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Elza [17]
2 years ago
7

Determine the molecular formula for each of the following compounds from the

Chemistry
1 answer:
bulgar [2K]2 years ago
7 0

The molecular formula for each of the following compound given the data is

  • C₂H₆
  • C₂H₂
  • C₄H₈Cl₂

<h3>A. How to determine the molecular formula </h3>

We'll begin by calculating the empirical formula. This can be obtained as follow:

  • C = 80%
  • H = 20%
  • Empirical formula =?

Divide by their molar mass

C = 80 / 12 = 6.67

H = 20 / 1 = 20

Divide by the smallest

C = 6.67 / 6.67 = 1

H = 20 / 6.67 = 3

Thus, the empirical formula of the compound CH₃

Thus, the molecular formula of the compound can be obtained as follow

  • Empirical formula = CH₃
  • Molar mass of compound = 30 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

Thus, the molecular formula of the compound is C₂H₆

<h3>B. How to determine the molecular formula </h3>

We'll begin by calculating the empirical formula. This can be obtained as follow:

  • C = 92.3%
  • H = 7.7%
  • Empirical formula =?

Divide by their molar mass

C = 92.3 / 12 = 7.7

H = 7.7 / 1 = 7.7

Divide by the smallest

C = 7.7 / 7.7 = 1

H = 7.7 / 7.7 = 1

Thus, the empirical formula of the compound CH

Thus, the molecular formula of the compound can be obtained as follow

  • Empirical formula = CH
  • Molar mass of compound = 26 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[CH]n = 26

[12 + 1]n = 26

13n = 26

Divide both side by 13

n = 26 / 13

n = 2

Molecular formula = [CH]n

Molecular formula = [CH]₂

Molecular formula = C₂H₂

Thus, the molecular formula of the compound is C₂H₂

<h3>C. How to determine the molecular formula </h3>

We'll begin by calculating the empirical formula. This can be obtained as follow:

  • C = 37.5%
  • H = 6.3%
  • Cl = 55.8%
  • Empirical formula =?

Divide by their molar mass

C = 37.5 / 12 = 3.125

H = 6.3 / 1 = 6.3

Cl = 55.8 / 35.5 = 1.572

Divide by the smallest

C = 3.125 / 1.572 = 2

H = 6.3 / 1.572 = 4

Cl = 1.572 / 1.572 = 1

Thus, the empirical formula of the compound C₂H₄Cl

Thus, the molecular formula of the compound can be obtained as follow

  • Empirical formula = C₂H₄Cl
  • Molar mass of compound = 127 g/mol
  • Molecular formula =?

Molecular formula = empirical × n = mass number

[C₂H₄Cl]n = 127

[(2×12) + (4×1) + 35.5]n = 127

63.5n = 127

Divide both side by 63.5

n = 127 / 63.5

n = 2

Molecular formula = [C₂H₄Cl]n

Molecular formula = [C₂H₄Cl]₂

Molecular formula = C₄H₈Cl₂

Thus, the molecular formula of the compound is C₄H₈Cl₂

Learn more about empirical formula:

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<h3>How to explain the reaction?</h3>

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3 years ago
How many moles of Cu(OH)2 are soluble in 1L of sodium hydroxide (NaOH) when the pH is 8.23?
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Answer:

4.96E-8 moles of Cu(OH)2

Explanation:

Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.

Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.

pH= -log[H]\\pH= -log (\frac{kw}{[OH]})

8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}

[OH]=1.69E-6

This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

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In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":

The expression for Kps is:

Kps= [Cu^{2+}] [OH]^2

The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

Kps= s*(2s+1.69E-6)^2

"s" is the soluble quantity of Cu(OH)2.

The solution for this third grade equation is s=4.96E-8 mol/L

Now, let us calculate the moles in 1 L:

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