Question

What is the molar mass of ammonium carbonate ${(N{H_4})_2}C{O_3}$ ?
A . $48.06\dfrac{g}{{mol}}$
B . $96.11\dfrac{g}{{mol}}$
C . $82.09\dfrac{g}{{mol}}$
D . $78.05\dfrac{g}{{mol}}$
E . $192.2\dfrac{g}{{mol}}$

Answer 1

- Hint: We know about the concept of moles where we have learnt that Molar mass of a compound is defined as the mass of the given compound (m) divided by the number of moles(n). Thus we know that Molar mass ($gmo{l^{ - 1}}$) $ = \dfrac{m}{n}$ .Here we will determine the molar mass of ammonium carbonate by using the formula that is ,formula mass=atomic weight of element X numbers of atoms.

Complete step-by-step solution -
Now recalling each atoms atomic masses from periodic table (we should remember the atomic masses of some important atoms) we have following atomic masses of atoms,N has molar mass $14.0\dfrac{g}{{mol}}$, H has molar mass $1.0\dfrac{g}{{mol}}$ , C has molar mass $12.0\dfrac{g}{{mol}}$ and O has molar mass $16.0\dfrac{g}{{mol}}$.We are assuming that we have one mole of ammonium carbonate as it is not given in the problem the amount or mass of the substance.In order to get molar mass now we have 2 Nitrogen,8 Hydrogen,1 carbon and 3 Oxygen so atomic mass or molar mass of each atom will be multiplied with number of atoms.so molar mass of ammonium carbonate ${(N{H_4})_2}C{O_3}$ is $2 \times 14.0 + 8 \times 1.0 + 1 \times 12.0 + 3 \times 16.0$$ = 96.09\dfrac{g}{{mol}}$ so option B . $96.11\dfrac{g}{{mol}}$ is the correct answer as it is very much close to $ = 96.09\dfrac{g}{{mol}}$ .

Note : To calculate molar mass of ammonium carbonate ${(N{H_4})_2}C{O_3}$ first we should know the atomic weight of each element then we count each atom’s number in the compound in the further processing of the problem we multiply the number of occurrence of each atom to their atomic masses and sum of each atom’s masses gives the atomic mass of compound which is the molar mass of ammonium carbonate.