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. 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. Samples of hydrated manganese perchlorate are heated to determine the amount of crystal water: Mn(ClO_4)_2 middot nH_2O rightarrow Mn(ClO_4)_2 + nH_2O Two samples of different weight are studied: m (before heating) = 1.629 g; m (after heating) = 1.142 g 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

4Al + 3O2 --------> 2Al2O3 MW Al = 27g/mol O = 16g/mol and Al2O3 = 102g/mol

so 1.5g of Al = 1.5g/27g/mol = 0.056mol of Al with stoichiometry we have to 4mol of Al gives 2 mol of Al2O3

so 0.056mol of Al x2/4 = 0.028mol of Al2O3 and 0.028mol x 102g/mol = 2.85g of Al2O3 formed

for the second reacction we have

6Al + 3O2 + 3N2 --------> 2Al2O3 + 2AlN3 i have than 1.5g of aluminun is 0.056mol of Al

we so have than 2.70g are 80% Al2O3 and 20% AlN3 so

mAlN3 = 0.2x2.70g = 0.54g and for Al2O3 m = 0.8 x 2.70g = 2.16g

nAl2O3 = 2.16g/102g/mol = 0.021mol of Al2O3 with stoichiometry we have to 2 mol of Al for one mol of Al2O3 so than mol of Al = 2x 0.021mol = 0.042mol

knowing its the mol of Al present in the nitride are Al total - Al Al2O3

0.056mol - 0.042mol = 0.014mol of Al

the m of Al present in the nitride is 0.014mol x 27g/mol = 0.38g of aluminun Al as the mass of aluminun nitride is 0.54g mass of nitride - mass of Al = m of nitrigen present

0.054g- 0.38g = 0.16g

and the N have a molecular mass MW of 14g/mol

the mol present in this compuest are 0.16g/14g/mol = 0.011mol and the mol of Al are 0.014mol

so the relation 1 to 1 as the nitride is AlN

2) for Mn(ClO4)2 we have a MW = 223.21g/mol

a) after heating m = 1.142g =) nMn(ClO4)2 = 1.142g/223/21g/mol = 5.12x10-3mol of Mn(ClO4)2

before heating m = 1.629g - 1.142g = 0.487g = m of H2O so n of H2O = 0.487g/18g/mol = 0.027mol

so n H2O/nMn(ClO4)2 are equal to number of water cristal per molecule of Mn(ClO4)2  

0.027mol/5.12x10-3mol = 5.27 the formule is Mn(ClO4)2.5H2O

b) after heating m = 6.645g =) nMn(ClO4)2 = 6.645g/223/21g/mol = 0.030mol of Mn(ClO4)2

before heating m = 9.048g - 6.645g = 2.403g = m of H2O so n of H2O = 2.403g/18g/mol = 0.13mol

so n H2O/nMn(ClO4)2 are equal to number of water cristal per molecule of Mn(ClO4)2  

0.13mol / 0.030mol = 4.33 the formule is Mn(ClO4)2.4H2O

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