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3.3.2 The mole, molar mass and gas volume calculations

The mole

Definition

Mole

The unit of amount of substance, symbol mol\text{mol}mol, where one mole of a substance contains 6.02×10236.02 \times 10^{23}6.02×1023 particles.

Definition

Avogadro constant

The number of particles, such as atoms, ions or molecules, in one mole of a substance, equal to 6.02×10236.02 \times 10^{23}6.02×1023.

  1. Atoms are far too small to count one by one, so chemists count them in very large, fixed-size groups called moles.
  2. The mole, symbol mol\text{mol}mol, is the unit of amount of substance.
  3. One mole of any substance contains 6.02×10236.02 \times 10^{23}6.02×1023 particles, and this number is the Avogadro constant.
  4. The particles being counted can be atoms, ions or molecules, so you must always say which particle you mean.
    1. One mole of helium contains 6.02×10236.02 \times 10^{23}6.02×1023 atoms, while one mole of carbon dioxide contains 6.02×10236.02 \times 10^{23}6.02×1023 molecules.
  5. The number of particles in a sample is found using number of particles=amount (mol)×6.02×1023\text{number of particles} = \text{amount (mol)} \times 6.02 \times 10^{23}number of particles=amount (mol)×6.02×1023.
  6. Rearranging gives amount (mol)=number of particles6.02×1023\text{amount (mol)} = \dfrac{\text{number of particles}}{6.02 \times 10^{23}}amount (mol)=6.02×1023number of particles​.
  7. One mole of a compound can contain more than one mole of a particular atom or ion, depending on its formula.
    1. One mole of H2O\text{H}_2\text{O}H2​O contains 2 mol2\ \text{mol}2 mol of hydrogen atoms and 1 mol1\ \text{mol}1 mol of oxygen atoms.
    2. One mole of CaCl2\text{CaCl}_2CaCl2​ contains 1 mol1\ \text{mol}1 mol of Ca2+\text{Ca}^{2+}Ca2+ ions and 2 mol2\ \text{mol}2 mol of Cl−\text{Cl}^-Cl− ions.
Example

Counting particles

  • A sample contains 0.250 mol0.250\ \text{mol}0.250 mol of water molecules.
number of molecules=0.250×6.02×1023=1.51×1023 \text{number of molecules} = 0.250 \times 6.02 \times 10^{23} = 1.51 \times 10^{23} number of molecules=0.250×6.02×1023=1.51×1023
  • Each water molecule contains three atoms, so the number of atoms is three times larger.
number of atoms=3×0.250×6.02×1023=4.52×1023 \text{number of atoms} = 3 \times 0.250 \times 6.02 \times 10^{23} = 4.52 \times 10^{23} number of atoms=3×0.250×6.02×1023=4.52×1023

Molar mass

Definition

Molar mass

The mass of one mole of a substance, measured in g/mol\text{g/mol}g/mol.

  1. The molar mass, MMM, of a substance is the mass of one mole of it, measured in g/mol\text{g/mol}g/mol.
  2. The molar mass has the same numerical value as the relative mass of the particles.
    1. For atoms, the molar mass equals ArA_rAr​, so 1 mol1\ \text{mol}1 mol of carbon atoms has a mass of 12 g12\ \text{g}12 g.
    2. For molecules and ionic compounds, the molar mass equals MrM_rMr​, so 1 mol1\ \text{mol}1 mol of water has a mass of 18 g18\ \text{g}18 g.
  3. The key relationship is amount of substance (mol)=mass (g)molar mass (g/mol)\text{amount of substance (mol)} = \dfrac{\text{mass (g)}}{\text{molar mass (g/mol)}}amount of substance (mol)=molar mass (g/mol)mass (g)​.
  4. In symbols this is n=mMn = \dfrac{m}{M}n=Mm​, which rearranges to m=n×Mm = n \times Mm=n×M and M=mnM = \dfrac{m}{n}M=nm​.
  5. Finding MMM from a measured mass and amount tells you the ArA_rAr​ or MrM_rMr​, because they have the same numerical value.
  6. The mass must be in grams, so convert a mass in kilograms by multiplying by 100010001000.
Example

