Chemical calculations

What you'll learn

How relative masses, isotopes and mass spectra connect to $A_r$ and $M_r$ .
How to use the mole, molar mass, concentration and gas equations confidently.
How to calculate empirical formulae, titration results, atom economy and percentage yield.
How to handle percentage uncertainty and sensible significant figures.

Why “amount of substance” matters

Atoms and molecules are far too small to count one by one in the lab. Instead, chemists weigh substances, measure volumes, or record spectra, then convert those measurements into amount of substance in mol.

Key Idea

The central strategy

Most chemical calculations become easier once you convert the given information into moles, use the balanced equation ratio, then convert into the unit asked for.

Relative mass terms

Chemistry uses a relative mass scale based on carbon-12. One atom of carbon-12 is assigned a mass of exactly 12, so one twelfth of a carbon-12 atom is the reference.

Definition

Relative mass terms

Relative isotopic mass is the mass of an atom of a particular isotope compared with one twelfth of the mass of a carbon-12 atom.
Relative atomic mass, $A_r$ , is the weighted mean mass of the atoms of an element compared with one twelfth of the mass of a carbon-12 atom.
Relative molecular mass, $M_r$ , is the sum of the $A_r$ values in a molecule.
Relative formula mass is the sum of the $A_r$ values in a formula unit, often used for ionic compounds such as NaCl.

$A_r$ and $M_r$ are relative, so they have no units. When the same number is used as a mass per mole, it becomes the molar mass, with units g mol⁻¹.

Mass spectrometry and isotopes

A mass spectrometer separates particles according to their mass-to-charge ratio, written as $m / z$ . For simple isotope questions, the ions usually have a charge of +1, so the $m / z$ value is effectively the mass number.

The basic stages are:

Ionisation: atoms or molecules form positive ions, for example Cl(g) → Cl⁺(g) + e⁻.
Acceleration: positive ions are accelerated by an electric field.
Separation: lighter ions, or ions with lower $m / z$ , are separated from heavier ones.
Detection: the detector records relative abundance, producing a mass spectrum.

Labelled schematic of a mass spectrometer showing ionisation, acceleration, flight tube, detector and mass spectrum output

For chlorine atoms, the main isotope peaks are at 35 and 37 in about a 3:1 ratio. For chlorine gas, Cl₂, molecular ion peaks appear at 70, 72 and 74:

70 from ³⁵Cl–³⁵Cl
72 from ³⁵Cl–³⁷Cl or ³⁷Cl–³⁵Cl
74 from ³⁷Cl–³⁷Cl

So their relative abundances are approximately 9:6:1.

Common Mistake

Peak height is not mass

The x-axis gives $m / z$ , while the peak height or area gives relative abundance. Use abundance as the weighting when calculating $A_r$ .

Example

Calculating relative atomic mass from isotope data

Chlorine has 75.8% ³⁵Cl and 24.2% ³⁷Cl. Estimate its relative atomic mass.

Treat the percentages as weightings, because $A_r$ is a weighted mean, not a simple average.
Substitute into the weighted mean calculation:
$Ar=(35×75.8)+(37×24.2)100A_r = \frac{(35 \times 75.8) + (37 \times 24.2)}{100}$
Calculate the value:
$Ar=2653+895.4100=35.5A_r = \frac{2653 + 895.4}{100} = 35.5$

The mole, Avogadro constant and molar mass

Definition

Mole and Avogadro constant

One mole is the amount of substance containing the Avogadro constant of particles. The Avogadro constant is $mol−1N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ .

The key equations are:

\frac{m}{M}

where $n$ is amount in mol, $m$ is mass in g, and $M$ is molar mass in g mol⁻¹.

N = nN_A

where $N$ is the number of particles.

Amount of substance calculation map linking moles to mass, particles, concentration, gas volume and the ideal gas equation

Example

Converting mass into particles

How many water molecules are in 9.00 g of H₂O?

