1.Calculating the number of C, H, O atoms in 9.74 grams of glucose.
Given that :
• Mass of C6H12O6 = 9.74g
,
• Molecular mass C6H12O6 =180.156g/mol
,
• Moles of C6H12O6 = mass/M.Mass
=9.74g/180.156g/mol
=0.054moles.
2. Calculate mole ratio of C, H, O , since we have 0.054moles of C6H12O6
(2.1) Carbon → 1 mole C6H12O6 : 6mol C
so, 0.054moles : x mol C
Therefore x mol C = 0.054 * 6 = 0.324moles of Carbon
• Number of atoms for C
No of C atoms = mole* Avogadro number
=0.324 * 6.022*10^23
=1.95x10^23 atoms
(2.2) Hydrogen → 1 mole of C6H12O6 : 12 mol of H
So; 0.054moles : x mol H
Therefore xmol H = 0.054*12 = 0.648 moles of Hydrogen
• Number of atoms for H
No of H atoms = 0.648 * 6.022*10^23
=3.90x10^23 atoms .
(2.3) Oxygen → 1 mole of C6H12O6 : 6 mol of O
So, 0.054 moles : xmol O
Therefore x mol O = 0.054 * 6 =0.324 moles
• Number of atoms for Oxygen =0.324 * 6.022*10^23
=1.95x10^23 atoms
Finally, number of C atoms =1.95x10^23 atoms. number of H atoms =3.90x10^23 atoms . number of O atoms =1.95x10^23 atoms