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sukhopar [10]
3 years ago
13

What is the empirical formula of a compound that contains 32.39 percent sodium, 22.53 percent sulfur, and 45.07 percent oxygen?

Chemistry
1 answer:
djverab [1.8K]3 years ago
5 0
To get the empirical formula of this compound, we take a basis of 100 grams which means each percentage is equivalent to 1 gram. Hence there is 32.39 grams sodium, 22. 53 grams sulfur and 45.07 grams oxygen. We convert each mass to their moles by dividing by their respective molar mass. Na: 1.408, S:0.704 and O:2.82. divide each with the lowest: Na: 2: S: 1 and O:4. Hence the formula is Na2SO4.
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According to Osmotic pressure equation:

π = i M R T

When π =0.307 atm & M = 0.01 mol & R (constant)= 0.0821 L-atom/mol-K &
T= 22+273 = 295 Kelvin

So Van't half vector i = π / (MRT)
                                   = 0.307 / (0.01 * 0.0821 * 295)
                                   = 1.27 

When there is no dissociation, i = no. of moles of Hf in 1 L of solution = (1-X) 
and when there is a complete dissociation so it is equal 2X according to this equation


HF(aq) + H2O (L) ⇆ H3O (aq) + F (aq)
(1-X)                                X              X

∴ i = (1-X) + (2x)

 1.27 = 1+X
∴X= 1.27 - 1 = 0.27 
∴ the percent ionization of the acid X = 27 %
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Answer:

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Hello!

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To solve this exercise, we are going to apply the common titration equation:

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Have a nice day!

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