Question

An organic compound has c 57.82% , h 3.6% and o 38.58% calculate its empirical formula

Answer 1

Hello Dear.

Here is the answer---

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Refers to the attachment for the Question,

The Calculations of the Empirical Formula is shown in the attachment.


Empirical Formula = C₄H₃O₂

Steps of this Calculations are ⇒

1) Make a Table which have 4 Rows and 6 Columns.

2) ⇒ In First Columns write the Name of the elements.
⇒ In second Columns write the Given % Composition,
⇒ In third column write the atomic mass of the elements,
⇒ In fourth columns, Calculate the Number of atoms by dividing the atomic masses of the elements with the  given % Composition.
⇒ In 5th columns, write the Simplest Ratio.

3) In the Final Column, Calculate the Whole number ratio which cannot be in Decimal. In the Given Question, Since, the Simplest Ratio of the Hydrogen is 1.5 thus we have to Multiply the whole ratio by 2.

4) Finally Calculate the Empirical Formula, by substituting the Whole number ratio into the Each elements.Thus, Empirical Formula of the Compound will be C₄H₃O₂.



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Hope it helps.


Have a Marvelous Day.

Answer 2

The Calculations of the Empirical Formula is shown in the attachment.

Empirical Formula = C₄H₃O₂

Steps of this Calculations are ⇒

1) Make a Table which have 4 Rows and 6 Columns.

2) ⇒ In First Columns write the Name of the elements.

⇒ In second Columns write the Given % Composition,

⇒ In third column write the atomic mass of the elements,

⇒ In fourth columns, Calculate the Number of atoms by dividing the atomic masses of the elements with the  given % Composition.

⇒ In 5th columns, write the Simplest Ratio.

3) In the Final Column, Calculate the Whole number ratio which cannot be in Decimal. In the Given Question, Since, the Simplest Ratio of the Hydrogen is 1.5 thus we have to Multiply the whole ratio by 2.

4) Finally Calculate the Empirical Formula, by substituting the Whole number ratio into the Each elements.Thus, Empirical Formula of the Compound will be C₄H₃O₂