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4 years ago

Describe how oxidation and reduction involve electrons, change oxidation numbers, and combine in oxidation-reduction reactions.

2 answers:

4 0

The process of oxidation is the loss of electrons or an increase in oxidation state by a molecule, atom or ion and reduction is the gain of electrons or a decrease in oxidation state. The oxidation number is the total number of electrons that an atom gains or loses in order to form a chemical bond with another atom. Redox reactions are electron transfer processes and include all chemical reactions in which atoms have their oxidation state changed.

adell [148]4 years ago

4 0

Answer:Oxidation is the loss of electrons.

Reduction is the gain of electrons.

Oxidation increases the oxidation number.

Reduction decreases the oxidation number.

In an oxidation-reduction reaction, the electrons from the oxidized atom or ion are transferred to the atom or ion that is reduced.

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3 years ago

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3 years ago

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3 years ago

A 5.325g sample of methyl benzoate, a compound in perfumes , was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.

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Answer: The empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

Explanation:

We are given:

Mass of C = 3.758 g

Mass of H = 0.316 g

Mass of O = 1.251 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{3.758g}{12g/mole}=0.313moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.316g}{1g/mole}=0.316moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.251g}{16g/mole}=0.078moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.078 moles.

For Carbon = $\frac{0.313}{0.078}=4.01\approx 4$

For Hydrogen = $\frac{0.316}{0.078}=4.05\approx 4$

For Oxygen = $\frac{0.078}{0.078}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 4 : 4 : 1

The empirical formula for the given compound is $C_4H_4O$

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

$n=\frac{\text{Molecular mass}}{\text{Empirical mass}}$

We are given:

Mass of molecular formula = 130 g/mol

Mass of empirical formula = 68 g/mol

Putting values in above equation, we get:

$n=\frac{130g/mol}{68g/mol}=1.9\approx 2$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_{(2\times 4)}H_{(2\times 4)}O_{(2\times 2)}=C_8H_8O_2$

Hence, the empirical and molecular formula of the compound is $C_4H_4O$ and $C_8H_8O_2$ respectively

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