Question #25b38

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Question #25b38

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Answer 1 · 2016-05-25T09:24:50

$A l_{2} {(S i O_{3})}_{3}$ $Al_2(SiO_3)_3$

Explanation:

We can proceed by calculating masses of elements X,Y and Z present in the molar mass of the compound $X_{2} {(Y Z_{3})}_{3}$ $X_2(YZ_3)_3$
using the given percentage composition of the compound.
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So mass of X present in one mole of the compound is $282.23 \times 19.12 % g = 53.96 g$ $282.23xx19.12%g=53.96g$
Since according to the formula of the comound its molecule contains 2 atoms of X ,
$the molar mass of X = \frac{53.96}{2} = 26.98 \frac{g}{mol}$ $"the molar mass of X"=53.96/2=26.98g/"mol"$
Now from periodic table it is found that X is nothing but Al.

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Similarly mass of Y in molar mass of the compound is
$282.23 \times 29.86 % = 84.27 g$ $282.23xx29.86%=84.27g$ .
As the molecule contains 3 atoms of Y,
$molar mass of Y = \frac{84.27}{3} = 28.09 \frac{g}{mol}$ $"molar mass of Y"=84.27/3=28.09g/"mol"$
According to periodic table it should be Si.

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Now finally 9 atoms of y are present in one molecule of the comound and its mass in one mole of the compound is
$282.23 \times 51.02 % \approx 144 g$ $282.23xx51.02%~~144g$
Hence $molar mass of Z = \frac{144}{9} = 16 \frac{g}{mol}$ $" ""molar mass of Z"= 144/9=16g/"mol"$
Consulting periodic table we can say the element Z is Oxygen O

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So the compound will have formula $A l_{2} {(S i O_{3})}_{3}$ $Al_2(SiO_3)_3$ which is Aluminium silicate

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Question #25b38

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