Question

The compound in which mass percentage of carbon is 75% and that of hydrogen is 25% is

Answer 1

Hello Dear.

Empirical Formula of the Compound is CH₄.

Empirical Formula is calculated in the table shown in the attachment.

Method to calculate the Empirical Formula,

1) 1) Draw the Table which have 3 Rows and 6 Columns.

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

3) In the Final Column, Calculate the Whole number ratio which cannot be in Decimal.
 
4) At last,Calculate the Empirical Formula, by substituting the Whole number ratio into the Each elements.Thus, Empirical Formula of the Compound will be CH₄.


Hope it helps.

Answer 2

Hello!

The compound in which mass percentage of carbon is 75% and that of hydrogen is 25% is .... ?

data:

Carbon (C) ≈ 12 a.m.u (g/mol)  

Hydrogen (H) ≈ 1 a.m.u (g/mol)

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100g solution)

C: 75 % = 75 g

H: 25 % = 25 g

The values ​​(in g) will be converted into quantity of substance (number of mols), dividing by molecular weight (g / mol) each of the values, we will see:

$C: \dfrac{75\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 6.25\:mol$

$H: \dfrac{25\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 25\:mol$

We realize that the values ​​found above are not integers, so we divide these values ​​by the smallest of them, so that the proportion does not change, let us see:

$C: \dfrac{6.25}{6.25}\to\:\:\boxed{C = 1}$

$H: \dfrac{25}{6.25}\to\:\:\boxed{H = 4}$

Thus, the minimum or empirical formula found for the compound will be:

$\boxed{\boxed{C_1H_4\:\:\:or\:\:\:CH_4}}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark$

Answer:

CH4 (Methane)

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I Hope this helps, greetings ... Dexteright02! =)