Find the molar mass of water:
$\times 1.0) + 16.0 = 18.0\ \text{g mol}^{-1}$
Convert mass into amount:
$\frac{9.00}{18.0} = 0.500\ \text{mol}$
Convert amount into number of molecules:
$\times 6.022 \times 10^{23} = 3.01 \times 10^{23}$

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms of each element in one molecule.

To find an empirical formula:

Convert masses or percentages into moles.
Divide all mole values by the smallest.
Multiply if needed to reach whole numbers.

Tip

Dealing with awkward ratios

If a ratio ends in about 0.5, multiply all ratios by 2. If it ends in about 0.33 or 0.67, multiply by 3.

Example

Finding empirical and molecular formulae

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its $M_r$ is 180. Find its empirical and molecular formulae.

Assume 100 g, so the masses are 40.0 g C, 6.7 g H and 53.3 g O. Convert each to moles:
$n(C)=40.012.0=3.33n(H)=6.71.0=6.7n(O)=53.316.0=3.33\begin{aligned} n(\text{C}) &= \frac{40.0}{12.0} = 3.33 \\ n(\text{H}) &= \frac{6.7}{1.0} = 6.7 \\ n(\text{O}) &= \frac{53.3}{16.0} = 3.33 \end{aligned}$
Divide by the smallest value, 3.33:
$C:H:O=1:2:1\text{C:H:O} = 1:2:1$
So the empirical formula is CH₂O.
Find the empirical formula mass and compare with the molecular mass:
$18030.0=6\frac{180}{30.0} = 6$
Therefore the molecular formula is C₆H₁₂O₆.

Concentration and solubility

Concentration tells you how much solute is present in a given volume of solution. In chemistry calculations, concentration is often measured in mol dm⁻³.

n = c V

Here, $V$ must be in dm³ if $c$ is in mol dm⁻³.

Concentration can also be expressed in g dm⁻³:

concentration=mV\text{mass concentration} = \frac{m}{V}

Solubility is the maximum concentration of a solute that dissolves in a solvent at a stated temperature, usually given in g dm⁻³ or mol dm⁻³.

Common Mistake

Forgetting the volume conversion

For solution calculations, convert cm³ to dm³ by dividing by 1000. So 250 cm³ is 0.250 dm³, not 250 dm³.

Example

Calculating concentration in mol dm⁻³ and g dm⁻³

5.85 g of NaCl is dissolved and made up to 250 cm³ with water. Calculate the concentration.

Calculate the molar mass of NaCl:
$58.5\ \text{g mol}^{-1}$
Convert mass into moles:
$\frac{5.85}{58.5} = 0.100\ \text{mol}$
Convert the volume into dm³ and calculate molar concentration:
$\frac{0.100}{0.250} = 0.400\ \text{mol dm}^{-3}$
Calculate mass concentration:
$dm−3\frac{5.85}{0.250} = 23.4\ \text{g dm}^{-3}$

Gases: molar volume and the ideal gas equation

The molar volume of a gas is the volume occupied by 1 mol of gas at specified temperature and pressure. At room temperature and pressure, a common approximation is 24.0 dm³ mol⁻¹, but only use this when the conditions match or the value is given.

If temperature or pressure changes, gas volume changes. For a fixed amount of gas:

V2=V1×T2T1×p1p2V_2 = V_1 \times \frac{T_2}{T_1} \times \frac{p_1}{p_2}

Temperature must be in K.

The more general equation is the ideal gas equation:

p V = n R T

Use $p$ in Pa, $V$ in m³, $n$ in mol, $T$ in K, and $8.314\ \text{J K}^{-1}\text{mol}^{-1}$ .

Common Mistake

Using the wrong gas volume unit

In $p V = n R T$ , volume must be in m³. In $n = c V$ , volume must be in dm³. These are different conversions.

Example

Using the ideal gas equation

A 250 cm³ sample of CO₂ is collected at 100 kPa and 298 K. Calculate the amount of CO₂.

