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3 years ago

a substance is analyzed to have a percent composition of 74.186% sodium and 25.814% fluorine. calculate the empirical formula

Chemistry

1 answer:

Novosadov [1.4K]3 years ago

5 0

Answer:

Na₂₆F₁₁

Explanation:

We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:

74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na

25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F

Divide each by smallest number of moles:

3.2255/1.3586 = 2.37

1.3586/1.3586 = 1

Multiply by common number to get a smallest whole number:

2.37*11 = 26,

1*11 = 11

The empirical formula is Na₂₆F₁₁

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Calculate [H3O+] and [OH−] for each of the following solutions at 25 ∘C given the pH. pH= 8.74, pH= 11.38, pH= 2.81

Gnom [1K]

Answer:

Explanation:

Given parameters;

pH = 8.74

pH = 11.38

pH = 2.81

Unknown:

concentration of hydrogen ion and hydroxyl ion for each solution = ?

Solution

The pH of any solution is a convenient scale for measuring the hydrogen ion concentration of any solution.

It is graduated from 1 to 14

pH = -log[H₃O⁺]

pOH = -log[OH⁻]

pH + pOH = 14

Now let us solve;

pH = 8.74

since pH = -log[H₃O⁺]

8.74 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{8.74}$

[H₃O⁺] = 1.82 x 10⁻⁹mol dm³

pH + pOH = 14

pOH = 14 - 8.74

pOH = 5.26

pOH = -log[OH⁻]

5.26 = -log[OH⁻]

[OH⁻] = 10 $^{-5.26}$

[OH⁻] = 5.5 x 10⁻⁶mol dm³

2. pH = 11.38

since pH = -log[H₃O⁺]

11.38 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{11.38}$

[H₃O⁺] = 4.17 x 10⁻¹² mol dm³

pH + pOH = 14

pOH = 14 - 11.38

pOH = 2.62

pOH = -log[OH⁻]

2.62 = -log[OH⁻]

[OH⁻] = 10 $^{-2.62}$

[OH⁻] =2.4 x 10⁻³mol dm³

3. pH = 2.81

since pH = -log[H₃O⁺]

2.81 = -log[H₃O⁺]

[H₃O⁺] = 10⁻ $^{2.81}$

[H₃O⁺] = 1.55 x 10⁻³ mol dm³

pH + pOH = 14

pOH = 14 - 2.81

pOH = 11.19

pOH = -log[OH⁻]

11.19 = -log[OH⁻]

[OH⁻] = 10 $^{-11.19}$

[OH⁻] =6.46 x 10⁻¹²mol dm³

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3 years ago

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