In the previous post, we discussed converting grams to moles. Remember, the first thing was to look up the molar mass in the periodic table, set up the correct conversion factor, and finally do the multiplication. Here is a short summary for converting the mass to moles:
Now, to calculate the number of molecules or atoms from mass, we need one extra step using the Avogadro’s number (6.02 x 1023). Remember, the Avogadro’s number shows how many particles, which can be atoms, molecules, ions, there are in one mole of a sample and because it is related to moles, we need to first convert the mass to moles.
So, there are two steps combined in this conversion and the plan is to first convert the mass to moles and then to the number of molecules using NA:
For example, how many molecules of PCl5 are there in a 54.0 g of sample?
First, we calculate the molar:
M (PCl5) = M (P) + 5 M (Cl) = 31.0 + 5 x 35.5 = 208.5 g/mol
The two conversion factors for calculating the moles would be:
\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{or}}\,\,\frac{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\]
We are going to use the first conversion factor because it has the units of grams on the denominator which allows us to cancel them with the initial amount given in grams:
\[{\rm{n}}\,{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\,{\rm{54}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{208}}{\rm{.5}}\;\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\;{\rm{ = }}\,{\rm{0}}{\rm{.260}}\,{\rm{mol}}\]
Once we have the number of moles, we use another conversion factor linking it with the Avogadro’s number.
\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{or}}\,\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\]
And this time, we need to pick the one that allows canceling the mols which is the second conversion factor:
\[{\rm{0}}{\rm{.26 }}\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\, \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\, = \;1.57\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{molecules PC}}{{\rm{l}}_{\rm{5}}}\]
You can always combine the conversions into a one-step process:
\[{\rm{N}}\,{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\,{\rm{54}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{208}}{\rm{.5}}\;\cancel{{{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\; \times \;\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\, = \;1.57\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{molecules}}\]
Another example: How many molecules of glucose, C6H12O6 are there in a 35.0 g sample?
The molar mass of glucose is:
M (C6H12O6) = 6 x 12 + 12 x 1.0 + 6 x 16 = 180 g/mol
Therefore, the two conversion factors for going mass → moles, and moles → molecules are:
\[\frac{{{\rm{1}}\,{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}{{{\rm{180}}{\rm{.}}\,{\rm{g}}}}\;,\,\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecules}}}}{{{\rm{1}}\,{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}\]
So, we can use these factors to convert the mass to the number of molecules:
\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{)}}\,{\rm{ = }}\;{\rm{35}}{\rm{.0}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}{{{\rm{180}}{\rm{.}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}\; = \;1.17\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]
Converting the Number of Molecules to Grams
Sometimes, you may be asked to determine the mass of a sample given the number of molecules. This is the reverse calculation, and we are going to use the other conversion factor that we obtain for linking the molar mass to moles, and the moles to the number of molecules.
For example, let’s say we were asked to determine the mass of a PCl5 sample that contains 7.58 x 1024 molecules. Remember, these were the conversion factors we got for finding the number of molecules for a PCl5 sample:
\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{or}}\,\,\frac{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\]
\[\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{or}}\,\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\]
This time, we are starting with the conversion factor containing the Avogadro’s number, and we need to pick the one that allows canceling the number of molecules, and then then the number of moles.
\[{\rm{m}}\,{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\,{\rm{7}}{\rm{.58}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\cancel{{{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{molecules}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\,{\rm{ \times }}\,\frac{{{\rm{208}}{\rm{.5}}\;{\rm{g}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\,}}\,{\rm{ = }}\;{\rm{2}}{\rm{.63}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{\rm{3}}}\,{\rm{g}}\]
Practice
Calculating Moles from Mass
Determine the number of moles in 59.7 grams of Al.
2.21 mol
The moles and mass are connected through the molar mass. The molar mass of Al is 27.0 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{27}}{\rm{.0}}\;{\rm{g}}\,{\rm{Al}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(Al)}}\,{\rm{ = }}\;{\rm{59}}{\rm{.7}}\cancel{{{\rm{g}}\,{\rm{Al}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{27}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{\rm{Al}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.21}}\;{\rm{mol}}\]
Determine the number of moles in 2.41 grams of FeO.
