that the unknown compound looked much like a salt and was odorless. The compound was white with a solid, crystal-like structure much like that of table salt (NaCl) just not as fine. The pH of the unknown compound proved to be 5. To test for pH, the unknown compound was dissolved in water and litmus paper was put into the solution. The litmus paper turned color and the color was compared to a pH chart. The method of investigating the identity of the unknown compound consisted of the completion of a series of tests in order to determine physical and chemical properties of the compound. The series of tests were performed to determine the solubility of the unknown compound, which cations and anions were present in the sample, its conductivity, and much more.

In order to determine the solubility of the unknown compound, qualitative and quantitative solubility tests were performed. A qualitative solubility test helps determine some initial chemical properties of the compound including its reactivity with some common natural solvents. To test the qualitative solubility, the unknown compound was added to five solvents to see which the compound dissolved in. These solvents included water (with a tested pH of 7), sodium hydroxide (NaOH), hydrochloric acid (HCl), toluene, and acetone. After stirring each mixture, observations were made. The compound dissolved in water because water molecules are polar, meaning the molecules interact through hydrogen bonds and dipole-dipole intermolecular forces, and the partially charged negative and positive areas of water molecules can break the bonds of other polar molecules or ionic molecules. Therefore, it was inferred the compound was either polar or ionic, meaning the compound is composed of anions, negatively charged ions, and cations, positively charged ions. The compound did not dissolve in the sodium hydroxide, a base, but formed a precipitate because the molecules making up the unknown compound did not dissociate in sodium hydroxide, meaning the bonds holding the molecules together were not broken due to the charges and sizes of the molecules. Therefore, it was inferred the compound was not an organic acid. An organic acid is a compound containing carbon with acidic properties, having a pH lower than 7. The compound dissolved in the hydrochloric acid indicating it could be an organic base, which is a compound containing carbon and having a pH greater than 7. When combined with acetone, the compound did not dissolve which further proved it was not nonpolar. Lastly, the compound was not soluble in acetone; it did not dissolve, so it had to be an ionic compound as stated in the title of the project and the report. See Table 2 in the group report for these results. From testing the solubility, the unknown sample was determined to be an ionic compound, containing positively and negatively charged ions and possibly an organic base.
Because the unknown compound appeared noticeably soluble in water, a quantitative solubility test was performed in order to accurately measure what amount of the compound dissolved in a given volume of a solution. 0.9 grams of the compound fully dissolved after adding 3mL of water. Using these results, the density was determined to be 0.3 g/mL. See Table 3 in the group report for these results. This provided an error. 0.3 g/mL actually verifies the solubility of the compound. The solubility is 300 g/L in 1 M H2O. To correctly measure the density of the compound, the water in the solution needed to be evaporated from the compound in order to determine how much of the compound dissolved. Then, the density could be calculated by subtracting the mass of the compound that dissolved form the initial mass, 0.9 grams.
In order to determine what cations were present in the compound, an analysis of cations was performed. A cation is a positively charged ion. One way cations can be identified is by a flame test. When electrons are heated, they become excited and jump up to a higher energy level. This excitation causes the electrons to become unstable, so they tend to try to fall back to the ground state energy level. When the electrons fall back to the ground state, energy is released in the form of light. The wavelength of this emission of light energy determines the flame coloration. The flame test showed a red flame coloration with medium to high intensity because red light is low in energy and therefore a larger wavelength than, for example, blue light. This red flame coloration meant the compound could contain calcium, strontium, or lithium as indicated in the lab manual1 on page sixty three. Strontium and lithium were ruled out because they were not options in the list given for possible identities of the compound. The flame test provided evidence for the prediction that the cation present in the unknown compound was calcium.
Another way cations can be identified is through a test for ammonium. A test for ammonium is performed to see if an ammonium ion is present in a compound. When ammonium is mixed with sodium hydroxide, the hydroxide ion releases a proton from the ammonium ion, producing ammonia and a fruity smell. The smell resulting from the mixture of the unknown compound and sodium hydroxide was odorless which indicates the absence of an ammonium ion. Ammonia was also able to be detected by heating an evaporating dish containing the unknown compound and sodium hydroxide over the flame of a bunsen burner then placing pH paper on a watch glass and placing the watch glass over the evaporating disk and observing any change in the color of the pH paper. If the pH paper darkens to blue, ammonium is present. The pH paper remained the same color indicating a pH of 7 and no ammonium ion present.
In order to determine what anions were present in the unknown compound, an analysis of anions was completed. Anion tests rely on the formation of an insoluble salt precipitate of the anion being tested for. Using five test tubes, chloride, sulfate, nitrate, carbonate, and acetate tests were performed. For the chloride test, it was observed a white precipitate had formed indicating the presence of a chloride ion (Cl-). For the sulfate test, it was observed the unknown compound had completely dissolved and no precipitate had formed. This indicated a sulfate ion (SO42-) was not present. For the nitrate test, it was observed that no precipitate had formed and the unknown compound dissolved completely and there was an absence of a nitrate ion (NO3-). The same thing happened for the carbonate test along with some bubbling indicating the absence of a carbonate ion (CO33-). For the acetate test, the mixture didn’t completely mix and a precipitate formed. However, there was no fruity smell, indicating the absence of an acetate ion. From the analysis of anions, the conclusion was made that the unknown compound contained a chloride ion. See Table 4 in the group report for these results.
