Experiment 3: When chemical reactions are carried out under well-defined conditi

Question

Experiment 3: When chemical reactions are carried out under well-defined conditions, mass differences provide an access to chemical compositions and empirical formulas. However, when no steps are included that correct for certain aspects of the reaction, or when the reaction conditions are not carefully controlled, one might obtain erroneous chemical formulas. One example is the synthesis of metal oxides. When the combustion of a metal is carried out in an oxygen atmosphere, the only products obtained are metal oxides. However, when carried out in air, part of the metal forms metal nitrides as well. Another example is the determination of crystal water. If carefully executed, the anhydrous compound is formed, and the hydration number is most often integer. If not heated sufficiently, not all crystal water is driven out. If overheated, the product might undergo thermal decomposition. Use the following examples to illustrate these ideas. 1. When a 1.50 g sample of aluminum metal is burned in an oxygen atmosphere, 2.83 g of aluminum oxide are produced. However, the combustion of 1.50 g ultrafine aluminum in air results in 2.70 g of a product, which is a mixture of 80% aluminum oxide and 20 % aluminum nitride (% by mass). Use this information to determine the empirical formulas of aluminum oxide and aluminum nitride. 2. Samples of hydrated manganese perchlorate are heated to determine the amount of crystal water: Two samples of different weight are studied: a) m (before heating) = 1.629 g ; m (after heating) = 1.142 g b) m (before heating) = 9.048 g ; m (after heating) = 6.645 g For both samples, calculate the number of moles of crystal water per formula unit of manganese perchlorate. Decide which sample is most likely to represent the correct result, and explain what might have gone wrong with the other.

Explanation / Answer

I will answer the first question:

Ok, we got first that for 1.5 g of Al with X oxygen we got 2,83 g of the oxide

First: how much O will we use?: 2.83 - 1.5= 1.33 g of Oxygen (remember that oxide is only metal plus oxygen)

Now a rule of three:

1.5 g of Al ------------------------------------------------------- 1.33 g of Oxygen

27 g (molecular weight of 1 Al) ----------------------- x g

X= 24 g / 16 g (molecular weight of 1 O)= 1.5

This means that for every Al there will be 1.5 O, we can´t use commas so we multiplied BOTH elements by 2 to make them entire:

2 Al for every 3 O, so our formula will be: Al2O3

for the second part we have the same aluminium but in air, so part of it will go to nitride:

2.70= total product

2.70 x 80 %= 2.16 g is oxide, let´s work here first:

Molecular weight of the oxide: 27 x 2 + 16 x 3= 102 g (Al2O3)

102 g of Al2O3 ---------------------------- 27x2 Al

2.16 g of Al2O3 -------------------------   X                

X= 1.14 g of Al is used for the Oxide

SO= 1.5 - 1.14= 0.36 g of Al is used for the nitride

2.70= total product

2.70 x 20 %= 0.54 g is nitride (metal plus nitrogen)

0.54 - 0.36= 0.18 g is Nitrogen only

and finally=

0.36 g of Al ------------------------------------------------------- 0.18 g of N

27 g (molecular weight of 1 Al) ------------------------ X g

X= 13.5 g of N/ 14 (molecular weight of 1 N)= 0.96 wich we can roun up to 1

this means that for every 1 Al there is 1 N= AlN

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