Question

The molecular weight of ${\left( {N{H_4}} \right)_2}C{O_3}$ is. Given atomic mass of $H = 1,{\text{ }}O = 16,{\text{ }}C = 12,{\text{ }}N = 14$.
A) $96\,g/mol.$
B) $98\,g/mol$.
C) $94\,g/mol.$
D) $80\,g/mol.$

Answer 1

We know. Molecular weight can be calculated by adding the mass of an atom multiplied by the number of atoms present in the molecule.
For Example, The molecular weight of the ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ is calculated as,
The atomic weight of carbon is $12.$
The atomic weight of hydrogen is $1.$
The molecular weight of the ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}} = 2 \times 12 + 6 = 30$

Complete step by step answer:
Here it is given that,
The atomic weight of carbon is $12g/mol$
The atomic weight of hydrogen is $1\,g/mol$
The atomic weight of nitrogen is $14g/mol$
The atomic weight of nitrogen is $16g/mol$
The molecular weight of ${\left( {N{H_4}} \right)_2}C{O_3} = {\left( {14 + 4} \right)_2} + 12 + \left( {3 \times 16} \right)$
The molecular weight of ${\left( {N{H_4}} \right)_2}C{O_3}$ is$96\,g/mol$.
Hence,
Therefore, the correct option is A.

Additional information:
The molecular weight of molecules is determined by various methods:
Mass spectrometry: This method is usually used in determining the mass of small molecules.
Hydrodynamic method: In this method weight is found out by Mark-Houwink relations.
Static Light scattering method: In this method molecular weight is determined from the amount of light scattered using the Zimm method.

Note:
If we want the weight in grams of lead, simply divide the molecular weight by the Avogadro’s number.
The mass of an atom can be calculated using the formula,
${\text{Mass of an atom = }}\dfrac{{{\text{Molar}}\,{\text{Mass}}}}{{{\text{Moles}}}}\,g$
We know that the molar mass is the mass of one mole of the substance. The unit of molar mass is ${\text{g/mol}}{\text{.}}$ a mole of a substance is equal to Avogadro’s number of that substance.
If the molecular weight of the substance is $207.2g/mol.$
Mass of an atom$= \dfrac{{207.2}}{{6.022 \times {{10}^{23}}}}g$
Mass of an atom$= 34.40 \times {10^{ - 23}}g$