Question

Which of the following contains the largest mass of hydrogen atoms?
A.$0.5{\text{ moles }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}$
B.${\text{1}}{\text{.1 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$
C.$1.5{\text{ moles }}{{\text{C}}_6}{{\text{H}}_8}{{\text{O}}_6}$
D.${\text{4}}{\text{.0 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$

Answer 1

A mole of a substance is the mass of a substance that contains exactly the same number of fundamental units, as in 12.0 g of C. The subscript of atom in the molecular formula indicates the number of moles of atom in one mole of formula.

Step by step answer: Calculate the mass of H atoms in$0.5{\text{ moles }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}$.

The molecular formula of the compound indicates that 1 mole ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ contains 2 moles of ${\text{H}}$ atoms.

So, ${\text{0}}{\text{.5 moles }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}{\text{ \times }}\dfrac{{{\text{2 mol H atom }}}}{{1{\text{ moles }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}}}{\text{ = 1}}{\text{.0 mol H atoms}}$

Now, convert these moles of ${\text{H}}$ atom to mass of ${\text{H}}$ atoms as follows:

${\text{Mass of atom = moles \times atomic mass}}$

The atomic mass of ${\text{H}}$ atom = $1.0{\text{ g/mol}}$

Thus,

${\text{Mass of H atoms = 1}}{\text{.0 mol H atoms \times 1}}{\text{.0 g/mol}}$

${\text{Mass of H atoms = 1}}{\text{.0 g}}$

Hence, $0.5{\text{ moles }}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}{{\text{O}}_{\text{4}}}$ contain ${\text{1}}{\text{.0 g}}$ of ${\text{H}}$ atoms.

Calculate the mass of H atoms in ${\text{1}}{\text{.1 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$.

The molecular formula of the compound indicates that 1 mole ${{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$ contains 8 moles of ${\text{H}}$ atoms.

So, ${\text{1}}{\text{.1 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}{\text{ \times }}\dfrac{{{\text{8 mol H atom }}}}{{1{\text{ moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}}}{\text{ = 8}}{\text{.8 mol H atoms}}$

Now, convert these moles of ${\text{H}}$ atom to mass of ${\text{H}}$ atoms as follows:

${\text{Mass of H atoms = 8}}{\text{.8 mol H atoms \times 1}}{\text{.0 g/mol}}$

${\text{Mass of H atoms = 8}}{\text{.8 g}}$

Hence, ${\text{1}}{\text{.1 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$ contain ${\text{8}}{\text{.8 g}}$ of ${\text{H}}$ atoms.

Calculate the mass of H atoms in $1.5{\text{ moles }}{{\text{C}}_6}{{\text{H}}_8}{{\text{O}}_6}$.

The molecular formula of the compound indicates that 1 mole ${{\text{C}}_6}{{\text{H}}_8}{{\text{O}}_6}$ contains 8 moles of ${\text{H}}$ atoms.

So, ${\text{1}}{\text{.5 moles }}{{\text{C}}_6}{{\text{H}}_8}{{\text{O}}_6}{\text{ \times }}\dfrac{{{\text{8 mol H atom }}}}{{1{\text{ moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}}}{\text{ = 12 mol H atoms}}$

Now, convert these moles of ${\text{H}}$ atom to mass of ${\text{H}}$ atoms as follows:

${\text{Mass of H atoms = 12 mol H atoms \times 1}}{\text{.0 g/mol}}$

${\text{Mass of H atoms = 12 g}}$

Hence, $1.5{\text{ moles }}{{\text{C}}_6}{{\text{H}}_8}{{\text{O}}_6}$ contain ${\text{12}}{\text{.0 g}}$ of ${\text{H}}$ atoms.

Calculate the mass of H atoms in ${\text{4}}{\text{.0 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$.

The molecular formula of the compound indicates that 1 mole ${{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$ contains 8 moles of ${\text{H}}$ atoms.

So, ${\text{4}}{\text{.0 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}{\text{ \times }}\dfrac{{{\text{8 mol H atom }}}}{{1{\text{ moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}}}{\text{ = 32 mol H atoms}}$

Now, convert these moles of ${\text{H}}$ atom to mass of ${\text{H}}$ atoms as follows:

${\text{Mass of H atoms = 32 mol H atoms \times 1}}{\text{.0 g/mol}}$

${\text{Mass of H atoms = 32 g}}$

Hence, ${\text{4}}{\text{.0 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$ contain ${\text{32 g}}$ of ${\text{H}}$ atoms.

Now, by comparing the mass of${\text{H}}$ atoms calculated for each compound we can say that ${\text{4}}{\text{.0 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$ contain the largest mass of hydrogen atoms.

Thus, the correct option is (D) ${\text{4}}{\text{.0 moles }}{{\text{C}}_3}{{\text{H}}_8}{{\text{O}}_3}$

Note: The mass of${\text{H}}$ atom in compounds containing the same number of moles of H atom depends on the moles of compounds. A compound having a greater number of moles will contain a greater number of ${\text{H}}$ atoms and thus greater mass of H atoms.