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The oxidation of glucose can result in several products. You perform combustion analysis on a 2.750 g sample that resulted from the oxidation of glucose and obtain 3.702 g CO2 and 1.514 g H2O. How many carbon atoms are in the empirical formula of the isolated sample? (Assume no other products present from the oxidation of glucose.)]

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Answer:

There is 1 carbon atom in the empirical formula.

Explanation:

First, let's use the given mass of CO₂ to calculate the moles of C, which will be the same amount of C moles present in the sample:

  • 3.702 g CO₂ ÷ 44g/mol = 0.0841 mol CO₂ = 0.0841 mol C

Then with the mass of H₂O we calculate the moles of H:

  • 1.514 g H₂O ÷ 18gH₂O/mol * [tex]\frac{2molH}{1molH_{2}O}[/tex] = 0.1682 mol H

Now we calculate the mass of C, H in the sample, in order to calculate the mass of O, which is calculated by substracting the mass of C and H from the total mass:

  • 0.0841 mol C * 12 g/mol = 1.0092 g C
  • 0.1682 mol H * 1 g/mol = 0.1682 g H
  • g O = 2.750 - (1.0092+0.1682) = 1.5726 g O

Then we convert g O to mol O

  • 1.5726 g O ÷ 16g/mol = 0.0983 mol O
  • In short, we have:

C ⇒ 0.0841 mol C

H ⇒ 0.1682 mol H

O ⇒ 0.0983 mol O

Now we divide those values by the lowest among them:

  • C ⇒ 0.0841 / 0.0841 = 1
  • H ⇒ 0.1682 / 0.0841 = 2
  • O ⇒ 0.0983 / 0.0841 = 1.1689 ≅ 1

So the empirical formula is CH₂O

There is 1 carbon atom in the empirical formula.

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