**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 ^{12}C isotope which happens to be 6.022 x 10^{23} atoms. This is called Avogadro’s number (*N*_{A} = 6.022 x 10^{23}).

One mole of any element contains 6.022 x 10^{23} of its atoms, and a mole of a compound contains 6.022 x 10^{23} molecules, ions-ions, and in general, 1 mol = 6.022 x 10^{23} 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 10^{23}.

**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:

Notice that the exact answer in the calculator is 0.5046728 and we round it off to three significant figures because both initial numbers contain three significant figures.

**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 0.0480 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:

\[{\rm{m}}\;{\rm{(C)}}\,{\rm{ = }}\;{\rm{0}}{\rm{.0480}}\;\cancel{{{\rm{mol}}\,{\rm{C}}}}\,{\rm{ \times }}\,\frac{{{\rm{12}}{\rm{.0}}\;{\rm{g}}\,{\rm{C}}}}{{{\rm{1}}\;\cancel{{{\rm{mol}}\;{\rm{C}}}}}}\;{\rm{ = }}\;{\rm{0}}{\rm{.576}}\,{\rm{g}}\]

**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 10^{23}, 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 *N*_{A}:

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

You can also do this in a one-step conversion by 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 Na_{2}CO_{3}?

In each unit of Na_{2}CO_{3}, there are 2 Na atoms, so we are going to determine the number of Na_{2}CO_{3} molecules and multiply it by two.

Let’s also break this down and do each step of the 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 Na_{2}CO_{3}:

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

**Step 3.** In the last step, we recognize that every molecule Na_{2}CO_{3 }has two sodium atoms, and therefore, we multiply the number of molecules by two to determine the number of Na atoms:

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

The little inconsistency in the final answer is due to rounding off the intermediate numbers. I do it here mainly to save space, but you can keep the numbers in the calculator and round them off at the end. This is the accurate way of rounding off to the correct number of significant figures in multistep calculations.

*Note:* Although Na_{2}CO_{3} 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**

- 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**

#### Practice

**Calculating Moles from Mass**

Determine the number of moles in 59.7 grams of Al.

Calculate the number of moles in 0.647 grams of Al_{2}O_{3}.

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

Determine the number of moles in 0.385 grams of N_{2}O_{3}.

Determine the number of moles in 165 grams of CaSO_{4}.

Calculate the molar mass of N_{2}O_{4} and determine how many moles of it are in a 23.9 g sample.

Calculate the number of moles in 165 grams of C_{3}H_{6}O.

Determine the number of moles in 452 grams of Co(NO_{3})_{3}.

**Calculating Mass from Moles **

Calculate the mass in grams of 0.598 moles of Fe.

Calculate the mass in grams of 0.987 moles of (NH_{4})_{2}S.

Calculate the mass in grams of 6.81 moles of Al_{2}(SO_{4})_{3}.

Calculate the mass in grams of 2.64 moles of methanol, CH_{3}OH.

Calculate the mass in grams of 9.42 moles of NiCl_{2}·6H_{2}O.

**Calculating the Number of Molecules from the Moles**

How many molecules are there in a 0.487 mol sample of PCl_{5}?

How many molecules (formula units) are there in a 5.84 mol sample of Na_{2}SO_{3}.

How many molecules of sucrose, C_{12}H_{22}O_{11} are there in a 0.684 mol sample?

Calculate the number of molecules in a 3.25-mol sample of propane, C_{3}H_{8}.

How many moles is 5.80 x 10^{25} molecules of POCl_{3}?

**Calculating the Number of Molecules from the Mass**

How many molecules are there in a 5.12-g sample of K_{2}O?

How many molecules of glucose, C_{6}H_{12}O_{6} are there in a 35.0 g sample?

Calculate the number of molecules of butane, C_{4}H_{10}, in its 2.40-gram sample.

How many Ethylene, C_{2}H_{4} molecules are present in a 46.2 g sample? The molar mass of C_{2}H_{4} is 28.1 g/mol.

**Calculating the Number of Atoms**

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

How many carbon atoms are there in a 0.590 mol sample of CCl_{4}.

How many carbon atoms are there in a 0.964 mol sample of C_{2}H_{6}.

Which sample contains more Cl atoms: a) 1.25 moles of CH_{2}Cl_{2} b) 2.15 moles of CH_{3}Cl

The molecular formula of morphine is C_{17}H_{19}NO_{3}. How many carbon atoms are in a 34.7-gram sample of morphine?

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 C_{3}H_{8}O.

a) How many moles of C_{3}H_{8}O are contained in a 12.0 g sample of the alcohol?

b) How many molecules of C_{3}H_{8}O 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 (C_{3}H_{8}O)?

d) How many atoms of carbon are contained in a 12.0 g sample of the isopropyl alcohol (C_{3}H_{8}O)?

why is the rounding inconsistent. sometimes they round up sometimes not, its within 3 significant figures but some r rounded up some down. if im to buy ur product its not very user friendly to have inconsistent answers. i will fail chem class because of this .

Hi there,

Rounding off to the correct number of significant figures does not always mean rounding up. Whether we round up or down depends on the number that follows the last significant figure. If that number is 5 or greater, we round up, if it is <5, then we round down. You can check this article for more details and examples: Significant Figures in Addition, Subtraction Multiplication, and Division

With this said, it is not impossible that I may overlook some things, and if this is the case, please let me know which calculation it is so I can check again.