Amount and mass

  • Calcium carbonate, CaCO3\text{CaCO}_3CaCO3​, has Mr=40+12+(3×16)=100M_r = 40 + 12 + (3 \times 16) = 100Mr​=40+12+(3×16)=100, so its molar mass is 100 g/mol100\ \text{g/mol}100 g/mol.
n=25.0100=0.250 mol n = \dfrac{25.0}{100} = 0.250\ \text{mol} n=10025.0​=0.250 mol
  • Sodium hydroxide, NaOH\text{NaOH}NaOH, has Mr=23+16+1=40M_r = 23 + 16 + 1 = 40Mr​=23+16+1=40, so 0.400 mol0.400\ \text{mol}0.400 mol of it has the mass below.
m=0.400×40=16.0 g m = 0.400 \times 40 = 16.0\ \text{g} m=0.400×40=16.0 g
Example

Identifying an element from its molar mass

  • A sample of a metal containing 0.150 mol0.150\ \text{mol}0.150 mol of atoms has a mass of 3.60 g3.60\ \text{g}3.60 g.
M=3.600.150=24.0 g/mol M = \dfrac{3.60}{0.150} = 24.0\ \text{g/mol} M=0.1503.60​=24.0 g/mol
  • The metal has Ar=24A_r = 24Ar​=24, so it is magnesium.
Example

Particles from a mass

  • A sample of carbon dioxide, Mr=44M_r = 44Mr​=44, has a mass of 4.40 g4.40\ \text{g}4.40 g.
n=4.4044=0.100 mol n = \dfrac{4.40}{44} = 0.100\ \text{mol} n=444.40​=0.100 mol number of molecules=0.100×6.02×1023=6.02×1022 \text{number of molecules} = 0.100 \times 6.02 \times 10^{23} = 6.02 \times 10^{22} number of molecules=0.100×6.02×1023=6.02×1022

Molar gas volume

Definition

Molar gas volume

The volume occupied by one mole of any gas, taken as 24 dm324\ \text{dm}^324 dm3 at room temperature and pressure.

  1. At the same temperature and pressure, one mole of any gas takes up the same volume, whichever gas it is.
  2. This is because gas particles are very far apart, so the volume depends on the number of particles and hardly at all on their size.
  3. At room temperature and pressure, r.t.p., one mole of any gas is taken to occupy 24 dm324\ \text{dm}^324 dm3, which is 24 000 cm324\,000\ \text{cm}^324000 cm3.
  4. For a gas volume in dm3\text{dm}^3dm3, use amount (mol)=volume (dm3)24\text{amount (mol)} = \dfrac{\text{volume (dm}^3\text{)}}{24}amount (mol)=24volume (dm3)​.
  5. For a gas volume in cm3\text{cm}^3cm3, use amount (mol)=volume (cm3)24 000\text{amount (mol)} = \dfrac{\text{volume (cm}^3\text{)}}{24\,000}amount (mol)=24000volume (cm3)​.
  6. Rearranging gives volume (dm3)=amount (mol)×24\text{volume (dm}^3\text{)} = \text{amount (mol)} \times 24volume (dm3)=amount (mol)×24.
  7. These relationships apply only to gases, and only at r.t.p.
Example

Gas volume calculations

  • A sample of 0.350 mol0.350\ \text{mol}0.350 mol of oxygen at r.t.p. has the volume below.
V=0.350×24=8.40 dm3 V = 0.350 \times 24 = 8.40\ \text{dm}^3 V=0.350×24=8.40 dm3
  • A volume of 600 cm3600\ \text{cm}^3600 cm3 of hydrogen at r.t.p. contains the amount below.
n=60024 000=0.0250 mol n = \dfrac{600}{24\,000} = 0.0250\ \text{mol} n=24000600​=0.0250 mol
  • A mass of 8.80 g8.80\ \text{g}8.80 g of carbon dioxide is 8.8044=0.200 mol\frac{8.80}{44} = 0.200\ \text{mol}448.80​=0.200 mol, so its volume at r.t.p. is found next.
V=0.200×24=4.80 dm3 V = 0.200 \times 24 = 4.80\ \text{dm}^3 V=0.200×24=4.80 dm3

