Question

How many moles are in $37.9$ grams of $NaHC{O_3}$?

Answer 1

In the question, we are asked to find the number of moles of the given compound. So, to find the number of moles and thus the number of atoms, we use the formula given below:
$N = \dfrac{m}{M}$
In the question we are provided by the given mass of the compound, we just need to get the molecular mass of the compound and divide it by the given mass of the compound.

Complete step by step answer:
In the question statement we are asked to find the number of moles. In order to calculate the number of moles of any substance present in the sample, we use the formula given below:
$N = \dfrac{m}{M}$
Here, N is the number of moles present in the sample
m is the given mass,
M is the molar mass of the compound.
In the question statement, we have the given mass of the compound i.e. $37.9$ grams. Now, to solve the question we need to get the value of the molar mass of the sample i.e. $NaHC{O_3}$
In the order to find the molecular mass of this compound, we need to add the atomic mass of the individual atoms as shown below:
Total molar mass = mass of Na + mass of H + mass of C + 3(mass of O)
Total molar mass= $\left( {22.99} \right) + \left( {1.01} \right) + \left( {12.{{ }}01} \right) + 3\left( {16.00} \right) = 84.01gmo{l^{ - 1}}$
Now, by using the formula given above, we can easily find the number of moles of the given compound.
Number of moles of $NaHC{O_3}$ = $\dfrac{{given{{ }}\;mass\;{{ }}\;of{{ }}\;NaHC{O_3}}}{{{{molar }}\;mass\;{{ }}of\;{{ }}\;NaHC{O_3}}}$

Number of moles of $NaHC{O_3}$ = $\dfrac{{37.9}}{{84.01}}$= $0.451moles$

Note: The mole is the unit to measure the quantity of substance present in any sample and number of atoms present in 1 mole is equal to $6.022 \times {10^{23}}$ where $6.022 \times {10^{23}}$ is NA called the Avogadro number.