Unknown A Lab Report Conclusion
Conclusion: (Identity, % error, other evidence, how to improve)
The identity of Unknown A is the compound cobalt(II) chloride hexahydrate (CoCl₂•6H₂O).This is justified by the similarities in color of red-purple crystal like grains, which is significant in identifying the product out of the eight suggested identities. Only two had similar characteristics which were CoCl₂•6H₂O and CoSO₄•6H₂O. Before unknown A was heated, there was a notable common physical property of the substance being red crystal grains. After heating the substance the sample turned into a light blue powder which was significant in identifying the compound. The average % H₂O Driven off was __% ,which closer to that of the __% of CoCl₂•6H₂O, rather than the __% of CoSO₄•6H₂O. The percent error in the calculations was ______% . Through the trials involving Unknown B, it was confidently determined to be Al(SO4)3 •16H2O. The substance appeared white and powder-like before heating, while it appeared white and crystal-like after heating. Al(SO4)3 •16H2O had very similar properties of the unknown substance. Also, in the first trial of the experiment the weight percent of water was found to be 44%, in the second trial it was found to be 43%. These two trials averaged out to a weight percent of water of __% which is similar to the weight …show more content…
The most plausible explanation for such an error would be due to some overheating on the substance while completing the experiment. There may have been grains of the powder escaping as the water molecules boiled out of the substance, causing it to change the mass. Due to this predicament, the percent error does not transmit a lot of confidence on the result of the experiment having the substance identified. If given the ability to do more trials in testing the substance, an improvement in confidence would result from a lower percent
The identity of Unknown A is the compound cobalt(II) chloride hexahydrate (CoCl₂•6H₂O).This is justified by the similarities in color of red-purple crystal like grains, which is significant in identifying the product out of the eight suggested identities. Only two had similar characteristics which were CoCl₂•6H₂O and CoSO₄•6H₂O. Before unknown A was heated, there was a notable common physical property of the substance being red crystal grains. After heating the substance the sample turned into a light blue powder which was significant in identifying the compound. The average % H₂O Driven off was __% ,which closer to that of the __% of CoCl₂•6H₂O, rather than the __% of CoSO₄•6H₂O. The percent error in the calculations was ______% . Through the trials involving Unknown B, it was confidently determined to be Al(SO4)3 •16H2O. The substance appeared white and powder-like before heating, while it appeared white and crystal-like after heating. Al(SO4)3 •16H2O had very similar properties of the unknown substance. Also, in the first trial of the experiment the weight percent of water was found to be 44%, in the second trial it was found to be 43%. These two trials averaged out to a weight percent of water of __% which is similar to the weight …show more content…
The most plausible explanation for such an error would be due to some overheating on the substance while completing the experiment. There may have been grains of the powder escaping as the water molecules boiled out of the substance, causing it to change the mass. Due to this predicament, the percent error does not transmit a lot of confidence on the result of the experiment having the substance identified. If given the ability to do more trials in testing the substance, an improvement in confidence would result from a lower percent