Reacting masses and gas volumes

  1. The coefficients in a balanced equation give the mole ratio of the reactants and products.
    1. N2+3H2→2NH3\text{N}_2 + 3\text{H}_2 \rightarrow 2\text{NH}_3N2​+3H2​→2NH3​ means that 1 mol1\ \text{mol}1 mol of nitrogen reacts with 3 mol3\ \text{mol}3 mol of hydrogen to make 2 mol2\ \text{mol}2 mol of ammonia.
  2. A stoichiometric calculation uses this mole ratio to link the amount of one substance to the amount of another.
  3. Work through these steps in order:
    1. Convert the quantity you know into moles.
    2. Use the mole ratio from the equation to find the moles of the substance you want.
    3. Convert these moles into the mass, gas volume or solution volume that is asked for.
  4. For gases at r.t.p., the mole ratio is also the volume ratio, because equal amounts of gas occupy equal volumes.
    1. In H2+Cl2→2HCl\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}H2​+Cl2​→2HCl, 30 cm330\ \text{cm}^330 cm3 of hydrogen reacts with 30 cm330\ \text{cm}^330 cm3 of chlorine to make 60 cm360\ \text{cm}^360 cm3 of hydrogen chloride.
Example

Mass of product from mass of reactant

2Mg(s)+O2(g)→2MgO(s) 2\text{Mg(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{MgO(s)} 2Mg(s)+O2​(g)→2MgO(s)
  • A mass of 6.00 g6.00\ \text{g}6.00 g of magnesium is 6.0024=0.250 mol\frac{6.00}{24} = 0.250\ \text{mol}246.00​=0.250 mol.
  • The mole ratio Mg:MgO\text{Mg}:\text{MgO}Mg:MgO is 2:22:22:2, which is 1:11:11:1, so 0.250 mol0.250\ \text{mol}0.250 mol of magnesium oxide forms.
m=0.250×40=10.0 g m = 0.250 \times 40 = 10.0\ \text{g} m=0.250×40=10.0 g
Example

Gas volume from mass of reactant

CaCO3(s)+2HCl(aq)→CaCl2(aq)+H2O(l)+CO2(g) \text{CaCO}_3\text{(s)} + 2\text{HCl(aq)} \rightarrow \text{CaCl}_2\text{(aq)} + \text{H}_2\text{O(l)} + \text{CO}_2\text{(g)} CaCO3​(s)+2HCl(aq)→CaCl2​(aq)+H2​O(l)+CO2​(g)
  • A mass of 5.00 g5.00\ \text{g}5.00 g of calcium carbonate is 5.00100=0.0500 mol\frac{5.00}{100} = 0.0500\ \text{mol}1005.00​=0.0500 mol.
  • The mole ratio CaCO3:CO2\text{CaCO}_3 : \text{CO}_2CaCO3​:CO2​ is 1:11:11:1, so 0.0500 mol0.0500\ \text{mol}0.0500 mol of carbon dioxide forms.
V=0.0500×24=1.20 dm3=1200 cm3 V = 0.0500 \times 24 = 1.20\ \text{dm}^3 = 1200\ \text{cm}^3 V=0.0500×24=1.20 dm3=1200 cm3
Example

Mass of reactant from a gas volume

2H2O2(aq)→2H2O(l)+O2(g) 2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} 2H2​O2​(aq)→2H2​O(l)+O2​(g)
  • A volume of 480 cm3480\ \text{cm}^3480 cm3 of oxygen at r.t.p. is 48024 000=0.0200 mol\frac{480}{24\,000} = 0.0200\ \text{mol}24000480​=0.0200 mol.
  • The mole ratio H2O2:O2\text{H}_2\text{O}_2 : \text{O}_2H2​O2​:O2​ is 2:12:12:1, so 0.0400 mol0.0400\ \text{mol}0.0400 mol of hydrogen peroxide decomposed.
  • Hydrogen peroxide has Mr=(2×1)+(2×16)=34M_r = (2 \times 1) + (2 \times 16) = 34Mr​=(2×1)+(2×16)=34.
m=0.0400×34=1.36 g m = 0.0400 \times 34 = 1.36\ \text{g} m=0.0400×34=1.36 g

Limiting reactants

Definition

Limiting reactant

The reactant that is completely used up first in a reaction, so it decides the maximum amount of product that can form.

