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…
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
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…
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