Stoichiometry

Grams to Molecules and Molecules to Mass

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 N_A:

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 10²⁴molecules. Remember, these were the conversion factors we got for finding the number of molecules for a PCl5 sample:

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.

answer

2.21 mol

Solution

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.

answer

0.0336 mol

Solution

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.

answer

0.00634 mol

Solution

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.

answer

0.0611 mol

Solution

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.

answer

0.00507 mol

Solution

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.

answer

1.21 mol

Solution

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.

answer

0.260 mol

Solution

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.

answer

2.84 mol

Solution

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.

answer

1.84 mol

Solution

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:

\[\frac{{{\rm{58}}{\rm{.1}}\;{\rm{g}}\,{\rm{Co}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{n}}\;{\rm{(Co}}{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{452}}\cancel{{{\rm{g}}\,{\rm{Co}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{245}}\,\cancel{{{\rm{g}}\,{\rm{Co}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.84}}\;{\rm{mol}}\]

10.

Calculating Mass from Moles

Calculate the mass in grams of 0.598 moles of Fe.

answer

33.4 g

Solution

The molar mass of Fe is 55.8 g/mol, therefore the conversion factor is:

\[\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}\,{\rm{Fe}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{m}}\;{\rm{(Fe)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.598}}\;\cancel{{{\rm{mol}}\,{\rm{Fe}}}}\,{\rm{ \times }}\,\frac{{{\rm{55}}{\rm{.8}}\;{\rm{g}}\,{\rm{Fe}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{Fe}}}}}}\;{\rm{ = }}\;{\rm{33}}{\rm{.4}}\;{\rm{g}}\]

11.

Calculate the mass in grams of 0.168 moles of NO.

answer

5.04 g

Solution

The molar mass of NO is 30.0 g/mol, therefore the conversion factor is:

\[\frac{{{\rm{30}}{\rm{.0}}\;{\rm{g}}\,{\rm{NO}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{m}}\;{\rm{(NO)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.168}}\;\cancel{{{\rm{mol}}\,{\rm{NO}}}}\,{\rm{ \times }}\,\frac{{{\rm{30}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{NO}}}}}}\;{\rm{ = }}\;{\rm{5}}{\rm{.04}}\;{\rm{g}}\]

12.

Calculate the mass in grams of 0.987 moles of (NH₄)₂S.

answer

67.3 g

Solution

The molar mass of (NH₄)₂S is 68.2 g/mol, therefore the conversion factor is:

\[\frac{{{\rm{68}}{\rm{.2}}\;{\rm{g}}\,{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{m}}\;{\rm{(}}{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)_{\rm{2}}}{\rm{S)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.987}}\;\cancel{{{\rm{mol}}\,{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}}}\,{\rm{ \times }}\,\frac{{{\rm{68}}{\rm{.2}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{{\left( {{\rm{N}}{{\rm{H}}_{\rm{4}}}} \right)}_{\rm{2}}}{\rm{S}}}}}}\;{\rm{ = }}\;{\rm{67}}{\rm{.3}}\;{\rm{g}}\]

13.

Calculate the mass in grams of 6.81 moles of Al₂(SO₄)₃.

answer

2.33 x 10³ g

Solution

The molar mass of Al₂(SO₄)₃is 342.2 g/mol, therefore the conversion factor is:

\[\frac{{{\rm{342}}{\rm{.2}}\;{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{m}}\;{\rm{(A}}{{\rm{l}}_{\rm{2}}}{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{6}}{\rm{.81}}\;\cancel{{{\rm{mol}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)}_{\rm{3}}}}}\,{\rm{ \times }}\,\frac{{{\rm{342}}{\rm{.2}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\left( {{\rm{S}}{{\rm{O}}_{\rm{4}}}} \right)}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.33}}\, \times \;{10^3}\;{\rm{g}}\]

14.

Calculate the mass in grams of 2.64 moles of methanol, CH₃OH.

answer

84.5 g

Solution

The molar mass of CH₃OH is 32.0 g/mol, therefore the conversion factor is:

\[\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{m}}\;{\rm{(C}}{{\rm{H}}_{\rm{3}}}{\rm{OH)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.64}}\;\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}\,{\rm{ \times }}\,\frac{{{\rm{32}}{\rm{.0}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{OH}}}}}}\;{\rm{ = }}\;{\rm{84}}{\rm{.5}}\;{\rm{g}}\]

15.

