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Morgarella [4.7K]
3 years ago
12

Determine the empirical formula of the following compound if a sample contains 0.104 molK, 0.052 molC, and 0.156 molO;?

Chemistry
1 answer:
faltersainse [42]3 years ago
3 0

Answer:

K₂CO₃    

Explanation:

Given parameters:

Number of moles of K = 0.104mol

Number of moles of C = 0.052mol

Number of moles of O = 0.156mol

Method

From the given parameters, to calculate the empirical formula of the elements K, C and O, we reduce the given moles to the simplest fraction.

Empirical formula is the simplest formula of a compound and it differs from the molecular formula which is the actual formula of a compound.

  • Divide the given moles through by the smallest which is C, 0.052mol.
  • Then approximate values obtained to the nearest whole number of multiply by a factor to give a whole number ratio.
  • This is the empirical formula

Solution

Elements                             K                       C                    O

Number of moles            0.104                0.052            0.156

Dividing by the

smallest                       0.104/0.052     0.052/0.052  0.156/0.052

                                            2                           1                     3

The empirical formula is K₂CO₃      

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The ksp of calcium carbonate, caco3, is 3.36 × 10-9 m2. calculate the solubility of this compound in g/l.
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CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
                       CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial                Y                   -                 -
Change           -X                  +X              +X
Equilibrium      Y-X                 X                X

Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²

                Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
                    X = 5.79 x 10⁻⁵ M

Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
                                                     = 5.79 x 10⁻⁵ mol/L

Molar mass of CaCO₃ = 100 g mol⁻¹

Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
                                                 = 5.79 x 10⁻³ g/L

7 0
3 years ago
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