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2 years ago

Please with this problem

Chemistry

1 answer:

olasank [31]2 years ago

7 0

<h3>Answer:</h3>

Empirical Formula = C₃H₈O₃

Molecular Formula = C₃H₈O₃

<h3>Solution:</h3>

Data Given:

Mass of Sample = 9.2 g

Mass of Carbon = 3.6 g

Mass of Hydrogen = 0.8 g

Mass of Oxygen = 9.2 - (3.6 + 0.8) = 4.8 g

Step 1: Calculate Moles of each Element;

Moles of C = Mass of C ÷ At.Mass of C

Moles of C = 3.6 ÷ 12.01

Moles of C = 0.2997 mol

Moles of H = Mass of H ÷ At.Mass of H

Moles of H = 0.8 ÷ 1.01

Moles of H = 0.7920 mol

Moles of O = Mass of O ÷ At.Mass of O

Moles of O = 4.8 ÷ 16.0

Moles of O = 0.3000 mol

Step 2: Find out mole ratio and simplify it;

C H O

0.2997 0.7920 0.3000

0.2997/0.2997 0.7920/0.2997 0.3000/0.2997

1 2.64 1.001

Multiply by 3,

3 7.92 ≈ 8 3

Hence, Empirical Formula = C₃H₈O₃

Step 3: Calculating Molecular Formula:

Molecular formula is calculated by using following formula,

Molecular Formula = n × Empirical Formula ---- (1)

Also, n is given as,

n = Molecular Weight / Empirical Formula Weight

Molecular Weight = 92 g.mol⁻¹

Empirical Formula Weight = 12 (C₃) + 1.01 (H₈) + 16 (O₃) = 92.08 g.mol⁻¹

So,

n = 92 g.mol⁻¹ ÷ 92 g.mol⁻¹

n = 1

Putting Empirical Formula and value of "n" in equation 1,

Molecular Formula = 1 × C₃H₈O₃

Molecular Formula = C₃H₈O₃

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Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

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First balance the main element in the reaction.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

Now balance oxygen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-+6H^+\rightarrow I^-+3H_2O$

Now balance the charge.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : $2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

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