A compound contains 64.3% carbon, 7.14% Hydrogen, and 28.6% Oxygen. What is the empirical formula for this substance? Оа
C₂H₂O
Oь
C9H₂O2
Ос
C₂H₂O
Od
CHO

Empirical formula is C₃H₄O

Explanation:

Empirical formula:

It is the simplest formula gives the ratio of atoms of different elements in small whole number .

Given data:

Percentage of hydrogen = 7.14%

Percentage of carbon = 64.3%

Percentage of oxygen = 28.6%

Empirical formula = ?

Solution:

Number of gram atoms of H = 7.14 / 1.01 = 7.1

Number of gram atoms of O = 28.6 / 16 = 1.8

Number of gram atoms of C = 64.3 / 12 = 5.4

Atomic ratio:

            C                  :         H            :         O

         5.4/1.8             :      7.1/1.8       :       1.8/1.8

            3                   :        4            :        1

C : H : O = 3 : 4 : 1

Empirical formula is C₃H₄O.

-be a crystalline solid

- be soluble in water

-conduct electricity

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2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

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