0.0336 mol
The moles and mass are connected through the molar mass. The molar mass of FeO is 71.8 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{71}}{\rm{.8}}\;{\rm{g}}\,{\rm{FeO}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(FeO)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.41}}\cancel{{{\rm{g}}\,{\rm{FeO}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{71}}{\rm{.8}}\,\cancel{{{\rm{g}}\,{\rm{FeO}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0336}}\;{\rm{mol}}\]
Calculate the number of moles in 0.647 grams of Al2O3.
0.00634 mol
The moles and mass are connected through the molar mass. The molar mass of Al2O3 is 102 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{71}}{\rm{.8}}\;{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.647}}\cancel{{{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{102}}\,\cancel{{{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00634}}\;{\rm{mol}}\]
Determine the number of moles in 3.56 grams of Mg(OH)2.
0.0611 mol
The moles and mass are connected through the molar mass. The molar mass of Mg(OH)2 is 58.3 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{58}}{\rm{.3}}\;{\rm{g}}\,{\rm{Mg}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(Mg}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}{\rm{)}}\,{\rm{ = }}\;{\rm{3}}{\rm{.56}}\cancel{{{\rm{g}}\,{\rm{Mg}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{58}}{\rm{.3}}\,\cancel{{{\rm{g}}\,{\rm{Mg}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0611}}\;{\rm{mol}}\]
Determine the number of moles in 0.385 grams of N2O3.
0.00507 mol
The moles and mass are connected through the molar mass. The molar mass of N2O3 is 76.0 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{76}}{\rm{.0}}\;{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.385}}\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{76}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00507}}\;{\rm{mol}}\]
Determine the number of moles in 165 grams of CaSO4.
1.21 mol
The moles and mass are connected through the molar mass. The molar mass of CaSO4 is 136.1 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{136}}{\rm{.1}}\;{\rm{g}}\,{\rm{CaS}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(CaS}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\,{\rm{ = }}\;{\rm{165}}\cancel{{{\rm{g}}\,{\rm{CaS}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{136}}{\rm{.1}}\,\cancel{{{\rm{g}}\,{\rm{CaS}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.21}}\;{\rm{mol}}\]
Calculate the molar mass of N2O4 and determine how many moles of it are in a 23.9 g sample.
0.260 mol
The moles and mass are connected through the molar mass. The molar mass of N2O4 is 92.0 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{92}}{\rm{.0}}\;{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\,{\rm{ = }}\;{\rm{23}}{\rm{.9}}\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{92}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.260}}\;{\rm{mol}}\]
Calculate the number of moles in 165 grams of C3H6O.
2.84 mol
The moles and mass are connected through the molar mass. The molar mass of C3H6O is 58.1 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{58}}{\rm{.1}}\;{\rm{g}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{165}}\cancel{{{\rm{g}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{58}}{\rm{.1}}\,\cancel{{{\rm{g}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.84}}\;{\rm{mol}}\]
Determine the number of moles in 452 grams of Co(NO3)3.
1.84 mol
The moles and mass are connected through the molar mass. The molar mass of Co(NO3)3 is 245 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{58}}{\rm{.1}}\;{\rm{g}}\,{\rm{Co}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{n}}\;{\rm{(Co}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{452}}\cancel{{{\rm{g}}\,{\rm{Co}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{245}}\,\cancel{{{\rm{g}}\,{\rm{Co}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.84}}\;{\rm{mol}}\]
Calculating Mass from Moles
Calculate the mass in grams of 0.598 moles of Fe.
33.4 g
The molar mass of Fe is 55.8 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}\,{\rm{Fe}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{m}}\;{\rm{(Fe)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.598}}\;\cancel{{{\rm{mol}}\,{\rm{Fe}}}}\,{\rm{ \times }}\,\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}\,{\rm{Fe}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Fe}}}}}}\;{\rm{ = }}\;{\rm{33}}{\rm{.4}}\;{\rm{g}}\]
Calculate the mass in grams of 0.168 moles of NO.
