Question

Data Table 1. Alum Data

Object	Mass (g)
Aluminum Cup (Empty)	2.4 g
Aluminum Cup + 2.0 grams of Alum	4.4 g
Aluminum Cup + Alum After 1st Heating	3.6 g
Aluminum Cup + Alum After 2nd Heating	3.4 g
Mass of Released H2O	1.0 g
Moles of Released H2O	18.006 g Not sure

Questions:

Calculate the moles of anhydrous (dry) KAl(SO₄)₂ that were present in the sample.

Calculate the ratio of moles of H₂O to moles of anhydrous KAl(SO₄)₂. Note: Report the ratio to the closest whole number.

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.

Describe any visual differences between the hydrated alum and the dried, anhydrous form.

How would the following errors affect the empirical formula for the compound? That is, will these errors cause the calculated number of moles of water in the hydrate to be artificially high or low?

The student ran out of time and did not do the second heating.

The student recorded the mass of the cup + sample incorrectly and started with 2.2 g of hydrated compound but used 2.0 g in the calculations.

Cu(II) sulfate exists as a hydrate. It has many practical uses including use as a fungicide and pesticide. When mixed with chromium and arsenic it forms the wood preservative called CCA. CCA was used in pressure treated wood to protect wood from rotting due to insects and microbial agents. Because CCA-treated wood contains toxic heavy metals, its use has been discontinued for home use and children’s play sets.

A chemist is given a sample of the CuSO₄ hydrate and asked to determine the empirical formula of it. The original sample weighed 42.75 g. After heating to remove the waters of hydration, the sample weighed 27.38 g. Determine the formula for this hydrate.

Answer 1

What you need to do here is find the weight of the water that was evolved then determine moles of water. Then find the weight of the dried (anhydrous) alum and determine moles of that, then to get the ratio of moles of alum to water.

Given 1 g of water
1/[(2 x 1.01) + 16.00] = 0.055 mole H2O

2g KAl(SO4)2
2/[39.1 + 27.0 + (2 x 32.0) + (8 x 16.0) = 0.0077 mols

0.055/0.0077 = 7.14
So it would be KAl(SO4)2.7H2O

The dried alum will be light blue in color as opposed to dark blue of hydrated alum.

If 2.2 g was taken instead of 2.0 g sample, their will be a -9 % error in the answer.

Given mass of sample (A), 42.75 g

Mass after water removal (B) 27.38 g

mass of H2O (A-B) = 15.37 g

moles of H2O = 15.37/18.01 = 0.853 mols

moles of CuSO4 = 27.38/69.559 = 0.394 mols

thus, 0.853/0.394 = 2.16 H2O

Empirical formula would be, CuSO4.2H2O