Question

A compound consists of 75% magnesium and 25% oxygen. Find the empirical formula.

Answer 1

Answer:

% Composition of elements:

• Magnesium = 75%
• Oxygen = 25%

Atomic mass of given elements:

• Magnesium = 24 g
• Oxygen = 16 g

Now, divide % composition by Atomic mass:

$\footnotesize\implies Mg = \dfrac{\% \: Composition}{Atomic \: mass} = \dfrac{75}{24} = \bf 3.125$

$\footnotesize\implies O = \dfrac{\% \: Composition}{Atomic \: mass} = \dfrac{25}{16} = \bf 1.5625$

Simplest Ratio:

$\footnotesize\implies Mg = \dfrac{3.125}{1.5625} = 2$

$\footnotesize\implies O = \dfrac{1.5625}{1.5625} = 1$

Empirical Formula:

$\footnotesize\implies \underline{ \boxed{ \red{ \bf Empirical \: Formula = Mg_2O}}}$

Answer 2

The empirical formula will be $Mg_{2}$ .

What is empirical formula?

The simplest entire number fraction of atoms contained in a chemical molecule is its empirical formula.

Calculation of empirical formula:

It is given that, composition of Mg = 75 %, composition of oxygen  = 25%.

It is known that, atomic mass of Mg = 24 g, atomic mass of oxygen = 16g.

By dividing % composition by atomic mass of given elements:

Mg = % composition / %atomic mass

= 75 / 24

= 3.125

Oxygen (O) = % Composition / atomic mass

= 25 / 1.56

Now, calculate simplest ratio of Mg and O.

For Mg = 3.125 / 1.56

= 2.

For O = 1.56 / 1.56

= 1

So, the empirical formula will be $Mg_{2}O$.

To know more about empirical formula.

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