Question

Calculate the number of atoms of hydrogen present in $5.6{\text{ g}}$ of urea, ${\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}}$. Also calculate the number of atoms of ${\text{N}}$, ${\text{C}}$ and ${\text{O}}$.

Answer 1

Moles is the ratio of the mass of substance in g to the molar mass of the substance in ${\text{g/mol}}$. The number of atoms of a compound is Avogadro’s number for 1 mole of compound. The number $6.022 \times {10^{23}}$ is known as Avogadro’s number.

Formula Used: ${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$

Complete step by step answer:

1. First we will calculate the number of moles of urea present in $5.6{\text{ g}}$ of urea using the formula as follows:
${\text{Number of moles}}\left( {{\text{mol}}} \right) = \dfrac{{{\text{Mass}}\left( {\text{g}} \right)}}{{{\text{Molar mass}}\left( {{\text{g/mol}}} \right)}}$
Substitute $5.6{\text{ g}}$ for the mass of urea, $60{\text{ g/mol}}$ for the molar mass of urea and solve for the number of moles of urea. Thus,
${\text{Number of moles of urea}} = \dfrac{{5.6{\text{ g}}}}{{60{\text{ g/mol}}}}$
${\text{Number of moles of urea}} = 0.0933{\text{ mol}}$
Thus, the number of moles of urea present in $5.6{\text{ g}}$ of urea are $0.0933{\text{ mol}}$.

2. The molecular formula for urea is ${\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}}$. From the molecular formula we can conclude that,
$1{\text{ mol }}{\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}} = 4{\text{ mol H}}$,
Thus,
${\text{Moles of H}} = 0.0933{\text{ mol urea}} \times \dfrac{{4{\text{ mol H}}}}{{1{\text{ mol urea}}}}$
${\text{Moles of H}} = 0.3732{\text{ mol H}}$
Thus, $0.0933{\text{ mol}}$ urea contains $0.3732{\text{ mol}}$ of hydrogen.
We know that ${\text{1 mol}}$ of hydrogen contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of H atoms}} = 0.3732{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 2.247 \times {10^{23}}{\text{ atoms}}$
Thus, $5.6{\text{ g}}$ of urea contains $2.247 \times {10^{23}}{\text{ atoms}}$ of hydrogen.

3. The molecular formula for urea is ${\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}}$. From the molecular formula we can conclude that,
$1{\text{ mol }}{\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}} = 2{\text{ mol N}}$,
Thus,
${\text{Moles of N}} = 0.0933{\text{ mol urea}} \times \dfrac{{2{\text{ mol N}}}}{{1{\text{ mol urea}}}}$
${\text{Moles of N}} = 0.1866{\text{ mol N}}$
Thus, $0.0933{\text{ mol}}$ urea contains $0.1866{\text{ mol}}$ of nitrogen.
We know that ${\text{1 mol}}$ of nitrogen contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of N atoms}} = 0.1866{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 1.123 \times {10^{23}}{\text{ atoms}}$
Thus, $5.6{\text{ g}}$ of urea contains $1.123 \times {10^{23}}{\text{ atoms}}$ of nitrogen.

4. The molecular formula for urea is ${\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}}$. From the molecular formula we can conclude that,
$1{\text{ mol }}{\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}} = 1{\text{ mol C}}$,
Thus,
${\text{Moles of C}} = 0.0933{\text{ mol urea}} \times \dfrac{{1{\text{ mol C}}}}{{1{\text{ mol urea}}}}$
${\text{Moles of C}} = 0.0933{\text{ mol C}}$
Thus, $0.0933{\text{ mol}}$ urea contains $0.0933{\text{ mol}}$ of carbon.
We know that ${\text{1 mol}}$ of carbon contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of C atoms}} = 0.0933{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 0.561 \times {10^{23}}{\text{ atoms}}$
Thus, $5.6{\text{ g}}$ of urea contains $0.561 \times {10^{23}}{\text{ atoms}}$ of carbon.

5. The molecular formula for urea is ${\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}}$. From the molecular formula we can conclude that,
$1{\text{ mol }}{\left( {{\text{N}}{{\text{H}}_{\text{2}}}} \right)_{\text{2}}}{\text{CO}} = 1{\text{ mol O}}$
Thus,
${\text{Moles of O}} = 0.0933{\text{ mol urea}} \times \dfrac{{1{\text{ mol O}}}}{{1{\text{ mol urea}}}}$
${\text{Moles of O}} = 0.0933{\text{ mol}}$
Thus, $0.0933{\text{ mol}}$ urea contains $0.0933{\text{ mol}}$ of oxygen.
We know that ${\text{1 mol}}$ of oxygen contains $6.022 \times {10^{23}}$ atoms. Thus,
${\text{Number of O atoms}} = 0.0933{\text{ mol}} \times \dfrac{{6.022 \times {{10}^{23}}{\text{ atoms}}}}{{1{\text{ mol}}}} = 0.561 \times {10^{23}}{\text{ atoms}}$
Thus, $5.6{\text{ g}}$ of urea contains $0.561 \times {10^{23}}{\text{ atoms}}$ of oxygen.
Thus, $5.6{\text{ g}}$ of urea contains $2.247 \times {10^{23}}{\text{ atoms}}$ of hydrogen, $1.123 \times {10^{23}}{\text{ atoms}}$ of nitrogen, $0.561 \times {10^{23}}{\text{ atoms}}$ of carbon and $0.561 \times {10^{23}}{\text{ atoms}}$ of oxygen.

Note:
The Avogadro’s number gives the number of atoms, ions or molecules present in one mole of any substance. The value of Avogadro’s number is $6.022 \times {10^{23}}$.