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3 years ago

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What is an empirical formula

2 answers:

yarga [219]3 years ago

8 0

A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

zlopas [31]3 years ago

6 0

<em>A</em><em>n empirical </em><em> </em><em>shows the simplest </em><em> </em><em>ratio among atoms in a compound</em><em> </em><em>.</em>

<h2><em>More</em><em> </em><em>information</em></h2>

<em>an empirical formula</em><em> </em><em>is the simplest formula for a compound </em><em>.</em><em>For</em><em> </em><em>example</em><em>,</em><em> </em><em>H2O2</em><em> </em><em>can be </em><em> </em><em>reduced to a </em><em>si</em><em>mpler</em><em> </em><em>formula</em><em>.</em><em> </em>
<em>Determining a Molecular Formula from an Empirical Formula The empirical formula for a compound is P 2 O 5 . Its experimental molar mass is 284 g/mol.</em>
<em>For example, the empirical formula of benzene (C5H5) is Qj, - because the mole ratio of carbon am'hydrogen is 1:1.</em>

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

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Explanation:

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Dividing the previous equation by x:

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Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

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