When chemical reactions are carried out under well-defined conditions, mass diff

Question

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:

Mn(ClO4)2·nH2OaMn(ClO4)2 +nH2O 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

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.

m = 1.5 g of Al(s)

m = 2.83 g of Aluminium oxide...

m = 1.5 g Al(fine)

m = 2.70 g of Al N

a)

for AlxOy

mol of Al = mass/MW = 1.5/26.98 = 0.05559 mol of Al

mass of O = 2.83-1.5 = 1.33 g of O

mol of O = mass/MW = 1.33/16 = 0.083125 mol of O

ratio --> 0.083125/ 0.05559 = 1.5

so

AlO1.5 = Al2O3

b)

m = 2.7 g of product

80 is Al2O3, 20 is AlN

2.7*0.2 = 0.54 g of AlN

ratio

mol of Al = 0.01111

mol of N = 0.01111

ratio is 1:1

so

AlN is the formula

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.