Definition

Excess reactant

A reactant present in more than the amount needed to react completely, so some of it is left over when the reaction stops.

  1. Reactants are often mixed in amounts that do not match the mole ratio in the equation.
  2. The reaction stops when the limiting reactant has been used up, and the leftover reactant is said to be in excess.
  3. The amount of product always depends on the limiting reactant and never on the reactant in excess.
  4. Making cheese sandwiches works the same way, because each sandwich needs two slices of bread and one slice of cheese.
    1. With 101010 slices of bread and 888 slices of cheese you can make only 555 sandwiches, so the bread is limiting and 333 slices of cheese are left over.
  5. Find the limiting reactant by following these steps:
    1. Convert the quantity of each reactant into moles.
    2. Divide each amount by the coefficient of that reactant in the balanced equation.
    3. The reactant with the smaller answer is the limiting reactant.
    4. Use the moles of the limiting reactant and the mole ratio to find the amount of product.
Example

Finding the limiting reactant

Mg(s)+2HCl(aq)→MgCl2(aq)+H2(g) \text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)} Mg(s)+2HCl(aq)→MgCl2​(aq)+H2​(g)
  • A 2.40 g2.40\ \text{g}2.40 g piece of magnesium is 2.4024=0.100 mol\frac{2.40}{24} = 0.100\ \text{mol}242.40​=0.100 mol, and it is added to acid containing 0.150 mol0.150\ \text{mol}0.150 mol of HCl\text{HCl}HCl.
  • Dividing by the coefficients gives 0.1001=0.100\frac{0.100}{1} = 0.10010.100​=0.100 for magnesium and 0.1502=0.0750\frac{0.150}{2} = 0.075020.150​=0.0750 for hydrochloric acid.
  • Hydrochloric acid gives the smaller value, so it is the limiting reactant and magnesium is in excess.
  • The mole ratio HCl:H2\text{HCl}:\text{H}_2HCl:H2​ is 2:12:12:1, so 0.0750 mol0.0750\ \text{mol}0.0750 mol of hydrogen forms.
V=0.0750×24=1.80 dm3 V = 0.0750 \times 24 = 1.80\ \text{dm}^3 V=0.0750×24=1.80 dm3

Concentration calculations

Definition

Concentration

The quantity of solute dissolved in each cubic decimetre of solution, measured in g/dm3\text{g/dm}^3g/dm3 or mol/dm3\text{mol/dm}^3mol/dm3.

  1. Concentration links the quantity of dissolved solute to the volume of the solution.
  2. The concentration in moles is concentration (mol/dm3)=amount of solute (mol)volume of solution (dm3)\text{concentration (mol/dm}^3\text{)} = \dfrac{\text{amount of solute (mol)}}{\text{volume of solution (dm}^3\text{)}}concentration (mol/dm3)=volume of solution (dm3)amount of solute (mol)​.
  3. The concentration by mass is concentration (g/dm3)=mass of solute (g)volume of solution (dm3)\text{concentration (g/dm}^3\text{)} = \dfrac{\text{mass of solute (g)}}{\text{volume of solution (dm}^3\text{)}}concentration (g/dm3)=volume of solution (dm3)mass of solute (g)​.
  4. The volume must be in dm3\text{dm}^3dm3, so convert a volume in cm3\text{cm}^3cm3 by dividing by 100010001000.
    1. A volume of 250 cm3250\ \text{cm}^3250 cm3 is 2501000=0.250 dm3\frac{250}{1000} = 0.250\ \text{dm}^31000250​=0.250 dm3.
    2. To change dm3\text{dm}^3dm3 back into cm3\text{cm}^3cm3, multiply by 100010001000.
  5. In symbols, c=nVc = \dfrac{n}{V}c=Vn​ rearranges to n=c×Vn = c \times Vn=c×V and V=ncV = \dfrac{n}{c}V=cn​.
  6. To change a concentration in mol/dm3\text{mol/dm}^3mol/dm3 into g/dm3\text{g/dm}^3g/dm3, multiply by the molar mass of the solute.
  7. To change a concentration in g/dm3\text{g/dm}^3g/dm3 into mol/dm3\text{mol/dm}^3mol/dm3, divide by the molar mass of the solute.
Example

