Respuesta :
These percentages are most likely mass percentages. Therefore, determine the mass of each element in 100g of the compound.
75g C, 25g H.
Divide these masses by their molar mass to find the ratios of the moles of element in the compound. Divide these values by the smallest of the two/three to get the ratios.
6.25:25 C:H --> 1:4
So the formulae are CH4 (methane).
75g C, 25g H.
Divide these masses by their molar mass to find the ratios of the moles of element in the compound. Divide these values by the smallest of the two/three to get the ratios.
6.25:25 C:H --> 1:4
So the formulae are CH4 (methane).
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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I Hope this helps, greetings ... Dexteright02! =)
