Question

Calculate by (a)% weight and (b) %mole each of the elements present in sugar

Answer 1

Explanation:

Molecular mass of sugar = $C_{12}H_{22}O_{11}$ : = 432 g/mol

Atomic mass of carbon atom = 12 g/mol

Atomic mass of hydrogen atom = 1 g/mol

Atomic mass of oxygen atom = 16 g/mol

a) Percentage of an element in a compound:

$\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{molecular mass of compound}}\times 100$

Percentage of carbon by weight in $C_{12}H_{22}O_{11}$:

$\frac{12\times 12 g/mol}{342 g/mol}\times 100=42.10\%$

Percentage of hydrogen by weight in $C_{12}H_{22}O_{11}$:

$\frac{22\times 1g/mol}{342 g/mol}\times 100=6.43\%$

Percentage of oxygen by weight in $C_{12}H_{22}O_{11}$:

$\frac{11\times 16g/mol}{342 g/mol}\times 100=51.46\%$

b) Percentage of mole each of the elements present in sugar:

=$\frac{\text{Moles of atoms of an element}}{\text{total moles of all types of atoms}}\times 100$

In mole of sugar we have 12 moles of carbon atom , 22 moles of hydrogen atoms and 11 moles of oxygen atoms.

Percentage of carbon by mole in $C_{12}H_{22}O_{11}$:

$\frac{12 mol}{45 mol}\times 100=26.66\%$

Percentage of hydrogen by mole in $C_{12}H_{22}O_{11}$:

$\frac{22 mol}{45 mol}\times 100=48.88\%$

Percentage of oxygen by mole in $C_{12}H_{22}O_{11}$:

$\frac{11 mol}{45 mol}\times 100=24.44\%$