A Hydrate’s Empirical Formula
Aaron Meehan
Ethan
11/06/17
Kotiba 09J
I have read and agree to the terms of the Academic Honesty Statement.
_Aaron Meehan____________11/06/2017_____________________
Student Signature Date
Purpose
This experiment conducted during lab was designed so that the students would be able to formulate the hydrate’s empirical formula after heating and observing the effects of heating a hydrate salt. By taking careful measurements and the usage of stichometry, the students should be able to accomplish their goals.
Introduction
By completing the procedure, the students should be able to determine the empirical formula of our given hydrate. (1) To do this the students must heat up some hydrate and record the mass before …show more content…
Subtract the two numbers. This is showing that the water has evaporated after being heated. To find the mass of chlorine, you take the final mass of the pure copper after it has reacted with the aluminum and has gone through the vacuum filtration process and subtract it from mass measured from the anhydride’s mass taken directly after being heated and cooled. This shows that not only has all the chlorine separated from the anhydride, but shows that the copper is by itself. To determine the empirical formula of the hydride, there are a few steps to take. First you need to find the percent by mass for copper, chlorine, and water (since it is an unbroken compound in this reaction). To find the percent by mass, you need to have the mass of either cooper, chlorine, or water and divide it by the mass of the hydride as a whole. The percentages should add up to 100% or close to it due to the possibilities of error. With these values multiply them each by 100. This 100 is used to represent 110 grams of a sample of the compound. Then divide each number by their respective molar mass, to find the number of moles present in the calculated sample. Then you divide each number by the lowest number of moles from chlorine, copper, or water. You are now left with a whole number ratio, which is what an empirical formula is. (2) The two trials had different empirical formulas. They are as follows: Trial 1: 〖Cu〗_2 Cl+2H_2 O, Trial 2: Cu〖Cl〗_2+H_2
Aaron Meehan
Ethan
11/06/17
Kotiba 09J
I have read and agree to the terms of the Academic Honesty Statement.
_Aaron Meehan____________11/06/2017_____________________
Student Signature Date
Purpose
This experiment conducted during lab was designed so that the students would be able to formulate the hydrate’s empirical formula after heating and observing the effects of heating a hydrate salt. By taking careful measurements and the usage of stichometry, the students should be able to accomplish their goals.
Introduction
By completing the procedure, the students should be able to determine the empirical formula of our given hydrate. (1) To do this the students must heat up some hydrate and record the mass before …show more content…
Subtract the two numbers. This is showing that the water has evaporated after being heated. To find the mass of chlorine, you take the final mass of the pure copper after it has reacted with the aluminum and has gone through the vacuum filtration process and subtract it from mass measured from the anhydride’s mass taken directly after being heated and cooled. This shows that not only has all the chlorine separated from the anhydride, but shows that the copper is by itself. To determine the empirical formula of the hydride, there are a few steps to take. First you need to find the percent by mass for copper, chlorine, and water (since it is an unbroken compound in this reaction). To find the percent by mass, you need to have the mass of either cooper, chlorine, or water and divide it by the mass of the hydride as a whole. The percentages should add up to 100% or close to it due to the possibilities of error. With these values multiply them each by 100. This 100 is used to represent 110 grams of a sample of the compound. Then divide each number by their respective molar mass, to find the number of moles present in the calculated sample. Then you divide each number by the lowest number of moles from chlorine, copper, or water. You are now left with a whole number ratio, which is what an empirical formula is. (2) The two trials had different empirical formulas. They are as follows: Trial 1: 〖Cu〗_2 Cl+2H_2 O, Trial 2: Cu〖Cl〗_2+H_2