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Flame Test Lab Report

951 Words4 Pages

Glucose (A) In order to identify glucose, a conductivity test was performed. Besides glucose, in the aqueous form, all of the possible substances create charged solutions due to the dissociation of a strong acid or ionic compound. However, since both glucose and water are polar molecules, they are able to mix together to form a non charged solution, leading to a low conductivity. Following this, a caramelization test was conducted. Using previous knowledge, it’s known that simple sugars caramelize with heat, which was exhibited by substance A. Silver nitrate (B) In order to identify silver nitrate, an unknown solution was mixed with a chlorine solution. Since a precipitate was formed, the unknown solution was confirmed as silver nitrate. …show more content…

The flame test showed a red flame, meaning the unknown solution was either lithium or strontium. Next, a precipitate test with carbonate was performed to show what the cation was. There was no reaction, meaning the cation in the unknown solution was lithium. Finally, a precipitate test with silver nitrate was conducted to figure out what the anion was. It formed a white precipitate, meaning the anion was chloride. Equation for carbonate precipitate: 2Li+ (aq) + CO32- (aq) → Li2CO3 (s) Equation for silver nitrate precipitate: Ag+ (aq) + Cl- (aq) → AgCl (s) Sodium hydroxide (D) In order to identify sodium hydroxide, a flame test was conducted. The flame test showed an orange flame, meaning the unknown solution could be sodium or possibly calcium. In order to differentiate between the two, a precipitate with carbonate test was performed. The carbonate test had no reaction, meaning the cation had to be sodium. Lastly, a precipitate test with silver nitrate was conducted, and the precipitate was brown, meaning the anion was hydroxide. Also, the solution turned the indicator dark blue, indicating a strong …show more content…

The dark violet color exemplified the fact that transition metals form colored compounds. Furthermore, the solubility of silver halides test was conducted, resulting in a milky white precipitate which signifies that the anion present was chlorine. Chemical Equation: Ag+ (aq) + Cl- (aq) → AgCl (s) Sodium sulfate (J) In order to identify sodium sulfate, a flame test was conducted. Orange flame was observed which meant that the cation was most likely sodium, but possibly calcium. To test which cation it was, a carbonate precipitate test was performed. No precipitate was formed which meant that the cation was sodium. Then, a barium precipitate test was conducted, and precipitate formed which meant it was either sodium sulfate or sodium carbonate. To differentiate between the two, a carbonate + strong acid test was conducted. Since it didn’t fizz, it was found that the anion was sulfate → J was sodium sulfate. Equation for carbonate precipitate (no reaction): CO32- (aq) + 2Na+ (aq) → Na2CO3 (aq) Equation for barium precipitate: Ba2+ (aq) + SO42- (aq) → BaSO4 (s) Oxalic acid

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