Mole Formula Lab
Shawn Dubbs Lab 1
The Mole Concept of and the Chemical Formula of a Hydrate
Abstract:
The objective of this lab was to conduct an experiment to analyze the molar components in alum. This was conducted by heating the alum till the water had evaporated and then determining the number of moles for each component. Then using the these amounts to figure out the empirical formula for alum.
Results:
In order to determine how much of each separate component a total mass was taken before hand for both the aluminum cup and the alum. These weights were shown to be 4.5g for the aluminum cup and 2.0g for the alum crystals. Once the alum was placed over the heat source it began to turn to a cleared liquid and started to bubble. Three trials were conducted and with a constant time over the heat source each time. (10 mins) The results of each trial are found in Table 1 below: Mass of Aluminum Cup | 2.5g | Mass of Aluminum cup+alum | 4.5g | Mass of Aluminum cup+alum 1st heat | 3.7 | Mass of Aluminum cup+alum 2nd heat | 3.5 | Mass of Aluminum cup+alum 3rd heat | 3.4 | …show more content…
Mass of alum | .9 | |
After the three trials were conducted and the amount of solid left was measured this was the anhydrous which was .9g.
Then from this number the amount of water that was in lost was measured at 1.1 g. After this then the number of moles of anhydrous were calculated from the molar mass of anhydrous. Then the moles of water were calculated. Then the stoichiometric coefficients for each component were calculated. The results found that there was two moles of anhydrous and 35 for water. Thus the empirical formula was found to be:
Error
/Discussion:
I found it very surprising that the amount of anhydrous was less than the amount of water. I hypothesized that since the natural state of water is liquid that the amount of moles of solid would be more than the amount of water. I do not believe that this is the best way to make sure that all the water in a sample of alum is out or not. TO be more exact maybe you would use an extremely hot oven to back the water out. I believe that also by extending the time the sample was under the heat, all the water would have been able to evaporate. There was error in this lab which may have caused the reason that our stoichiometric coefficients were different than the actual empirical formula for the sample. When adding the alum crystals the whole bag spilt out into the aluminum cup, this may have caused the amount of alum that I began with to differ from the given 2.0 g. Another cause was the scale, the aluminum cup mass fluctuated between 2.5-2.7 and never stopped. It was recorded that it was 2.5g thus this could have thrown the actual amount of anhydrous to be lower or higher than expected. Finally, there was a bit of calculation error. This was caused to some minor rounding errors that the calculator that was being used had. It rounded to 8 significant places; however, the amount of moles was only accurate to one decimal place. Thus rounding and calculation error could have caused a mild fluctuation of the final results.
Questions: A) Determine the moles of anhydrous?
B) Determine the moles of Water lost?
C) Calculate the ratio of water to moles of anhydrous
D) Calculate the percent water:mass of water lost/mass of hydrated salt x100%
E) Write the empirical formulat for hydrated alum, based on your experiment.
F) How do you know that all the water of hydration has been removed from an unkown hydrate? Explain
We can never be totally sure if all the water is out of the hydrate unless the hydrate is completely burned up. There is always going to be water in the hydrate, you can only get so much out.
G) How does the experimental empirical formula of hydrated alum compare to the empirical formula of hydrated alum.
The empirical formula for the hydrated alum is different than the actual hydrated alum empirical formula in that the actual empirical formula was where the experimental empirical formula was . Source: http://webmineral.com/data/Alum-(K).shtml
The Mole Concept of and the Chemical Formula of a Hydrate
Abstract:
The objective of this lab was to conduct an experiment to analyze the molar components in alum. This was conducted by heating the alum till the water had evaporated and then determining the number of moles for each component. Then using the these amounts to figure out the empirical formula for alum.
Results:
In order to determine how much of each separate component a total mass was taken before hand for both the aluminum cup and the alum. These weights were shown to be 4.5g for the aluminum cup and 2.0g for the alum crystals. Once the alum was placed over the heat source it began to turn to a cleared liquid and started to bubble. Three trials were conducted and with a constant time over the heat source each time. (10 mins) The results of each trial are found in Table 1 below: Mass of Aluminum Cup | 2.5g | Mass of Aluminum cup+alum | 4.5g | Mass of Aluminum cup+alum 1st heat | 3.7 | Mass of Aluminum cup+alum 2nd heat | 3.5 | Mass of Aluminum cup+alum 3rd heat | 3.4 | …show more content…
Mass of alum | .9 | |
After the three trials were conducted and the amount of solid left was measured this was the anhydrous which was .9g.
Then from this number the amount of water that was in lost was measured at 1.1 g. After this then the number of moles of anhydrous were calculated from the molar mass of anhydrous. Then the moles of water were calculated. Then the stoichiometric coefficients for each component were calculated. The results found that there was two moles of anhydrous and 35 for water. Thus the empirical formula was found to be:
Error
/Discussion:
I found it very surprising that the amount of anhydrous was less than the amount of water. I hypothesized that since the natural state of water is liquid that the amount of moles of solid would be more than the amount of water. I do not believe that this is the best way to make sure that all the water in a sample of alum is out or not. TO be more exact maybe you would use an extremely hot oven to back the water out. I believe that also by extending the time the sample was under the heat, all the water would have been able to evaporate. There was error in this lab which may have caused the reason that our stoichiometric coefficients were different than the actual empirical formula for the sample. When adding the alum crystals the whole bag spilt out into the aluminum cup, this may have caused the amount of alum that I began with to differ from the given 2.0 g. Another cause was the scale, the aluminum cup mass fluctuated between 2.5-2.7 and never stopped. It was recorded that it was 2.5g thus this could have thrown the actual amount of anhydrous to be lower or higher than expected. Finally, there was a bit of calculation error. This was caused to some minor rounding errors that the calculator that was being used had. It rounded to 8 significant places; however, the amount of moles was only accurate to one decimal place. Thus rounding and calculation error could have caused a mild fluctuation of the final results.
Questions: A) Determine the moles of anhydrous?
B) Determine the moles of Water lost?
C) Calculate the ratio of water to moles of anhydrous
D) Calculate the percent water:mass of water lost/mass of hydrated salt x100%
E) Write the empirical formulat for hydrated alum, based on your experiment.
F) How do you know that all the water of hydration has been removed from an unkown hydrate? Explain
We can never be totally sure if all the water is out of the hydrate unless the hydrate is completely burned up. There is always going to be water in the hydrate, you can only get so much out.
G) How does the experimental empirical formula of hydrated alum compare to the empirical formula of hydrated alum.
The empirical formula for the hydrated alum is different than the actual hydrated alum empirical formula in that the actual empirical formula was where the experimental empirical formula was . Source: http://webmineral.com/data/Alum-(K).shtml