Question

What is the empirical formula of an oxide of element
(E) containing 40% element by mass? (Atomic mass
of E = 32)
(1) EO2
(2) EO
(3) EO3
(4) E2O3
​

Answer 1

Answer:

(3) EO3

Explanation:

At. mass:

E = 32

O = 16

% Wt.:

E = 40

O = 60

No. of moles:

E = 40/32 = 1.25

O = 60/16 = 3.75

Simplest ratio:

E = 1.25/1.25 = 1

O = 3.75/1.25 = 3

So, Empirical Formula is EO3.

Answer 2

Empirical formula of the oxide will be EO3 i.e Option 3.

Empirical formula is defined as the simplest whole number ratio of constituent atoms of any compound. Example-Ethane has molecular formula of C2H6 hence the empirical formula will be CH3 ,Butane has molecular formula of C4H10 so the empirical formula will be C2H5 etc.

There is a relationship between empirical and molecular formulas. There exists a number n which when multiplied to empirical formula gives molecular formula.

Molecular formula=n*Empirical formula

When you calculate the weight of the molecular formula it is called molecular weight and the weight of the empirical formula is called empirical weight.

Molecular weight=n*Empirical weight

Given in the question we have 40% element by mass so the rest of the 60% will be of oxygen.

We are given an atomic mass of E=32 and we know that the atomic mass of oxygen is 16.Using this information we prepare a table. We already know the percentage and atomic weight so we calculate the atomic ratio and then using atomic ratios we calculate the simplest ratios. The ratios should not be in decimal and must be in whole numbers.

For more knowledge refer to these links.

https://brainly.in/question/9099308?msp_srt_exp=4

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