Question

How many moles of sodium bicarbonate, \[NaHC{O_3}\], are in \[508g\], of \[NaHC{O_3}\]?

Answer 1

Hint: Sodium bicarbonate is an inorganic salt of sodium cation and bicarbonate anion. A mole of a compound is the weight of a compound equal to the molar mass of the compound.

Complete step by step answer:
A mole of a substance is the ratio of the amount of the substance and the molar mass of the substance
\[Moles = \dfrac{{amount{\text{ }}of{\text{ }}substance}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}substance}}\]
In order to determine the number of moles of any substance we need the molar mass of that particular substance. The molar mass is generally calculated using the atomic mass of individual atoms present in the compound. The molar mass for a compound with molecular formula \[{A_x}{B_y}\] = \[x{\text{ }} \times \] atomic mass of \[A\] + \[y{\text{ }} \times \] atomic mass of \[B\]. The unit of molar mass is expressed in \[g/mol\].
The given compound is sodium bicarbonate. It is an inorganic compound. It is sparingly soluble in water. The molar mass of sodium carbonate (\[NaHC{O_3}\]) = atomic mass of \[Na\]+ atomic mass of \[H\]+ atomic mass of \[C\]+ \[3{\text{ }} \times \] atomic mass of \[O\].
\[ = 23 + 1 + 12 + 3 \times 16 = 84g/mol\].
The given amount of \[NaHC{O_3}\] is \[508g\]. Thus the moles of sodium bicarbonate is calculated as
\[Moles = \dfrac{{amount{\text{ }}of{\text{ }}NaHC{O_3}}}{{molar{\text{ }}mass{\text{ }}of{\text{ }}NaHC{O_3}}}\]
\[ = \dfrac{{508g}}{{84g/mol}}\]
\[ = 6.05moles.\] .

Note: The mole concept is necessary for conveniently expressing the amount of a substance. The mole of a substance is equal to the amount of a substance which contains exactly\[6.023 \times {10^{23}}\] numbers of particles of the substance. The mole of substance gives the information about the amount of substance and also the number of elements present in it.