To find the empirical formula you would first need to find the moles of each element:

58.8g/ 12.0g = 4.9 mol C

9.9g/ 1.0g = 9.9 mol H

31.4g/ 16.0g = 1.96 O

Then you divide by the smallest number of moles of each:

4.9/1.96 = 2.5

9.9/1.96 = 6

1.96/1.96 = 1

Since there is 2.5, you find the least number that makes each moles a whole number which is 2.

So the empirical formula is C5H12O2.

6 0

4 years ago

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction. half-reaction i

Natasha2012 [34]

Answer: Oxidation reaction: $Al(s)\rightarrow Al^{3+}(aq)+3e^-$

Reduction reaction: $Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$

Overall redox reaction : $2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)$

Explanation:

Oxidation-reduction reaction or redox reaction is defined as the reaction in which oxidation and reduction reactions occur simultaneously.

Oxidation reaction is defined as the reaction in which a substance looses its electrons. The oxidation state of the substance increases.

$Al(s)\rightarrow Al^{3+}(aq)+3e^-$

Reduction reaction is defined as the reaction in which a substance gains electrons. The oxidation state of the substance gets reduced.

$Sn^{2+}(aq)+2e^-\rightarrow Sn(s)$

Overall redox reaction : $2Al(s)+3Sn^{2+}(aq)\rightarrow 2Al^{3+}(aq)+3Sn(s)$

5 0

3 years ago