5.04 g
The molar mass of NO is 30.0 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{30}}{\rm{.0}}\;{\rm{g}}\,{\rm{NO}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{m}}\;{\rm{(NO)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.168}}\;\cancel{{{\rm{mol}}\,{\rm{NO}}}}\,{\rm{ \times }}\,\frac{{{\rm{30}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{NO}}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.04}}\;{\rm{g}}\]
Calculate the mass in grams of 0.987 moles of (NH4)2S.
67.3 g
The molar mass of (NH4)2S is 68.2 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{68}}{\rm{.2}}\;{\rm{g}}\,{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{m}}\;{\rm{(}}{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)_{\rm{2}}}{\rm{S)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.987}}\;\cancel{{{\rm{mol}}\,{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}}}\,{\rm{ \times }}\,\frac{{{\rm{68}}{\rm{.2}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}}}}}\;{\rm{ = }}\;{\rm{67}}{\rm{.3}}\;{\rm{g}}\]
Calculate the mass in grams of 6.81 moles of Al2(SO4)3.
2.33 x 103 g
The molar mass of Al2(SO4)3 is 342.2 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{342}}{\rm{.2}}\;{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{m}}\;{\rm{(A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{6}}{\rm{.81}}\;\cancel{{{\rm{mol}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)}_{\rm{3}}}}}\,{\rm{ \times }}\,\frac{{{\rm{342}}{\rm{.2}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.33}}\, \times \;{10^3}\;{\rm{g}}\]
Calculate the mass in grams of 2.64 moles of methanol, CH3OH.
84.5 g
The molar mass of CH3OH is 32.0 g/mol, therefore the conversion factor is:
\[\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{m}}\;{\rm{(C}}{{\rm{H}}_{\rm{3}}}{\rm{OH)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.64}}\;\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}\,{\rm{ \times }}\,\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}}}\;{\rm{ = }}\;{\rm{84}}{\rm{.5}}\;{\rm{g}}\]
Calculate the mass in grams of 9.42 moles of NiCl2·6H2O.
2.24 x 103 g
The molar mass of NiCl2·6H2O is 219.6 g/mol. Do not forget to add the mass of 6 moles of H2O (6 x 18 = 108 g/mol) to the molar mass of NiCl2. Therefore, the conversion factor is:
\[\frac{{{\rm{237}}{\rm{.6}}\;{\rm{g}}\,{\rm{NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{1}}\;{\rm{mol}}}}\]
\[{\rm{m}}\;{\rm{(NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{9}}{\rm{.42}}\;\cancel{{{\rm{mol}}\,{\rm{NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\,{\rm{ \times }}\,\frac{{{\rm{237}}{\rm{.6}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.24}}\, \times \,{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{g}}\]
Calculating the Number of Molecules from the Moles
How many molecules are there in a 0.487 mol sample of PCl5?
2.93 x 1023
To determine the number of molecules, we use the Avogadro’s number which shows the number of particles (molecules) in one mole of the sample.
\[\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}}}{{{\rm{1}}\,{\rm{mol}}}}\,\,or\,\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}}}\;\;\;\;\]
\[{\rm{N}}\;{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.487}}\,\cancel{{{\rm{mol}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\; = \;2.93\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]
How many molecules (formula units) are there in a 5.84 mol sample of Na2SO3.
3.52 x 1024
\[{\rm{N}}\;{\rm{(N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{5}}{\rm{.84}}\,\cancel{{{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}}}}}\; = \;3.52\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]
How many molecules of sucrose, C12H22O11 are there in a 0.684 mol sample?
4.12 x 1023
\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.684}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}}}}}\; = \;4.12\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]
Calculate the number of molecules in a 3.25-mol sample of propane, C3H8.
1.96 x 1024
\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{)}}\,{\rm{ = }}\;{\rm{3}}{\rm{.25}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}}}\; = \;1.96\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]
How many moles is 5.80 x 1025 molecules of POCl3?
96.3 moles
This is a conversion in reverse directions, therefore, we use the other conversion factor linking the number of moles and number of molecules.
\[{\rm{n}}\;{\rm{(POC}}{{\rm{l}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;5.80\; \times \;{10^{25}}\,\cancel{{{\rm{molecules}}\,{\rm{POC}}{{\rm{l}}_{\rm{3}}}\,}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}\,{\rm{POC}}{{\rm{l}}_{\rm{3}}}{\rm{ }}}}{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,\cancel{{{\rm{molecules}}\;{\rm{POC}}{{\rm{l}}_{\rm{3}}}}}}}\; = \;96.3\,{\rm{moles}}\;\;\;\]
Calculating the Number of Molecules from the Mass
How many molecules are there in a 5.12-g sample of K2O?
3.27 x 1022
There is an extra step here which is to first convert the mass to moles. After this, it is the same conversion as we have seen in the previous section for converting moles to the number of molecules using the Avogadro’s number. You can solve these types of problems by step-by-step or one-step conversion.
The first step is to convert the mass to moles:
\[{\rm{n}}\;{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{5}}{\rm{.12}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}}}{{\rm{K}}_{\rm{2}}}{\rm{O}}}}{{{\rm{94}}{\rm{.2}}\cancel{{\rm{g}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0543524}}\;{\rm{mol}}\;\]
Keep the decimals and round them off in the last step.
Once we have the number of moles, we can calculate the number of molecules.
\[{\rm{N}}\;{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.0543524}}\,\cancel{{{\rm{mol}}\,{{\rm{K}}_{\rm{2}}}{\rm{O}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{K}}_{\rm{2}}}{\rm{O}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{K}}_{\rm{2}}}{\rm{O}}}}}}\; = \;3.27\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]
A quicker method is to combine these two steps in one. Just make sure to use the correct conversion factor that allows canceling the units.
\[{\rm{N}}\;{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{5}}{\rm{.12}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{K}}_{\rm{2}}}{\rm{O}}}}}}{{{\rm{94}}{\rm{.2}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{K}}_{\rm{2}}}{\rm{O}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{K}}_{\rm{2}}}{\rm{O}}}}}}\; = \;3.27\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]
Calculate the number of molecules of butane, C4H10, in its 2.40-gram sample.
2.49 x 1022
\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.40}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}}}{{{\rm{58}}{\rm{.1}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}}}\; = \;2.49\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]
How many Ethylene, C2H4 molecules are present in a 46.2 g sample? The molar mass of C2H4 is 28.0 g/mol.
9.90 x 1023
\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}\,{\rm{ = }}\;{\rm{46}}{\rm{.2}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}}}{{{\rm{28}}{\rm{.1}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}}}\; = \;9.90\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]
Calculating the Number of Atoms
Calculate the number of atoms in a 2.56-mol sample of Ca.
3.84 x 1022
We have seen how to convert the number of moles to the number of molecules using the Avogadros’ number. Finding the number of atoms follows the same steps when working with atoms. Because one mole contains 6.022 x 1023 atoms of Ca, we can calculate the number of atoms by multiplying the number of moles by the Avogadro’s number:
\[{\rm{N}}\;{\rm{(Ca)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.56}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{\rm{Ca}}}}}}{{{\rm{40}}{\rm{.1}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{Ca}}\,{\rm{atoms}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{Ca}}}}}}\; = \;3.84\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]
How many carbon atoms are there in a 0.590 mol sample of CCl4.
3.55 x 1023
To find the number of atoms/ions in a molecule, multiply the number of molecules by the subscript of that atom. In this case, the subscript of carbon is one, so the number of carbon atoms is going to be equal to the number of molecules.
\[{\rm{N}}\;{\rm{(C atoms)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.590}}\,\cancel{{{\rm{mol}}\,{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{CC}}{{\rm{l}}_{\rm{4}}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{C}}}}{{{\rm{1}}\,\cancel{{{\rm{CC}}{{\rm{l}}_{\rm{4}}}\,{\rm{molecule}}}}\,}}\; = \;3.55\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]
How many carbon atoms are there in a 0.964 mol sample of C2H6.