In order to test for conductivity, each of the mixtures produced during the anion test were tested for using a volt meter. If a solution conducts electricity well, it is ionic, meaning the compound is made of positive and negative ions. Charged ions conduct electricity. The unknown solution dissolved in water conducted electricity well. From this, it was confirmed that the unknown sample was an ionic compound. See Table 5 in the group report for these results.
Based on the results of these series of tests, it was concluded that the identity of the unknown compound was calcium chloride (CaCl2). Calcium was the only option for the cation present, and chloride was the only option for the anion present.
To further confirm the conclusion, five double displacement reactions were carried out for both the initially unknown sample and calcium chloride. A double displacement reaction is a chemical reaction in which the anions and cations from each reacting molecule are exchanged to form different products. The reactions yielded the same products for both samples.
CaCl2 (s) + 2AgNO3 (l)  Ca(NO3)2 (l) + 2AgCl (s)
The reactants calcium chloride (CaCl2), the unknown substance, and silver nitrate (AgNO3) undergo a double displacement reaction to yield products calcium nitrate (CaNO3) and solid precipitate salt silver chloride (AgCl). The precipitate salt silver chloride wad formed because the silver ion and the chlorine ion did not dissociate.
CaCl2 (s) + 2NaOH (l)  Ca(OH)2 (s) + 2NaCl (aq)
The reactants calcium chloride (CaCl2), the unknown substance, and sodium hydroxide (NaOH), a base, undergo a double displacement reaction to yield products sodium chloride (NaCl) and solid precipitate calcium hydroxide (Ca(OH)2).
CaCl2 (s) + NaC2H3CO2  CaCO2+ NaC2H3Cl2
The reactants calcium chloride (CaCl2), the unknown substance, and sodium acetate (NaC2H3CO2), a salt of acetic acid, undergo a reaction to yield products calcium carbonate (CaCO2) and NaC2H3Cl2).
CaCl2 (s) + K2SO4 (l)  2KCl (l) + CaSO4 (s)
The reactants calcium chloride (CaCl2), the unknown substance, and potassium sulfate (K2SO4), a salt, undergo a double displacement reaction to yield products potassium chloride (KCl) and a solid precipitate calcium sulfate (CaSO4).
CaCl2 (s) + Na2C2O4 (l)  CaC2O4 (s) + 2NaCl (aq)
The reactants calcium chloride (CaCl2), the unknown substance, and sodium oxalate (Na2C2O4), a salt, undergo a double displacement reaction to yield products sodium chloride (NaCl) and a solid precipitate calcium oxalate (CaC2O4).
See Table 6 in the group lab report for these results.
CaCl2 is an ionic compound not commonly found in nature. It is fairly harmless with a LD50 of 1000 mg/kg (for a rat) according to “Calcium Chloride – Typical MSDS”2. It is not necessary nor cost effective to remove the substance from the landfill. Calcium chloride is not highly reactive and therefore not hazardous to the environment. The MSDS of calcium chloride advises avoiding contact with eyes and skin and avoiding inhalation or ingestion of the substance. If the compound becomes in contact with skin, rinse the affected area immediately. If the compound gets into the eyes, flush thoroughly with water. Calcium chloride is a low toxicity chemical and can cause can cause gastrointestinal problems and other conditions when ingested. If ingested, induce vomiting and drink plenty of water. Be sure to seek medical help. (“Calcium Chloride”)3.
After determining the identity of the compound and completing the comparison reactions, two syntheses of the compound calcium chloride were developed. To synthesize calcium chloride, combine hydrochloric acid (HCl) with either calcium hydroxide (Ca(OH)2) or calcium carbonate (CaCO3), and separate the water from the precipitate through evaporation.
Ca(OH)2 (s) + 2HCl (l)  CaCl2 (s) + 2H2O (l)
CaCO3 (s) + 2HCl (l)  CaCl2 (s) + H2O (l) + CO2 (g)
Gravimetric analysis, or vacuum filtration, is used to separate liquid from a solid precipitate in order to determine the percent yield, the ratio of an expected amount of a compound to a theoretical amount of a compound. In this case, mixing silver nitrate with the compound calcium chloride yielded calcium nitrate and precipitate silver chloride. After filtering the liquid from the precipitate and evaporating any remaining moisture, the mass of the precipitate was measured. This was done three time with the unknown sample and twice using calcium chloride. The average mass of the precipitate produced in the reaction with the unknown sample was 0.492 grams, and the average mass of the precipitate produced using calcium chloride was 0.418 grams. These were precise calculations, verifying the identity of the compound. In addition, the percent yields read 128.57% and 109.23%. Percent yield should read less than 100%. The reason for the percent yields larger than 100% could be a result of impurities in the precipitate or moisture that was not completely evaporated from the solid.
During the experiments, it is probable that measurements and calculations were off. For example, the actual density of calcium chloride is 2.15 g/cm3, not 0.3g/mL (where 1 mL = 1 cm3) and the percent yields being greater than 100% as explained previously in the report. For future investigations, it would be beneficial to perform multiple trials of the tests to obtain more accurate results.
ay in here…

About Essay Sauce

Essay Sauce is the free student essay website for college and university students. We've got thousands of real essay examples for you to use as inspiration for your own work, all free to access and download.

...(download the rest of the essay above)

About this essay:

If you use part of this page in your own work, you need to provide a citation, as follows:

Essay Sauce, Calcium chloride. Available from:<> [Accessed 25-01-22].

These Science essays have been submitted to us by students in order to help you with your studies.

* This essay may have been previously published on at an earlier date.

Review this essay:

Please note that the above text is only a preview of this essay.

Review Content

Latest reviews: