8. A 0.1000 g sample of grain alcohol, known to contain only carbon, hydrogen an

Question

8. A 0.1000 g sample of grain alcohol, known to contain only carbon, hydrogen and oxygen, was allowed react completely with oxygen to form products CO2 and H20. These products were trapped separately and weighed. 0.1910 g of CO2 and 0.1172 of H20 were found. What is the empirical formula of the compound? .2 9. The human body needs at least 1.03 x 10 mol O, every minute. If all of this oxygen is used for the cellular respiration reaction that breaks down glucose, how many moles of glucose does the human body consume each minute? CoH 206(s)602)6co6 H200)

Explanation / Answer

8.

Combustion of 0.10 g sample of the alcohol produced 0.1910 g of CO2 and 0.1172 g of H2O.
Mass C in 0.1910 g of CO2 = (12 / 44) x 0.1910 = 0.0521 g of C
Mass H in 0.1172 g of H2O = (2 / 18) x 0.1172   = 0.0130 g of H
So, mass of O = 0.1000 - (0.0521 + 0.0130)      = 0.0349 g of O

Divide through by respective atomic mass
C = 0.0521 / 12 = 0.0043 mol
H = 0.0130 / 1   = 0.0130 mol
O = 0.0349 / 16 = 0.0022 mol

Divide through by smallest value
C = 0.0043 / 0.0022 = 2
H = 0.0130 / 0.0022 = 6
O = 0.0022 / 0.0022 = 1

Empirical formula = C2H6O (ethanol C2H5OH)

9.

C6H12O6(s)   +   6O2(g)   ------->   6CO2(g)   +   6H2O(g)

In the above reaction equation

      6 moles of O2(g) reacts with = 1 mol of C6H12O6(s)

Or, 1 mol of O2(g) reacts with = (1/6) mol of C6H12O6(s)

Or, (1.03 x 10-2) mol of O2(g) reacts with = (1.03 x 10-2/6) mol of C6H12O6(s)

Or, (1.03 x 10-2) mol of O2(g) reacts with = 0.0017 mol of C6H12O6(s)

So, each minute human body consumes 0.0017 moles of C6H12O6(s).

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