1.16 x 1024
This is very similar to the previous example with the difference that there are 2 carbon atoms in each molecule of C2H6. Therefore, there is going to be twice more carbon atoms than C2H6 molecules.
\[{\rm{N}}\;{\rm{(C atoms)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.964}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{C}}}}{{{\rm{1}}\,\cancel{{{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{6}}}\,{\rm{molecule}}}}\,}}\; = \;1.16\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]
So, if we stop the conversion before the last step, we’d see that there are 5.805 x 1023 molecules of C2H6 and therefore, we multiply that number by two to determine the number of carbon atoms.
Which sample contains a more Cl atoms: a) 1.25 moles of CH2Cl2 b) 2.15 moles of CH3Cl
a) 1.25 moles of CH2Cl2
Sometimes, you may need to save time and solve the problem using a shortcut. For example, here, if we figure which sample has a greater number of moles of Cl, we can conclude that it also has more Cl atoms because we only multiply both numbers by the Avogadros’ number.
So, 1.25 mol CH2Cl2 contains 2 x 1.25 = 2.50 mol Cl atoms. 2.15 mol CH3Cl, on the other hand, contains the same number of Cl atoms because each molecule contains one atom of Cl.
Therefore, 1.25 mol CH2Cl2 contains more Cl atoms than 2.15 mol CH3Cl.
Let’s confirm this with a complete calculation to practice it a little more too:
\[{\rm{N}}\;{\rm{(Cl atoms)}}\,{\rm{ = }}\;{\rm{1}}{\rm{.25}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{Cl}}}}{{{\rm{1}}\,\cancel{{{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\,{\rm{molecule}}}}\,}}\; = \;1.51\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]
\[{\rm{N}}\;{\rm{(Cl atoms)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.15}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{Cl}}}}{{{\rm{1}}\,\cancel{{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\,{\rm{molecule}}}}\,}}\; = \;1.29\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]
The molecular formula of morphine is C17H19NO3. How many carbon atoms are in a 34.7-gram sample of morphine?
1.25 x 1024
One molecule of C17H19NO3 contains 17 carbon atoms, so once we find the number of molecules, we multiply it by 17:
\[{\rm{N}}\;{\rm{(C}}\,{\rm{atoms)}}\,{\rm{ = }}\;{\rm{34}}{\rm{.7}}\;\cancel{{{\rm{g}}\;{\rm{Mor}}{\rm{.}}\,}}{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{\rm{Mor}}{\rm{.}}}}}}{{{\rm{285}}{\rm{.3}}\,\cancel{{{\rm{g}}\,{\rm{Mor}}{\rm{.}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{Mor}}{\rm{.}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{Mor}}{\rm{.}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{17}}\;{\rm{C}}\,{\rm{atoms}}}}{{{\rm{1}}\,\cancel{{{\rm{molecule}}\,{\rm{Mor}}{\rm{.}}}}}}{\rm{ = }}\;{\rm{1}}{\rm{.25}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]
Isopropyl alcohol, also known as isopropanol, has found a widespread application in the preparation of pharmaceutical products. Answer the following questions considering that the molecular formula of isopropanol is C3H8O.
a) How many moles of C3H8O are contained in a 12.0 g sample of the alcohol?
b) How many molecules of C3H8O are contained in a 12.0 g sample of the alcohol?
c) How many atoms of oxygen are contained in a 12.0 g sample of the isopropyl alcohol (C3H8O)?
d) How many atoms of carbon are contained in a 12.0 g sample of the isopropyl alcohol (C3H8O)?
a) 0.200 moles of C3H8O
b) 1.20 x 1023 molecules of C3H8O
c) 1.20 x 1023 oxygen atoms
d) 3.60 x 1023 carbon atoms
Check Also
- Subatomic particles and Isotopes
- Naming Monatomic and Polyatomic Ions
- Naming Ionic Compounds
- Naming Covalent Compounds
- Naming Acids and Bases
- Atomic and Molecular Masses
- The Mole and Molar Mass
- Percent Composition and Empirical Formula
- Stoichiometry of Chemical Reactions
- Limiting Reactant
- Limiting Reactant Practice Problems
- Reaction/Percent Yield
- Stoichiometry Practice Problems