Concentration of a prepared solution

  • A mass of 5.85 g5.85\ \text{g}5.85 g of sodium chloride is dissolved in water and made up to 500 cm3500\ \text{cm}^3500 cm3, which is 0.500 dm30.500\ \text{dm}^30.500 dm3.
concentration=5.850.500=11.7 g/dm3 \text{concentration} = \dfrac{5.85}{0.500} = 11.7\ \text{g/dm}^3 concentration=0.5005.85​=11.7 g/dm3
  • Sodium chloride has Mr=23+35.5=58.5M_r = 23 + 35.5 = 58.5Mr​=23+35.5=58.5, so the amount dissolved is 5.8558.5=0.100 mol\frac{5.85}{58.5} = 0.100\ \text{mol}58.55.85​=0.100 mol.
c=0.1000.500=0.200 mol/dm3 c = \dfrac{0.100}{0.500} = 0.200\ \text{mol/dm}^3 c=0.5000.100​=0.200 mol/dm3
Example

Amount and volume from a concentration

  • A 25.0 cm325.0\ \text{cm}^325.0 cm3 portion of 0.200 mol/dm30.200\ \text{mol/dm}^30.200 mol/dm3 sodium hydroxide has a volume of 0.0250 dm30.0250\ \text{dm}^30.0250 dm3.
n=0.200×0.0250=0.00500 mol n = 0.200 \times 0.0250 = 0.00500\ \text{mol} n=0.200×0.0250=0.00500 mol
  • The volume of 0.250 mol/dm30.250\ \text{mol/dm}^30.250 mol/dm3 solution that contains 0.0500 mol0.0500\ \text{mol}0.0500 mol of solute is found by rearranging.
V=0.05000.250=0.200 dm3=200 cm3 V = \dfrac{0.0500}{0.250} = 0.200\ \text{dm}^3 = 200\ \text{cm}^3 V=0.2500.0500​=0.200 dm3=200 cm3

Titration calculations

Definition

Titre

The volume of solution added from the burette to reach the end-point of a titration.

Definition

Concordant titres

Titres from repeated titrations that agree closely with one another and are used to calculate the mean titre.

  1. A titration finds the exact volume of one solution that reacts with a measured volume of another.
  2. The titre is the volume added from the burette, found by subtracting the initial burette reading from the final reading.
  3. The titration is repeated, and the mean titre is calculated using only the concordant titres.
    1. From titres of 24.3024.3024.30, 23.6523.6523.65 and 23.75 cm323.75\ \text{cm}^323.75 cm3, the first value is left out because it does not agree with the others.
    2. The mean titre is 23.65+23.752=23.70 cm3\frac{23.65 + 23.75}{2} = 23.70\ \text{cm}^3223.65+23.75​=23.70 cm3.
  4. With the mean titre, the calculation follows these steps:
    1. Use n=c×Vn = c \times Vn=c×V to find the moles in the solution whose concentration and volume are both known.
    2. Use the mole ratio from the balanced equation to find the moles of the other solute.
    3. Use c=nVc = \dfrac{n}{V}c=Vn​ to find an unknown concentration, or V=ncV = \dfrac{n}{c}V=cn​ to find an unknown volume.
    4. Multiply by the molar mass if the concentration is needed in g/dm3\text{g/dm}^3g/dm3.
  5. Every volume must be converted from cm3\text{cm}^3cm3 into dm3\text{dm}^3dm3 before it goes into n=c×Vn = c \times Vn=c×V.
Example