Convert into ideal gas equation units:
$100000\ \text{Pa}, \quad V = 2.50 \times 10^{-4}\ \text{m}^3$
Rearrange the equation:
$\frac{pV}{RT}$
Substitute and calculate:
$\frac{100000 \times 2.50 \times 10^{-4}}{8.314 \times 298} = 0.0101\ \text{mol}$

Stoichiometry and titrations

Stoichiometry means using the mole ratios in a balanced chemical equation. The coefficients in the equation give the reacting ratio in moles.

For example:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

This is a 1:1 reaction, so the moles of HCl equal the moles of NaOH at the equivalence point.

In an acid-base titration, you usually:

use a pipette to measure a fixed volume into a conical flask
add a suitable indicator
use a burette to add the other solution
do a rough titration first
repeat until you obtain concordant titres, then calculate a mean titre

Tip

Using the rough titre

If your rough titre is about 24 cm³, you can add solution quickly to around 22 cm³ in accurate runs, then add dropwise near the endpoint.

Example

Calculating concentration from titration data

25.00 cm³ of NaOH is titrated with 0.1000 mol dm⁻³ HCl. The titres are: rough 24.20 cm³, then 23.60 cm³, 23.55 cm³ and 23.90 cm³. Calculate the NaOH concentration.

Select concordant titres. 23.60 cm³ and 23.55 cm³ are close; 23.90 cm³ is not used.
$cm3\text{mean titre} = \frac{23.60 + 23.55}{2} = 23.575\ \text{cm}^3$
Convert the mean titre into dm³ and calculate moles of HCl:
$moln(\text{HCl}) = 0.1000 \times 0.023575 = 0.0023575\ \text{mol}$
Use the 1:1 equation, so $moln(\text{NaOH}) = 0.0023575\ \text{mol}$ in 25.00 cm³.
Calculate the NaOH concentration:
$dm−3c(\text{NaOH}) = \frac{0.0023575}{0.02500} = 0.0943\ \text{mol dm}^{-3}$

Atom economy and percentage yield

Atom economy measures how much of the reactant atoms end up in the desired product.

products×100\text{atom economy} = \frac{M_r\ \text{of desired product}}{\text{sum of } M_r\ \text{values of all products}} \times 100

Remember to include balancing coefficients.

Percentage yield compares the actual mass made with the theoretical maximum mass.

yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Example

Calculating atom economy and percentage yield

Glucose ferments to ethanol:

C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)

18.0 g of glucose produces 7.80 g of ethanol. Calculate the atom economy for ethanol and the percentage yield.

Calculate atom economy using product formula masses:
$economy=2×46.0(2×46.0)+(2×44.0)×100=51.1%\text{atom economy} = \frac{2 \times 46.0}{(2 \times 46.0) + (2 \times 44.0)} \times 100 = 51.1\%$
Find moles of glucose:
$\frac{18.0}{180} = 0.100\ \text{mol}$
Use the equation ratio: 1 mol glucose forms 2 mol ethanol, so 0.100 mol glucose forms 0.200 mol ethanol.
Calculate theoretical ethanol mass and percentage yield:
$yield=7.809.20×100=84.8%\begin{aligned} m &= 0.200 \times 46.0 = 9.20\ \text{g} \\ \text{percentage yield} &= \frac{7.80}{9.20} \times 100 = 84.8\% \end{aligned}$

Percentage error and significant figures

Every measurement has uncertainty. A burette reading might be ±0.05 cm³, but a titre uses two readings, so the titre uncertainty is often ±0.10 cm³.

value×100\text{percentage uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100

When combining measurements in a calculation, percentage uncertainties are often added to estimate the overall percentage uncertainty.

Example

Estimating percentage uncertainty in a titre

A titre is 23.58 cm³. Each burette reading has an uncertainty of ±0.05 cm³. Estimate the percentage uncertainty in the titre.

