amount of substance Avogadro constant and molecular mass

Learning objectives

Outcome
Calculate the amount of substance $n$ from a known number of particles $N$ and the Avogadro constant $N_A$ .
Outcome
Calculate the mass of one molecule $m_0$ from the molar mass $M$ and vice versa.
Outcome
Apply the formula $n=\dfrac{m}{M}$ to find the mass of a sample or the amount of substance.
Outcome
Distinguish molar mass from molecular mass, stating the units.

Understand the idea

Amount of Substance, the Avogadro Constant, and Molecular Mass

Text

Matter consists of atoms and molecules, so it is convenient to count particles not individually but in portions of a definite size. The amount of substance $n$ is a physical quantity showing how many such portions (moles) are contained in a sample: one mole always contains the same number of particles, equal to the Avogadro constant $N_A \approx 6{,}022\times10^{23}\ \text{mol}^{-1}$ . The relation between the number of particles $N$ and the amount of substance is given by $n=\dfrac{N}{N_A}$ . The mass of one molecule $m_0$ is related to the molar mass $M$ (the mass of one mole) by $m_0=\dfrac{M}{N_A}$ , and the molar mass of a substance equals the sum of the molar masses of the atoms that make up the molecule. If the sample mass $m$ is known, the amount of substance can also be found from $n=\dfrac{m}{M}$ ; the units are $[n]=\text{mol}$ , $[M]=\text{g/mol}$ or $\text{kg/mol}$ , and $[m_0]=\text{kg}$ .

Worked examples

Direct

Problem

A sample contains $N=3{,}011\times10^{24}$ molecules. Find the amount of substance $n$ if $N_A=6{,}022\times10^{23}\ \text{mol}^{-1}$ .

Solution

Use the relation $n=\dfrac{N}{N_A}$ . Substitute the numerical values:

n=\dfrac{3{,}011\times10^{24}}{6{,}022\times10^{23}}=5{,}00\ \text{mol}.

Answer: $n=5{,}00$ mol.

Application

Problem

The molar mass of water is $M(\text{H}_2\text{O})=18{,}0\ \text{g/mol}$ . Find the mass of one water molecule in kilograms. Take $N_A=6{,}022\times10^{23}\ \text{mol}^{-1}$ .

Solution

Convert the molar mass to $\text{kg/mol}$ :

M=18{,}0\times10^{-3}\ \text{kg/mol}=0{,}0180\ \text{kg/mol}.

The mass of one molecule is:

m_0=\dfrac{M}{N_A}=\dfrac{0{,}0180}{6{,}022\times10^{23}}\approx2{,}99\times10^{-26}\ \text{kg}.

Answer: $m_0\approx2{,}99\times10^{-26}$ kg.

Transfer

Problem

A water drop has volume $V=0{,}050\ \text{mL}$ and density $\rho=1{,}0\ \text{g/cm}^3$ . Taking the molar mass of water as $M=18{,}0\ \text{g/mol}$ , find the amount of substance and the number of molecules in the drop.

Solution

Mass of the drop:

m=\rho V=1{,}0\ \text{g/cm}^3\times0{,}050\ \text{cm}^3=0{,}050\ \text{g}.

Amount of substance:

n=\dfrac{m}{M}=\dfrac{0{,}050}{18{,}0}\approx2{,}8\times10^{-3}\ \text{mol}.

Number of molecules:

N=nN_A\approx2{,}8\times10^{-3}\times6{,}022\times10^{23}\approx1{,}7\times10^{21}.

Answer: $n\approx2{,}8\times10^{-3}$ mol, $N\approx1{,}7\times10^{21}$ molecules.

Practice

Text

Safe self-study activity. Open a trusted educational simulation of matter or a spreadsheet. Choose a substance (for example, $\text{H}_2\text{O}$ , $\text{CO}_2$ , or $\text{NaCl}$ ), enter the atomic masses of the elements, and calculate the molar mass $M$ . Then set a sample mass $m$ , calculate $n=m/M$ and $N=nN_A$ , and also the mass of one molecule $m_0=M/N_A$ . Compare the results for equal-mass samples of different substances and explain the differences. Do not use an open flame or heat substances; perform all calculations on a computer or in your notebook.

Visual reference

Purpose

Show how the number of particles, amount of substance, Avogadro constant, sample mass, molar mass, and mass of one molecule are related.

Representation

Converter diagram: in the centre is the node $n$ (mol), around it are the nodes $N$ , $N_A$ , $m$ , $M$ , $m_0$ ; the arrows are labelled with the formulas $n=N/N_A$ , $N=nN_A$ , $n=m/M$ , $M=m/n$ , $M=m_0N_A$ , $m_0=M/N_A$ .

Real-world examples

A glass of water in the kitchen

Scenario

A glass is filled with $250\ \mathrm{mL}$ of water. The density of water is $\rho=1.00\ \mathrm{g/mL}$ , so the mass of water is $m=250\ \mathrm{g}$ . The molecular mass of water is $M_r(\mathrm{H_2O})=18.0$ , so its molar mass is $M=18.0\ \mathrm{g/mol}$ . Then the amount of substance is $n=m/M=250/18.0=13.9\ \mathrm{mol}$ , and the number of molecules is $N=nN_A=13.9\cdot6.022\times10^{23}=8.37\times10^{24}$ .

Connection

This uses the link between the mass of a substance and the number of particles: the molar (molecular) mass converts grams into moles, and the Avogadro constant converts moles into a number of molecules.

Breathing after a sprint

Scenario

After a short run, a student takes a deep breath of air with volume $V=0.50\ \mathrm{L}$ . At $20\,^\circ\mathrm{C}$ and $1\ \mathrm{atm}$ , the molar volume of air is about $V_m=24.0\ \mathrm{L/mol}$ . The average molar mass of air is $M=29\ \mathrm{g/mol}$ . Then $n=V/V_m=0.50/24.0=0.0208\ \mathrm{mol}$ , the mass of air is $m=nM=0.0208\cdot29=0.604\ \mathrm{g}$ , and the number of molecules is $N=nN_A=0.0208\cdot6.022\times10^{23}=1.25\times10^{22}$ .

Connection

The example shows how the amount of substance is linked to the number of molecules and to mass: even a small portion of air contains a huge number of particles, and molar mass allows a transition from moles to grams.

Steel cable of a bridge

Scenario

One section of a steel bridge cable has mass $m=1200\ \mathrm{kg}$ . Assuming the steel is mainly iron with molar mass $M(\mathrm{Fe})=56\ \mathrm{g/mol}=0.056\ \mathrm{kg/mol}$ , we get $n=m/M=1200/0.056=2.14\times10^{4}\ \mathrm{mol}$ . The number of iron atoms in this section is $N=nN_A=2.14\times10^{4}\cdot6.022\times10^{23}=1.29\times10^{28}$ .

Connection

This is a direct application of $n=m/M$ and multiplication by the Avogadro constant: the macroscopic mass of the cable is converted into moles and then into a number of atoms.