Concentration of sodium hydroxide

  • A 25.0 cm325.0\ \text{cm}^325.0 cm3 portion of sodium hydroxide is neutralised by a mean titre of 20.0 cm320.0\ \text{cm}^320.0 cm3 of 0.100 mol/dm30.100\ \text{mol/dm}^30.100 mol/dm3 nitric acid.
HNO3(aq)+NaOH(aq)→NaNO3(aq)+H2O(l) \text{HNO}_3\text{(aq)} + \text{NaOH(aq)} \rightarrow \text{NaNO}_3\text{(aq)} + \text{H}_2\text{O(l)} HNO3​(aq)+NaOH(aq)→NaNO3​(aq)+H2​O(l) n(HNO3)=0.100×0.0200=0.00200 mol n(\text{HNO}_3) = 0.100 \times 0.0200 = 0.00200\ \text{mol} n(HNO3​)=0.100×0.0200=0.00200 mol
  • The mole ratio is 1:11:11:1, so the sodium hydroxide also contains 0.00200 mol0.00200\ \text{mol}0.00200 mol.
c(NaOH)=0.002000.0250=0.0800 mol/dm3 c(\text{NaOH}) = \dfrac{0.00200}{0.0250} = 0.0800\ \text{mol/dm}^3 c(NaOH)=0.02500.00200​=0.0800 mol/dm3
  • Sodium hydroxide has Mr=40M_r = 40Mr​=40, so the concentration by mass is 0.0800×40=3.20 g/dm30.0800 \times 40 = 3.20\ \text{g/dm}^30.0800×40=3.20 g/dm3.
Example

Titration with a 1 to 2 mole ratio

  • A 25.0 cm325.0\ \text{cm}^325.0 cm3 portion of 0.200 mol/dm30.200\ \text{mol/dm}^30.200 mol/dm3 sodium hydroxide is neutralised by a mean titre of 20.0 cm320.0\ \text{cm}^320.0 cm3 of sulfuric acid.
H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l) \text{H}_2\text{SO}_4\text{(aq)} + 2\text{NaOH(aq)} \rightarrow \text{Na}_2\text{SO}_4\text{(aq)} + 2\text{H}_2\text{O(l)} H2​SO4​(aq)+2NaOH(aq)→Na2​SO4​(aq)+2H2​O(l) n(NaOH)=0.200×0.0250=0.00500 mol n(\text{NaOH}) = 0.200 \times 0.0250 = 0.00500\ \text{mol} n(NaOH)=0.200×0.0250=0.00500 mol
  • The mole ratio H2SO4:NaOH\text{H}_2\text{SO}_4 : \text{NaOH}H2​SO4​:NaOH is 1:21:21:2, so the acid contains 0.005002=0.00250 mol\frac{0.00500}{2} = 0.00250\ \text{mol}20.00500​=0.00250 mol.
c(H2SO4)=0.002500.0200=0.125 mol/dm3 c(\text{H}_2\text{SO}_4) = \dfrac{0.00250}{0.0200} = 0.125\ \text{mol/dm}^3 c(H2​SO4​)=0.02000.00250​=0.125 mol/dm3
Common Mistake
  • Do not divide a gas volume in cm3\text{cm}^3cm3 by 242424, because 242424 is the molar gas volume in dm3\text{dm}^3dm3.
  • Do not use the molar gas volume for solids, liquids or solutions, because it applies only to gases.
  • Do not skip the mole ratio, because the moles of two substances are equal only when the ratio is 1:11:11:1.
  • Do not include a rough or anomalous titre in the mean.
  • Do not round numbers partway through a calculation, because early rounding can change the final answer.
Exam technique

Mole calculations in the exam

  • Set out each calculation as a chain of labelled steps, because a mistake in one step can still let you earn the marks for the steps after it.
  • Write the unit after every value, because a missing or wrong unit on the final answer usually loses that mark.
  • Give the final answer to the same number of significant figures as the least precise data in the question, which is usually three.
  • Write the mole ratio out in words, such as "1 mol1\ \text{mol}1 mol of CaCO3\text{CaCO}_3CaCO3​ gives 1 mol1\ \text{mol}1 mol of CO2\text{CO}_2CO2​", so you never forget to use it.
Self review
  • How many particles are there in one mole of a substance?
  • Which equation links amount of substance, mass and molar mass?
  • What volume does one mole of any gas occupy at r.t.p.?
  • How do you decide which reactant is the limiting reactant?
  • How many moles of solute are in 50.0 cm350.0\ \text{cm}^350.0 cm3 of a 0.200 mol/dm30.200\ \text{mol/dm}^30.200 mol/dm3 solution?
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