Calculate the mass in grams of 9.42 moles of NiCl₂·6H₂O.

answer

2.24 x 10³ g

Solution

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 NiCl_2. Therefore, the conversion factor is:

\[\frac{{{\rm{237}}{\rm{.6}}\;{\rm{g}}\,{\rm{NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{{{\rm{1}}\;{\rm{mol}}}}\]

\[{\rm{m}}\;{\rm{(NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{9}}{\rm{.42}}\;\cancel{{{\rm{mol}}\,{\rm{NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\,{\rm{ \times }}\,\frac{{{\rm{237}}{\rm{.6}}\;{\rm{g}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{NiC}}{{\rm{l}}_{\rm{2}}}{\rm{\cdot6}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.24}}\, \times \,{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{g}}\]

16.

Calculating the Number of Molecules from the Moles

How many molecules are there in a 0.487 mol sample of PCl₅?

answer

2.93 x 10²³

Solution

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.

\[\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}}}{{{\rm{1}}\,{\rm{mol}}}}\,\,or\,\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecules}}}}\;\;\;\;\]

\[{\rm{N}}\;{\rm{(PC}}{{\rm{l}}_{\rm{5}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.487}}\,\cancel{{{\rm{mol}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{PC}}{{\rm{l}}_{\rm{5}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{PC}}{{\rm{l}}_{\rm{5}}}}}}}\; = \;2.93\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]

17.

How many molecules (formula units) are there in a 5.84 mol sample of Na₂SO₃.

answer

3.52 x 10²⁴

Solution

\[{\rm{N}}\;{\rm{(N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{5}}{\rm{.84}}\,\cancel{{{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{3}}}}}}}\; = \;3.52\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]

18.

How many molecules of sucrose, C₁₂H₂₂O₁₁ are there in a 0.684 mol sample?

answer

4.12 x 10²³

Solution

\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.684}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{{\rm{12}}}}{{\rm{H}}_{{\rm{22}}}}{{\rm{O}}_{{\rm{11}}}}}}}}\; = \;4.12\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]

19.

Calculate the number of molecules in a 3.25-mol sample of propane, C₃H₈.

answer

1.96 x 10²⁴

Solution

\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{)}}\,{\rm{ = }}\;{\rm{3}}{\rm{.25}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}}}}}\; = \;1.96\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]

20.

How many moles is 5.80 x 10²⁵ molecules of POCl₃?

answer

96.3 moles

Solution

This is a conversion in reverse directions, therefore, we use the other conversion factor linking the number of moles and number of molecules.

\[{\rm{n}}\;{\rm{(POC}}{{\rm{l}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;5.80\; \times \;{10^{25}}\,\cancel{{{\rm{molecules}}\,{\rm{POC}}{{\rm{l}}_{\rm{3}}}\,}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}\,{\rm{POC}}{{\rm{l}}_{\rm{3}}}{\rm{ }}}}{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,\cancel{{{\rm{molecules}}\;{\rm{POC}}{{\rm{l}}_{\rm{3}}}}}}}\; = \;96.3\,{\rm{moles}}\;\;\;\]

21.

Calculating the Number of Molecules from the Mass

How many molecules are there in a 5.12-g sample of K₂O?

answer

3.27 x 10²²

Solution

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.

\[{\rm{N}}\;{\rm{(}}{{\rm{K}}_{\rm{2}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{5}}{\rm{.12}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{K}}_{\rm{2}}}{\rm{O}}}}}}{{{\rm{94}}{\rm{.2}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{K}}_{\rm{2}}}{\rm{O}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{K}}_{\rm{2}}}{\rm{O}}}}}}\; = \;3.27\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]

22.

Calculate the number of molecules of butane, C₄H₁₀, in its 2.40-gram sample.

answer

2.49 x 10²²

Solution

\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}{\rm{)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.40}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}}}{{{\rm{58}}{\rm{.1}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{4}}}{{\rm{H}}_{{\rm{10}}}}}}}}\; = \;2.49\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]

23.

How many Ethylene, C₂H₄ molecules are present in a 46.2 g sample? The molar mass of C₂H₄ is 28.0 g/mol.

answer

9.90 x 10²³

Solution

\[{\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{)}}\,{\rm{ = }}\;{\rm{46}}{\rm{.2}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}}}{{{\rm{28}}{\rm{.1}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}}}}}\; = \;9.90\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]

24.