A titre is calculated from final reading minus initial reading, so two readings are involved:
$cm3\text{absolute uncertainty} = 0.05 + 0.05 = 0.10\ \text{cm}^3$
Calculate percentage uncertainty:
$0.1023.58×100=0.424%\frac{0.10}{23.58} \times 100 = 0.424\%$
Use this to judge precision: quoting many extra digits in a final concentration would not be meaningful.

Exam technique

In the exam

Start by converting the given data into moles using the correct equation and units.
Use the balanced equation ratio before converting into the final requested quantity.
For titrations, ignore rough titres and obvious outliers; average concordant titres only.
Check units carefully: dm³ for concentration, m³ for $p V = n R T$ , and K for gas temperature.

Self review

Check yourself

Why does chlorine gas show molecular ion peaks at 70, 72 and 74?
When should you use $n = c V$ rather than $\frac{m}{M}$ ?
How would you decide which titres to include in a mean titre?

What you'll learn

How relative masses, isotopes and mass spectra connect to $A_r$ and $M_r$ .
How to use the mole, molar mass, concentration and gas equations confidently.
How to calculate empirical formulae, titration results, atom economy and percentage yield.
How to handle percentage uncertainty and sensible significant figures.

Why “amount of substance” matters

Key Idea

The central strategy

Most chemical calculations become easier once you convert the given information into moles, use the balanced equation ratio, then convert into the unit asked for.

Relative mass terms

Chemistry uses a relative mass scale based on carbon-12. One atom of carbon-12 is assigned a mass of exactly 12, so one twelfth of a carbon-12 atom is the reference.

Definition

Relative mass terms

Relative isotopic mass is the mass of an atom of a particular isotope compared with one twelfth of the mass of a carbon-12 atom.
Relative atomic mass, $A_r$ , is the weighted mean mass of the atoms of an element compared with one twelfth of the mass of a carbon-12 atom.
Relative molecular mass, $M_r$ , is the sum of the $A_r$ values in a molecule.
Relative formula mass is the sum of the $A_r$ values in a formula unit, often used for ionic compounds such as NaCl.

$A_r$ and $M_r$ are relative, so they have no units. When the same number is used as a mass per mole, it becomes the molar mass, with units g mol⁻¹.

Mass spectrometry and isotopes

The basic stages are:

Ionisation: atoms or molecules form positive ions, for example Cl(g) → Cl⁺(g) + e⁻.
Acceleration: positive ions are accelerated by an electric field.
Separation: lighter ions, or ions with lower $m / z$ , are separated from heavier ones.
Detection: the detector records relative abundance, producing a mass spectrum.

Labelled schematic of a mass spectrometer showing ionisation, acceleration, flight tube, detector and mass spectrum output

For chlorine atoms, the main isotope peaks are at 35 and 37 in about a 3:1 ratio. For chlorine gas, Cl₂, molecular ion peaks appear at 70, 72 and 74:

70 from ³⁵Cl–³⁵Cl
72 from ³⁵Cl–³⁷Cl or ³⁷Cl–³⁵Cl
74 from ³⁷Cl–³⁷Cl

So their relative abundances are approximately 9:6:1.

Common Mistake

Peak height is not mass

The x-axis gives $m / z$ , while the peak height or area gives relative abundance. Use abundance as the weighting when calculating $A_r$ .

Example

Calculating relative atomic mass from isotope data

Chlorine has 75.8% ³⁵Cl and 24.2% ³⁷Cl. Estimate its relative atomic mass.

Treat the percentages as weightings, because $A_r$ is a weighted mean, not a simple average.
Substitute into the weighted mean calculation:
$Ar=(35×75.8)+(37×24.2)100A_r = \frac{(35 \times 75.8) + (37 \times 24.2)}{100}$
Calculate the value:
$Ar=2653+895.4100=35.5A_r = \frac{2653 + 895.4}{100} = 35.5$

The mole, Avogadro constant and molar mass

Definition

Mole and Avogadro constant

One mole is the amount of substance containing the Avogadro constant of particles. The Avogadro constant is $mol−1N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ .

The key equations are:

\frac{m}{M}

where $n$ is amount in mol, $m$ is mass in g, and $M$ is molar mass in g mol⁻¹.

