Question

Calculate the number of atoms of hydrogen present in 5.6 g of urea, \[\ce{(NH2)2CO}\]. Also calculate the number of atoms of N, C, and O.

Answer 1

Given: Mass of urea = 5.6 g

To find: The number of atoms of hydrogen, nitrogen, carbon, and oxygen

Calculation: Molecular formula of urea is [\ce{(NH2)2CO}\] Molar mass of urea = 60 g mol⁻¹

Number of moles = `"Mass of a substance"/"Molar mass of a substance" = (5.6"g")/(60"g mol"^-1)` = 0.0933 mol
∴ Moles of urea = 0.0933 mol

Number of atoms = Number of moles × Avogadro's constant

Now, 1 molecule of urea has a total of 8 atoms, out of which 4 atoms are of H, 2 atoms are of N, 1 of C, and 1 of O.

∴ Number of H atoms in 5.6 g of urea = (4 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 2.247 × 10²³ atoms of hydrogen

∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 1.124 × 10²³ atoms of nitrogen

∴ Number of C atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of carbon

∴ Number of O atoms in 5.6 g of urea = (1 × 0.0933) mol × 6.022 × 10²³ atoms/mol
= 0.562 × 10²³ atoms of oxygen

∴ 5.6 g of urea contains 2.247 × 10²³ atoms of H.

1.124 × 10²³ atoms of N,

0.562 × 10²³ atoms of C,

and 0.562 × 10²³ atoms of O.