a student paid little attention to the size of his burner flame while heating a 0.979 gram sample of calcium oxalate monohydrate. The final mass of the cooled solid was 0.670 gram. Was the reaction product anhydrous calcium oxalate?
You want to convert 0.979 g CaC2O4.H2O to mass CaC2O4 an compar with 0.670g.
Convert 0.979 g CaC2O4.H2O to mols CaC2O4.H2O, then to mols CaC2O4, then to grams CaC2O4.
0.979g CaC2O4.H2O x (1 mol CaC2O4.H2O/molar mass CaC2O4.H2O) x (1 mol CaC2O4/1 mol C2O4.H2O) x (molar mass CaC2O4/1 mol CaC2O4) = ? g CaC2O4. Compare with 0.670 g.
Well, it seems like that student's attention went up in smoke! 🙃 Given the decrease in mass upon cooling, it suggests that the reaction product is indeed anhydrous calcium oxalate. Looks like the student accidentally conducted a bit of a disappearing act! 🎩💨
To determine whether the reaction product is anhydrous calcium oxalate, we need to compare the initial and final masses of the sample.
1. Calculate the mass of water lost:
Initial mass of sample = 0.979 grams
Final mass of solid = 0.670 grams
Mass of water lost = Initial mass - Final mass
= 0.979 grams - 0.670 grams
= 0.309 grams
2. Calculate the moles of water lost:
The molar mass of water is 18.015 g/mol.
Moles of water lost = Mass of water lost / Molar mass of water
= 0.309 grams / 18.015 g/mol
= 0.017 moles
3. Calculate the moles of calcium oxalate monohydrate:
The molar mass of calcium oxalate monohydrate is 146.113 g/mol.
Moles of calcium oxalate monohydrate = Mass of sample / Molar mass of calcium oxalate monohydrate
= 0.979 grams / 146.113 g/mol
= 0.0067 moles
4. Compare the moles of water lost to the moles of calcium oxalate monohydrate:
The ratio of moles of water lost to moles of calcium oxalate monohydrate should be 1:1 for hydrated calcium oxalate.
Moles of water lost / Moles of calcium oxalate monohydrate = 0.017 moles / 0.0067 moles
≈ 2.54
Since the moles of water lost to the moles of calcium oxalate monohydrate ratio is approximately 2.54 (not 1:1), the reaction product is not anhydrous calcium oxalate. The hydrated form of calcium oxalate monohydrate decomposed during heating, resulting in the loss of water.
To determine if the reaction product was anhydrous calcium oxalate, we need to compare the masses of the hydrated calcium oxalate with the final mass of the cooled solid.
First, let's calculate the mass of water lost during the reaction:
Mass of water lost = Initial mass of sample - Final mass of cooled solid
= 0.979 g - 0.670 g
= 0.309 g
Now, we can calculate the molar mass of water (H2O) and calcium oxalate monohydrate (CaC2O4·H2O) to determine the theoretical mass of water in the hydrated compound.
Molar mass of H2O = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Molar mass of CaC2O4·H2O = 40.08 g/mol + 2(12.01 g/mol) + 4(16.00 g/mol) + 18.02 g/mol = 146.12 g/mol
Next, we can calculate the theoretical mass of water in 0.979 g of calcium oxalate monohydrate:
Theoretical mass of water = (18.02 g/mol / 146.12 g/mol) * 0.979 g
= 0.120 g
Comparing the mass of water lost (0.309 g) with the theoretical mass of water (0.120 g), we can see that the mass of water lost is greater than the theoretical mass, suggesting that the product is not anhydrous calcium oxalate.
Therefore, we can conclude that the reaction product is not anhydrous calcium oxalate.