Calculating the Number of Atoms

Calculate the number of atoms in a 2.56-mol sample of Ca.

answer

3.84 x 10²²

Solution

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:

\[{\rm{N}}\;{\rm{(Ca)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.56}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{\rm{Ca}}}}}}{{{\rm{40}}{\rm{.1}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{Ca}}\,{\rm{atoms}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{Ca}}}}}}\; = \;3.84\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{22}}}}\;\;\;\]

25.

How many carbon atoms are there in a 0.590 mol sample of CCl₄.

answer

3.55 x 10²³

Solution

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.

\[{\rm{N}}\;{\rm{(C atoms)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.590}}\,\cancel{{{\rm{mol}}\,{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{CC}}{{\rm{l}}_{\rm{4}}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{CC}}{{\rm{l}}_{\rm{4}}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{C}}}}{{{\rm{1}}\,\cancel{{{\rm{CC}}{{\rm{l}}_{\rm{4}}}\,{\rm{molecule}}}}\,}}\; = \;3.55\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;\]

26.

How many carbon atoms are there in a 0.964 mol sample of C₂H₆.

answer

1.16 x 10²⁴

Solution

This is very similar to the previous example with the difference that there are 2 carbon atoms in each molecule of C₂H_6.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.

27.

Which sample contains a more Cl atoms: a) 1.25 moles of CH₂Cl₂ b) 2.15 moles of CH₃Cl

answer

a) 1.25 moles of CH₂Cl₂

Solution

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 CH₂Cl₂contains 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 CH₂Cl₂contains more Cl atoms than 2.15 mol CH₃Cl.

Let’s confirm this with a complete calculation to practice it a little more too:

\[{\rm{N}}\;{\rm{(Cl atoms)}}\,{\rm{ = }}\;{\rm{1}}{\rm{.25}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{Cl}}}}{{{\rm{1}}\,\cancel{{{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{l}}_{\rm{2}}}\,{\rm{molecule}}}}\,}}\; = \;1.51\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]

\[{\rm{N}}\;{\rm{(Cl atoms)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.15}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{atom}}\,{\rm{of}}\,{\rm{Cl}}}}{{{\rm{1}}\,\cancel{{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{Cl}}\,{\rm{molecule}}}}\,}}\; = \;1.29\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]

28.

The molecular formula of morphine is C₁₇H₁₉NO₃. How many carbon atoms are in a 34.7-gram sample of morphine?

answer

1.25 x 10²⁴

Solution

One molecule of C₁₇H₁₉NO₃contains 17 carbon atoms, so once we find the number of molecules, we multiply it by 17:

\[{\rm{N}}\;{\rm{(C}}\,{\rm{atoms)}}\,{\rm{ = }}\;{\rm{34}}{\rm{.7}}\;\cancel{{{\rm{g}}\;{\rm{Mor}}{\rm{.}}\,}}{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{\rm{Mor}}{\rm{.}}}}}}{{{\rm{285}}{\rm{.3}}\,\cancel{{{\rm{g}}\,{\rm{Mor}}{\rm{.}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\cancel{{{\rm{Mor}}{\rm{.}}\,{\rm{molecules}}}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{Mor}}{\rm{.}}}}}}\;{\rm{ \times }}\,\frac{{{\rm{17}}\;{\rm{C}}\,{\rm{atoms}}}}{{{\rm{1}}\,\cancel{{{\rm{molecule}}\,{\rm{Mor}}{\rm{.}}}}}}{\rm{ = }}\;{\rm{1}}{\rm{.25}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;\]

29.

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)?

answer

a) 0.200 moles of C₃H₈O

answer

b) 1.20 x 10²³ molecules of C₃H₈O

answer

c) 1.20 x 10²³oxygen atoms

answer

d) 3.60 x 10²³ carbon atoms

video

30.

Check Also

Atoms, Molecules, and Ions

Stoichiometry

Converting the Number of Molecules to Grams

Practice

Calculating Moles from Mass

Calculating Mass from Moles

Calculating the Number of Molecules from the Moles

Calculating the Number of Molecules from the Mass

Calculating the Number of Atoms

Leave a Comment Cancel reply