A. Calculate the moles of anhydrous (dry) KAl(SO4)2 that were present in the sample. Step 1: Find the mass of the dry anhydrous salt show work: Final weighing - mass of pan = mass of anhydrous salt Step 2: Find the molar mass of KAl(SO4)2 and show work here Step 3: Find the moles with work shown: grams of the dry salt divided by the molar mass = moles B. Calculate the ratio of moles of H2O to moles of anhydrous KAl(SO4)2. Note: Report the ratio to the closest whole number. Show work here: Take the value of the moles of water and divide by the moles of the salt. Round to the nearest whole number C. Write the empirical formula for the hydrated alum, based on your experimental results and answer to Question B. Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(S04)2•4H20. Show your work.
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
A. Calculate the moles of anhydrous (dry) KAl(SO4)2 that were present in the sample.
- Step 1: Find the mass of the dry anhydrous salt show work:
Final weighing - mass of pan = mass of anhydrous salt - Step 2: Find the molar mass of KAl(SO4)2 and show work here
- Step 3: Find the moles with work shown: grams of the dry salt divided by the molar mass = moles
B. Calculate the ratio of moles of H2O to moles of anhydrous KAl(SO4)2. Note: Report the ratio to the closest whole number.
- Show work here: Take the value of the moles of water and divide by the moles of the salt. Round to the nearest whole number
C. Write the empirical formula for the hydrated alum, based on your experimental results and answer to Question B.
- Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(S04)2•4H20. Show your work.
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13 Subtract the water mass from the original hydrate mass of 2.00 g to obtain the mass of anhydrous KAI(SO
14 Calculate the molecular mass of H,0 and record in Data Table 2.
Experiment 2
EEE Data Table 2
Photo 2
Photo 3
Photo 4
Data Table 2: Alum Data
Object
Mass (g)
Aluminum
2.4 g
Cup
(Empty)
Aluminum
4.4 g
Cup + 2.0
grams of
Alum
Aluminum
3.6
Cup +
Alum
After 1st
Heating
Aluminum
3.4
Cup +
Alum
After 2nd
Heating
Mass of
10g
Released
H2O
Molar
2 not sure
I
mass of
H2O
Moles of
18.006 g
Released
H2O](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F18aea133-e89a-46b3-a739-31e710ff6855%2F95a92264-e96b-49a1-85fb-ae0f5665ef99%2Fqcg6rft_processed.jpeg&w=3840&q=75)
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