Question

I synthesized Common Alum in my lab.

KAl(SO₄)₂ $\cdot$ 12H₂O(s)

We then had to find the water content based on the initial mass of the Common Alum subtracted from the anhydrous mass after heating.

I had 1g of the Common Alum and after heating it, I had 0.57g and I was able to find the percent of water content: 43%.

1g - 0.57 = 0.43g

0.43g / 1g = 0.43 x 100 = 43%

Now I have to calculate the percentage error for the value I got but I do not know how to do that. I know I have to subtract the approximate value from the exact value then divide by the exact value and finally multiply by 100 but I'm not sure how to find the approximate value. Does it involve the empirical formula of Common Alum and finding the molar mass of KAl(SO₄)₂ and 12H₂O?

I would appreciate the assistance and explanation! Thank you.

Answer 1

To find the percentage error, the exact mass of waters of crystallization contained in 1g of potash alum is required, from which the error is calculated as per the formula:

Percentage error = Eract mass - Experimental mass 2 x 100 Fract mass

From this, it can be inferred that the error can be both positive and negative.

To find the exact mass of water molecules in the hydrated alum sample, the moles of alum contained in 1g is required. Moles = Mass/Molar mass. So, from the molar mass of alum (= 474.37 g/mol) the moles in 1g is found to be 2.1080 mmol.

Next, since we know that each mole of potash alum contains 12 moles of water, the millimoles of water in the sample taken should be 12*2.1080 = 25.296 mmol.

The molar mass of water is 18.0153 g/mol. It will therefore occupy a mass of 25.296*18.0153 = 0.4557g, which is the exact mass of water contained in the 1g sample of common alum taken.

Thus, the percentage error for the experiment is

which amounts to 5.6396%.

Percentage error = 0.4557 – 0.43 0.4557 - X 100