Chemistry, asked by pandyajimmy273, 1 month ago

In an organic compound, hydrogen is 2/9 times of C present and oxygen is 4 times of hydrogen present by mass. The empirical formula of the compound is ....​

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Answers

Answered by dzeo225
0

Answer:

C3H8O2

Explanation:

Its the right answer

Answered by juhipandey1000
0

Answer:

C3H802

Explanation:

To find the Empirical Formula of any compound, we just need to apply basic unitary method whenever there are fractions of the compounds provided in the question.

As it's given, Hydrogen Present in the Compound is 2/9 times the amount of Carbon Present by Mass. From this we can deduce how much Hydrogen is present for 1 atom of carbon.

Hydrogen Present = 2/9 times of Carbon by mass

Hydrogen Present= 2/9 × 12

Hydrogen Present= 8/3 g per atom of Carbon

So for each Atom of carbon present, there's 8/3 atoms of Hydrogen. When finding the Empirical Formula, we remove all the fractions present by simply multiplying the number of atoms present for each element.

So for 3 Carbon atoms present, there will be 8 Hydrogen atoms in the compound.

It's also given that Oxygen is 4 times of Hydrogen Present by mass.

Oxygen Present = 4 times of Hydrogen Present by Mass

Oxygen Present = 4 × 8 = 32g

Molar mass of Oxygen is 16g, therefore 2 atoms of Oxygen must be present in the compound's Empirical Formula.

So the Empirical Formula should have 3 atoms of Carbon (C), 8 atoms of Hydrogen (H) and 2 atoms of Oxygen (O). Therefore the Empirical Formula is C3H802.

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