Question

100 Points!

A student reacted 0.3000 grams of aluminum with oxygen gas and obtained 0.5333 grams of aluminum oxide. Based upon the experimental data, determine the experimental empirical formula of this aluminum oxide.

Answer 1

Answer:

Empirical Formula = Al3O4 .

Explanation:

Mass of Aluminium = 0.3000 g

Mass of Oxygen in oxide = 0.5333 g - 0.3000 g = 0.2333 g

Moles of Aluminium = Mass / MOlar Mass = 0.3000 g / 27 g = 0.01112 moles

Moles of Oxygen = Mass / Molar Mass =0.2333 g / 16 g = 0.01458 moles

Now we will divide each of the molar quantities by the smallest number of moles.

Aluminium : 0.01112 moles / 0.01112 moles = 1

Oxygen : 0.01458 moles/ 0.01112 moles =1.31

Since, the ratio between Al and O is 1:1.31 which is not the smallest whole numebr ratio. SO, we will multiply it by the smallest number to get whole number ratios.

So, Here on multiplying the ratio by 3 ,we get Al : O = 3:4.

Answer 2

AI304 IS YOUR ANSWERRR