# What is the Mole?

The mole is just a number like a dozen, a hundred, or a million. By definition, a mole is the number of atoms in a 12-g sample of 12C isotope which happens to be 6.022 x 1023 atoms. This is called Avogadro’s number (NA = 6.022 x 1023).

One mole of any element contains 6.022 x 1023 of its atoms, and a mole of a compound contains 6.022 x 1023 molecules, ions-ions, and in general, 1 mol = 6.022 x 1023 particles.

# Molar Mass

The mass of one mole of atoms/molecules/ions is called its molar mass (M) expressed in g/mol.

Numerically, the molar mass is equal to the atomic mass of a given atom or a molecule, so we can look up the molar mass of an element in the periodic table.

For example, we can say that the mass of Cu is 63.55 amu or 63.55 g/mol. The molecular mass of water is 18.0 amu, and the molar mass is 18.0 g/mol.

For the molar mass, keeping one decimal is usually an acceptable approximation, and for the Avogadro’s number, you can use 6.02 x 1023.

# Calculating Moles from the Mass

To calculate the moles from a given mass (m), the molar mass of the component is used. The given mass is the mass of the sample, and it can be any number, for example, we can have 10 g of salt, 15 g, or 100 g. The molar mass, on the other hand, is a constant number for a given atom or a molecule as it is for a specific amount of it.

So, to calculate the moles from a given mass, we make a conversion factor correlating 1 mole with the molar of the given component.

For example, how many moles of sulfur are there in a 16.2 g sample?

The conversion factors are:

$\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}{{{\rm{32}}{\rm{.1}}\;{\rm{g}}\,{\rm{S}}}}\;\,and\,\,\frac{{{\rm{32}}{\rm{.1}}\;{\rm{g}}\,{\rm{S}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{S}}}}$

These two simply indicate that 1 mol of S weighs 32.1 g, which is the molar mass of the sulfur, and we can find it in the periodic table.

To calculate the number of moles, write the given mass and multiply by the conversion factor that has the mass in the denominator so that they can be canceled:

You can also do this by using the formula:

So, for 13.2 g of sulfur, we’d have:

# Mass from the Moles

If the question asks us to calculate the mass from the moles, we will need to use the other conversion factor so that the moles now can be canceled.

For example, how many moles of carbon are there in a 0.048 mol sample?

The molar mass of carbon is 12.0 g/mol, so the conversion factors are:

$\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}}}{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{S}}}}\;\,and\,\,\frac{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{C}}}}$

Using the second conversion factor, we can calculate the mass as:

${\rm{m}}\;{\rm{(C)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.048}}\;\cancel{{{\rm{mol}}\,{\rm{C}}}}\,{\rm{ \times }}\,\frac{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{C}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.576}}\,{\rm{g}}$

# Number of Atoms from the Moles

Whether it is the number of atoms, molecules, or ions, use Avogadro’s number to calculate it from the moles.

For example, calculate the number of iron atoms in a 3.5-mole sample.

This time, we are going to use a conversion factor correlating 1 mole and the Avogadro’s number. Since 1 mole = 6.022 x 1023, we can write these two conversion factors:

$\frac{{{\rm{1}}\;{\rm{mol}}\;{\rm{Fe}}}}{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Fe}}\;{\rm{atoms}}}}\;\;\;\,and\;\;\,\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\,{\rm{Fe}}\;{\rm{atoms}}}}{{{\rm{1}}\;{\rm{mol}}\;{\rm{Fe}}}}$

# Number of Atoms from the Mass

This is going to be one additional step compared to the conversion from the number of moles. 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 atoms using NA:

For example, calculate the number of copper atoms in its 44.52 g sample.

You can also do this in a one-step conversion combining these two together:

# Number of Atoms in a Molecule

This adds another step because there may be more than one atom in a given molecule. In this case, to find the number of atoms/ions in a molecule, multiply the number of molecules by the subscript of that atom.

For example, how many Na ions are there in a 658 g sample of Na2CO3?

In each unit of Na2CO3, there are 2 Na atoms, so we are going to determine the number of Na2CO3 molecules and multiply it by two.

Let’s also break this down and do each step of calculation separately.

Step 1. In the first step, determine the number of moles:

${\rm{n}}\;{\rm{(N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{658}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}}}{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{106}}\cancel{{\rm{g}}}}}\;{\rm{ = }}\;{\rm{6}}{\rm{.20}}\;{\rm{mol}}\;$

Step 2. Calculate the number of molecules in 6.20 moles of Na2CO3:

${\rm{N}}\;{\rm{(N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{6}}{\rm{.20}}\cancel{{{\rm{mol}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.02}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{\rm{molecule}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{1}}\cancel{{{\rm{mol}}}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}\; = \;3.73\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;\;\;$

Step 3. In the last step, we recognize that every molecule Na2CO3 has two sodium atoms, and therefore, we multiply the number of molecules by two to determine the number of Na atoms:

${\rm{N}}\;{\rm{(Na)}}\,{\rm{ = }}\;3.73\cancel{{{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}{\rm{molecule}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{2}}\;{\rm{Na}}\;{\rm{ions}}}}{{\cancel{{{\rm{1}}\;{\rm{molecule}}\;{\rm{N}}{{\rm{a}}_{\rm{2}}}{\rm{C}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{7}}{\rm{.46}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{24}}}}\;{\rm{Na}}\;{\rm{ions}}$

Note: Although Na2CO3 is an ionic compound and sodium is present in an ionic form, it is a habitual way of referring to an element in these types of calculations.

