Molar Mass

• Molar mass is the mass of one mole of a substance.

• One mole corresponds to a group of |6.022\times10^{23}| particles.

What is the molar mass of ammonia |\text{(NH}_3)|?

3. Let’s replace the variables with the known values and solve the problem.
   ||\begin{align}M_{\text{NH}_3} &=14.01\ \text{g/mol}+3\times1.01\ \text{g/mol}\\M_{\text{NH}_3} &=17.04\ \text{g/mol}\end{align}||

4. The final answer is: 
   The molar mass of ammonia is |17.04\ \text{g/mol.}|

What is the molar mass of aluminum sulfate |(\text{Al}_2(\text{SO}_4)_3)|?

where

|M\!:| molar mass in grams per mole |\text{(g/mol)}|
|m\!:| mass in grams |\text{(g)}|
|n\!:| number of moles in moles |\text{(mol)}|

The molar mass of calcium |\text{(Ca)}| corresponds to the mass of |40.08\ \text{g}| per |1\ \text{mol}.| By dividing |40.08\ \text{g}| by |1\ \text{mol},| the |\bf\text{g/mol}| unit is obtained for mole mass.
||M_{\text{Ca}}=\dfrac{m}{n}=\dfrac{40.08\ \bf\text{g}}{1\ \bf\text{mol}}=40.08\ \bf\text{g/mol}||

1. Let's identify the given and missing values.
   ||\begin{align}n &=\ ?\ \text{mol}\\m &=97.02\ \text{g}\end{align}||

   The molar mass of sodium chloride |(\text{NaCl})| can be calculated as follows:
   ||\begin{align}\\M_{\text{NaCl}} &=M_{\text{Na}}+M_{\text{Cl}}\\M_{\text{NaCl}} &=22.99\ \text{g/mol}+35.45\ \text{g/mol}\\M_{\text{NaCl}} &=58.44\ \text{g/mol}\end{align}||

2. Since the number of moles |(n)| is unknown, the formula where it is isolated should be used.
   ||M=\dfrac{m}{n} \ \Rightarrow\ n=\dfrac{m}{M}||

3. Let’s replace the variables with the known values and solve the problem.
   ||\begin{align}n &=\dfrac{97.02\ \cancel{\text{g}}}{58.44\ \cancel{\text{g}}\text{/mol}}\\\\n &\approx1.660\ \text{mol}\end{align}||

4. The final answer is:
   The salt shaker contains approximately |1.660\ \text{mol}| of sodium chloride.

What is the mass of |2.25\ \text{mol}| of water |(\text{H}_2\text{O})|?

1. Let's identify the given and missing values.
   ||\begin{align}m &=\ ?\ \text{g}\\n &=2.25\ \text{mol}\end{align}||

   The molar mass of water |(\text{H}_2\text{O})| can be calculated as follows:
   ||\begin{align}\\M_{\text{H}_2\text{O}} &=2\ M_{\text{H}}+M_{\text{O}}\\M_{\text{H}_2\text{O}} &=2\times1.01\ \text{g/mol}+16.00\ \text{g/mol}\\M_{\text{H}_2\text{O}} &=18.02\ \text{g/mol}\end{align}||

2. Since the mass |(m)| is unknown, the formula where it is isolated should be used.
   ||M=\dfrac{m}{n} \ \Rightarrow\ m=nM||

3. Let’s replace the variables with the known values and solve the problem.
   ||\begin{align}m &=2.25\ \cancel{\text{mol}}\times18.02\ \text{g/}\cancel{\text{mol}}\\m &\approx40.5\ \text{g}\end{align}||

4. The final answer is:
   The mass of |2.25\ \text{mol}| of water is approximately |40.5\ \text{g}.|

|n=\dfrac{N}{N_{\text{A}}}|

where

|n\!:| number of moles in moles |(\text{mol})|
|N\!:| number of particles |(\text{particules})|
|N_{\text{A}}\!:| Avogadro’s number |(6.022\times10^{23}\ \text{particules/mol})|

a) Determine the number of moles in the sample.

1. Let’s identify the given and missing values.
   ||\begin{align}n &=\ ?\ \text{mol}\\N &=1.402\times10^{24}\ \text{atoms}\\N_{\text{A}} &=6.022\times10^{23}\ \text{atoms/mol}\end{align}||

2. Choose the correct formula to use.
   ||n=\dfrac{N}{N_{\text{A}}}||

3. Let’s replace the variables with the known values and solve the problem:
   ||\begin{align}n &=\dfrac{1.402\times10^{24}\ {\cancel{\text{atoms}}}}{6.022\times10^{23}\ {\cancel{\text{atoms}}}\text{/mol}}\\\\n &\approx2.328\ \text{mol}\end{align}||

4. The answer is:
   |1.402\times10^{24}| carbon atoms correspond to |2.328\ \text{mol}| of carbon.

b) Determine the mass of the sample.

1. Let’s identify the given and missing data. The molar mass of carbon |\text{(C)}| can be determined using the periodic table.
   ||\begin{align}m &=\ ?\ \text{g}\\n &=2.328\ \text{mol}\\M_{\text{C}} &=12.01\ \text{g/mol}\end{align}||

2. Since the mass |(m)| is unknown, the formula where it is isolated should be used.
   ||M=\dfrac{m}{n} \ \Rightarrow\ m=nM||

3. Let’s replace the variables with the known values and solve the problem:
   ||\begin{align}m &=2.328\ \cancel{\text{mol}}\times12.01\ \text{g/}\cancel{\text{mol}}\\m &\approx27.96\ \text{g}\end{align}||

4. The answer is:
   |1.402\times10^{24}| carbon atoms correspond to approximately |27.96\ \text{g}| of carbon.