A compound contains 75% carbon and 25%hydrogen by mass. What is the empirical formulafor this compound?CHCHCH2 18DONE

To get the empirical formula of the compound, we convert each mass percentage to mole percentage by dividing each to corresponding molar mass. 75% mass of carbon is 6.25 while 25% hydrogen is 25. We divide 25 by 6.25, we get 4. Hence, the empirical formula is CH4.

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dexteright02 dexteright02 · Answer 2 · 2019-01-13T20:37:29+01:00

Hello!

A compound contains 75% carbon and 25% hydrogen by mass. What is the empirical formula for this compound ?

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:

[tex]C: \dfrac{75\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} = 6.25\:mol[/tex]

[tex]H: \dfrac{25\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 25\:mol[/tex]

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:

[tex]C: \dfrac{6.25}{6.25}\to\:\:\boxed{C = 1}[/tex]

[tex]H: \dfrac{25}{6.25}\to\:\:\boxed{H = 4}[/tex]

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

[tex]\boxed{\boxed{C_1H_4\:\:\:or\:\:\:CH_4}}\Longleftarrow(Empirical\:Formula)\end{array}}\qquad\checkmark[/tex]

Answer:

CH4 (Methane)

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