Check Also

1.

## Calculating Moles from Mass

Determine the number of moles in 59.7 grams of Al.

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:

$\frac{{{\rm{27}}{\rm{.0}}\;{\rm{g}}\,{\rm{Al}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(Al)}}\,{\rm{ = }}\;{\rm{59}}{\rm{.7}}\cancel{{{\rm{g}}\,{\rm{Al}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{27}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{\rm{Al}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.21}}\;{\rm{mol}}$

2.

Determine the number of moles in 2.41 grams of FeO.

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:

$\frac{{{\rm{71}}{\rm{.8}}\;{\rm{g}}\,{\rm{FeO}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(FeO)}}\,{\rm{ = }}\;{\rm{2}}{\rm{.41}}\cancel{{{\rm{g}}\,{\rm{FeO}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{71}}{\rm{.8}}\,\cancel{{{\rm{g}}\,{\rm{FeO}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0336}}\;{\rm{mol}}$

3.

Calculate the number of moles in 0.647 grams of Al2O3.

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:

$\frac{{{\rm{71}}{\rm{.8}}\;{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.647}}\cancel{{{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{102}}\,\cancel{{{\rm{g}}\,{\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00634}}\;{\rm{mol}}$

4.

Determine the number of moles in 3.56 grams of Mg(OH)2.

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:

$\frac{{{\rm{58}}{\rm{.3}}\;{\rm{g}}\,{\rm{Mg}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(Mg}}{\left( {{\rm{OH}}} \right)_{\rm{2}}}{\rm{)}}\,{\rm{ = }}\;{\rm{3}}{\rm{.56}}\cancel{{{\rm{g}}\,{\rm{Mg}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{58}}{\rm{.3}}\,\cancel{{{\rm{g}}\,{\rm{Mg}}{{\left( {{\rm{OH}}} \right)}_{\rm{2}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.0611}}\;{\rm{mol}}$

5.

Determine the number of moles in 0.385 grams of N2O3.

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:

$\frac{{{\rm{76}}{\rm{.0}}\;{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}{\rm{)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.385}}\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{76}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.00507}}\;{\rm{mol}}$

6.

Determine the number of moles in 165 grams of CaSO4.

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:

$\frac{{{\rm{136}}{\rm{.1}}\;{\rm{g}}\,{\rm{CaS}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(CaS}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\,{\rm{ = }}\;{\rm{165}}\cancel{{{\rm{g}}\,{\rm{CaS}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{136}}{\rm{.1}}\,\cancel{{{\rm{g}}\,{\rm{CaS}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{1}}{\rm{.21}}\;{\rm{mol}}$

7.

Calculate the molar mass of N2O4 and determine how many moles of it are in a 23.9 g sample.

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:

$\frac{{{\rm{92}}{\rm{.0}}\;{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(}}{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}{\rm{)}}\,{\rm{ = }}\;{\rm{23}}{\rm{.9}}\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{92}}{\rm{.0}}\,\cancel{{{\rm{g}}\,{{\rm{N}}_{\rm{2}}}{{\rm{O}}_{\rm{4}}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.260}}\;{\rm{mol}}$

8.

Calculate the number of moles in 165 grams of C3H6O.

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:

$\frac{{{\rm{58}}{\rm{.1}}\;{\rm{g}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}}}{{{\rm{1}}\;{\rm{mol}}}}$

${\rm{n}}\;{\rm{(}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O)}}\,{\rm{ = }}\;{\rm{165}}\cancel{{{\rm{g}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}}}{{{\rm{58}}{\rm{.1}}\,\cancel{{{\rm{g}}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{\rm{O}}}}}}\;{\rm{ = }}\;{\rm{2}}{\rm{.84}}\;{\rm{mol}}$

9.

Determine the number of moles in 452 grams of Co(NO3)3.

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.

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.

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.

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.

2.33 x 103 g

Solution

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

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

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

14.

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

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.

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 NiCl2. Therefore, the conversion factor is:

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

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

16.

## 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

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.

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?

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.

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?

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?

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.

How many molecules of glucose, C6H12O6 are there in a 35.0 g sample?

1.17 x 1023

Solution

Refer to the first problem in “Calculating the number of molecules from the mass” for a detailed explanation.

${\rm{N}}\;{\rm{(}}{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}{\rm{)}}\,{\rm{ = }}\;{\rm{35}}{\rm{.0}}\cancel{{\rm{g}}}\;{\rm{ \times }}\,\frac{{{\rm{1}}\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}{{{\rm{180}}{\rm{.2}}\cancel{{\rm{g}}}}}\;{\rm{ \times }}\,\frac{{{\rm{6}}{\rm{.022}}\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}\,{\rm{molecules}}}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{{\rm{C}}_{\rm{6}}}{{\rm{H}}_{{\rm{12}}}}{{\rm{O}}_{\rm{6}}}}}}}\; = \;1.17\;{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{23}}}}\;\;\;$

23.

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

2.49 x 1022

Solution

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

24.

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

Solution

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

25.

## Calculating the Number of Atoms

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

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:

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

26.

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

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.

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

27.

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

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 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.

28.

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

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 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:

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

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

29.

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

1.25 x 1024

Solution

One molecule of C17H19NO3 contains 17 carbon atoms, so once we find the number of molecules, we multiply it by 17:

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

30.

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)

a) 0.200 moles of C3H8O

b)

b) 1.20 x 1023 molecules of C3H8O

c)