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steposvetlana [31]
3 years ago
7

Diazomethane has the following composition by mass: 28.57% C, 4.80% H, and 66.64% N. The molar mass of diazomethane is 42.04 g/m

ol. Find the molecular formula of diazomethane.
Chemistry
1 answer:
Alex777 [14]3 years ago
8 0

Answer:

CH2N2

Explanation:

To find the molecular formula, we must first find the empirical formula as follows:

28.57% C - 28.57g of Carbon

4.80% H - 4.80g of Hydrogen

66.64% N - 66.64g of Nitrogen

Next, we convert this mass values to mole by dividing by their respective atomic mass.

C = 28.57/12 = 2.38mol

H = 4.80/1 = 4.80mol

N = 66.64/14 = 4.76mol

Next, we divide each mole value by the smallest mole value (2.38mol)

C = 2.38mol ÷ 2.38 = 1

H = 4.80mol ÷ 2.38 = 2.01

N = 4.76mol ÷ 2.38 = 2

The empirical ratio of C, H and N is therefore 1:2:2. Hence, the empirical formula is CH2N2

To calculate the molecular formula;

(CH2N2)n = 42.04 g/mol

{12 + 1(2) + 14(2)}n = 42.04

{12 + 2 + 28}n = 42.04

{42}n = 42.04

n = 42.04/42

n = 1.00009

Since n = 1, molecular formula is CH2N2

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The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
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Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

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