Question

calculate number of atoms of hydrogen present in 5.6g of urea and also number of atoms of N C O

Answer 1

Urea = NH2CONH2

M.wt = 60

5.6 g urea = 5.6/60 = 0.093 moles

1 mole of urea contains 4N atoms of H

0.093 moles contains 0.093 x 4N atoms of H

= 2.24 x 10^23 atoms of H

No of atoms of Nitrogen = half of the number of atoms of H

= 1.12 x 10^23 atoms of N

No of atoms of C or O = half of the number of atoms of N

= 0.56 x 10^23 atoms of C or O

= 5.6 x 10^22 atoms of C or O

Answer 2

Answer: Number of hydrogen atoms is $2.16\times 10^{21}$ atoms

Solution:

Chemical formula of urea is $NH_2CONH_2$.

Number of moles= $\frac{\text{given weight}}{\text{molecular weight}}$

Moles of urea in 5.6 g =

$\frac{5.6 g}{60 g/mol}=0.09 mol$

1 mole of urea contains =$6.022\times 10^{23}$ molecules

$N_A=\text{Avogadro Number}=6.022\times 10^{23}$

Then total number of molecules of urea in 0.09 moles will be:

=$0.09\times (6.022\times 10^{23})=5.4\times 10^{22}$ molecules.

Total number of atom of an element in a molecule= $\text{total number of atoms of that element}\times \text{total number of molecules of the compound}$

Number of hydrogen atoms =$4\times 5.4\times 10^{22}=2.16\times 10^{21}$ atoms

Number of carbon atoms =$1\times 5.4\times 10^{22}=5.4\times 10^{22}$ atoms

Number of nitrogen atoms =$2\times 5.4\times 10^{22}=1.08\times 10^{21}$ atoms

Number of oxygen atoms =$1\times 5.4\times 10^{22}=5.4\times 10^{22}$ atoms