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3 years ago

1) Chemical analysis of a gaseous compound show its composition to be 36.4% carbon, 57.5% fluorine, and 6.1%

Chemistry

1 answer:

Serga [27]3 years ago

5 0

Answer:

1. Empirical formula = CFH₂

Molecular formula = C₉F₉H₁₈

2. Empirical formula is CH₂Cl

Molecular formula = C₂H₄Cl₂

3. Empirical formula = CFH₂

Molecular formula = C₉F₉H₁₈

Explanation:

1) Given data

Percentage of carbon= 36.4%

Percentage of fluorine = 57.5%

Percentage of hydrogen = 6.1%

mass = 296 g

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of C = 36.4/12 = 3.03

Number of gram atoms of F = 57.5/19 = 3.03

Number of gram atoms of H = 6.1/ 1 = 6.1

Atomic ratio:

C : F : H

3.03/3.03 : 3.03/3.03 : 6.1/3.03

1 : 1 : 2

C : F : H = 1 : 1 : 2

Empirical formula is CFH₂

Now we calculate Molecular formula

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 296 / 32

n = 9.25 we take it as 9

Molecular formula = n (empirical formula)

Molecular formula = 9 ( CFH₂)

Molecular formula = C₉F₉H₁₈

Check the answer by calculating the molecular mass, it should be almost 296 g.

Molecular formula = C₉F₉H₁₈

molecular mass = 12 (9) + 18.998 (9) + 1 (18)

molecular mass = 108 + 170.98 + 18

molecular mass = 296.98 g/mol

2) Given data

Percentage of carbon= 24.3%

Percentage of hydrogen = 4.1%

Percentage of chlorine = 71.6%

molecular mass = 99.8 g/mol

Molecular formula = ?

Solution:

Number of gram atoms of C = 24.3/12 = 2

Number of gram atoms of H = 4.1/1 = 4.1

Number of gram atoms of Cl = 71.6/35.5 = 2

Atomic ratio:

C : H : Cl

2/2 : 4/2 : 2/2

1 : 2 : 1

C : H : Cl = 1 : 2 : 1

Empirical formula is CH₂Cl

Now we calculate Molecular formula

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 99.8 / 49.5

n = 2.02 we take it as 2

Molecular formula = n (empirical formula)

Molecular formula = 2 ( CH₂Cl)

Molecular formula = C₂H₄Cl₂

Check the answer by calculating the molecular mass, it should be almost 99.8 g.

Molecular formula = C₂H₄Cl₂

molecular mass = 12 (2) + 1 (4) + 2 (35.5)

molecular mass = 24 + 4 + 71

molecular mass = 99 g/mol

3) Given data

Percentage of carbon= 36.4%

Percentage of fluorine = 57.5%

Percentage of hydrogen = 6.1%

mass = 296 g

Empirical formula = ?

Molecular formula = ?

Solution:

Number of gram atoms of C = 36.4/12 = 3.03

Number of gram atoms of F = 57.5/19 = 3.03

Number of gram atoms of H = 6.1/ 1 = 6.1

Atomic ratio:

C : F : H

3.03/3.03 : 3.03/3.03 : 6.1/3.03

1 : 1 : 2

C : F : H = 1 : 1 : 2

Empirical formula is CFH₂

Now we calculate Molecular formula

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 296 / 32

n = 9.25 we take it as 9

Molecular formula = n (empirical formula)

Molecular formula = 9 ( CFH₂)

Molecular formula = C₉F₉H₁₈

Check the answer by calculating the molecular mass, it should be almost 296 g.

Molecular formula = C₉F₉H₁₈

molecular mass = 12 (9) + 18.998 (9) + 1 (18)

molecular mass = 108 + 170.98 + 18

molecular mass = 296.98 g/mol

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Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10.

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Answer:

[Ag⁺] = 1.3 × 10⁻⁵M

s = 3.3 × 10⁻¹⁰ M

Because the common ion effect.

Explanation:

1. Value of [Ag⁺] in a saturated solution of AgCl in distilled water.

The value of [Ag⁺] in a saturated solution of AgCl in distilled water is calculated by the dissolution reaction:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

The ICE (initial, change, equilibrium) table is:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0

C -s +s +s

E X - s s s

Since s is very small, X - s is practically equal to X and is a constant, due to which the concentration of the solids do not appear in the Ksp equation.

Thus, the Ksp equation is:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × s

Ksp = s²

By substitution:

1.8 × 10⁻¹⁰ = s²

s = 1.34 × 10⁻⁵M

Rounding to two significant figures:

[Ag⁺] = 1.3 × 10⁻⁵M ← answer

2. Molar solubility of AgCl(s) in seawater

Since, the conentration of Cl⁻ in seawater is 0.54 M you must introduce this as the initial concentration in the ECE table.

The new ICE table will be:

AgCl(s) ⇄ Ag⁺ (aq) + Cl⁻ (aq)

I X 0 0.54

C -s +s +s

E X - s s s + 0.54

The new equation for the Ksp equation will be:

Ksp = [Ag⁺] [Cl⁻]

Ksp = s × ( s + 0.54)

Ksp = s² + 0.54s

By substitution:

1.8 × 10⁻¹⁰ = s² + 0.54s

s² + 0.54s - 1.8 × 10⁻¹⁰ = 0

Now you must solve a quadratic equation.

Use the quadratic formula:

$s=\dfrac{-0.54\pm\sqrt{0.54^2-4(1)(-1.8\times 10^{-10})}}{2(1)}$

The positive and valid solution is s = 3.3×10⁻¹⁰ M ← answer

3. Why is AgCl(s) less soluble in seawater than in distilled water.

AgCl(s) is less soluble in seawater than in distilled water because there are some Cl⁻ ions is seawater which shift the equilibrium to the left.

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<h3>Further explanation</h3>

Given

Reaction(balanced)

2Na + Cl₂⇒ 2NaCl

20 g Na

10 g Cl₂

Required

Excess reactant

NaCl produced

Solution

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mol Cl₂(MW=71 g/mol):

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mol : coefficient :

Na = 0.87 : 2 = 0.435

Cl₂ = 0.141 : 1 = 0.141

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Excess reactant : Na

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