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What is the empirical formula of a substance that contains 0.100 mol of carbon, 0.200 mol of hydrogen, and 9.99×10−2 mol of oxygen?

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dexteright02
Hello!

We have the following data:

C: (carbon) = 0.100 mol
H: (hydrogen) = 0.200 mol
O: (oxygen) = 9.99*10^-2 = 0.0999 mol

We note that the values ​​found above are not integers, so let's divide these values ​​by the smallest of them, so that the proportion is not changed, let's see:

[tex]C: \dfrac{0.100}{0.0999} \to\:\boxed{C \approx 1}[/tex]

[tex]H: \dfrac{0.200}{0.0999} \to\:\boxed{H \approx 2}[/tex]

[tex]O: \dfrac{0.0999}{0.0999} \to\:\boxed{O = 1}[/tex]

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

[tex]\boxed{\boxed{C_1H_2O_1\:\:or\:\:CH_2O}}\end{array}}\qquad\checkmark[/tex]

I Hope this helps, greetings ... DexteR!

Answer:

Empirical formula of the compound = [tex]CH_2O[/tex]

Explanation:

Empirical formula is the mole ratio of each element in a chemical compound.

Given:

No. of mol of C = 0.100 mol

No. of mol of H = 0.200

No. of mol of O = [tex]9.99 \times 10^-2\ mol = 0.099\ mol[/tex]

For the determination of empirical formula, no. of moles of each element should be in whole number.

If the no. of moles of elements are not in whole number, then divide moles of each element by the smallest mole of the element.

So, ratio of each element is:

C = [tex]\frac{0.100}{0.099}\approx 0.100\ mol[/tex]

H = [tex]\frac{0.100}{0.099}\approx 0.200\ mol[/tex]

O = [tex]\frac{0.0.099}{0.099} = 0.100\ mol[/tex]

Mole ratio is

C : H : O = 1 : 2 : 1

So, the empirical formula = [tex]CH_2O[/tex]

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