N = nN_A

where $N$ is the number of particles.

Amount of substance calculation map linking moles to mass, particles, concentration, gas volume and the ideal gas equation

Example

Converting mass into particles

How many water molecules are in 9.00 g of H₂O?

Find the molar mass of water:
$\times 1.0) + 16.0 = 18.0\ \text{g mol}^{-1}$
Convert mass into amount:
$\frac{9.00}{18.0} = 0.500\ \text{mol}$
Convert amount into number of molecules:
$\times 6.022 \times 10^{23} = 3.01 \times 10^{23}$

Empirical and molecular formulae

The empirical formula is the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms of each element in one molecule.

To find an empirical formula:

Convert masses or percentages into moles.
Divide all mole values by the smallest.
Multiply if needed to reach whole numbers.

Tip

Dealing with awkward ratios

If a ratio ends in about 0.5, multiply all ratios by 2. If it ends in about 0.33 or 0.67, multiply by 3.

Example

Finding empirical and molecular formulae

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its $M_r$ is 180. Find its empirical and molecular formulae.

Assume 100 g, so the masses are 40.0 g C, 6.7 g H and 53.3 g O. Convert each to moles:
$n(C)=40.012.0=3.33n(H)=6.71.0=6.7n(O)=53.316.0=3.33\begin{aligned} n(\text{C}) &= \frac{40.0}{12.0} = 3.33 \\ n(\text{H}) &= \frac{6.7}{1.0} = 6.7 \\ n(\text{O}) &= \frac{53.3}{16.0} = 3.33 \end{aligned}$
Divide by the smallest value, 3.33:
$C:H:O=1:2:1\text{C:H:O} = 1:2:1$
So the empirical formula is CH₂O.
Find the empirical formula mass and compare with the molecular mass:
$18030.0=6\frac{180}{30.0} = 6$
Therefore the molecular formula is C₆H₁₂O₆.

Concentration and solubility

Concentration tells you how much solute is present in a given volume of solution. In chemistry calculations, concentration is often measured in mol dm⁻³.

n = c V

Here, $V$ must be in dm³ if $c$ is in mol dm⁻³.

Concentration can also be expressed in g dm⁻³:

concentration=mV\text{mass concentration} = \frac{m}{V}

Solubility is the maximum concentration of a solute that dissolves in a solvent at a stated temperature, usually given in g dm⁻³ or mol dm⁻³.

Common Mistake

Forgetting the volume conversion

For solution calculations, convert cm³ to dm³ by dividing by 1000. So 250 cm³ is 0.250 dm³, not 250 dm³.

Example

Calculating concentration in mol dm⁻³ and g dm⁻³

5.85 g of NaCl is dissolved and made up to 250 cm³ with water. Calculate the concentration.

Calculate the molar mass of NaCl:
$58.5\ \text{g mol}^{-1}$
Convert mass into moles:
$\frac{5.85}{58.5} = 0.100\ \text{mol}$
Convert the volume into dm³ and calculate molar concentration:
$\frac{0.100}{0.250} = 0.400\ \text{mol dm}^{-3}$
Calculate mass concentration:
$dm−3\frac{5.85}{0.250} = 23.4\ \text{g dm}^{-3}$

Gases: molar volume and the ideal gas equation

If temperature or pressure changes, gas volume changes. For a fixed amount of gas:

V2=V1×T2T1×p1p2V_2 = V_1 \times \frac{T_2}{T_1} \times \frac{p_1}{p_2}

Temperature must be in K.

The more general equation is the ideal gas equation:

p V = n R T

Use $p$ in Pa, $V$ in m³, $n$ in mol, $T$ in K, and $8.314\ \text{J K}^{-1}\text{mol}^{-1}$ .

Common Mistake

Using the wrong gas volume unit

In $p V = n R T$ , volume must be in m³. In $n = c V$ , volume must be in dm³. These are different conversions.

Example

Using the ideal gas equation

A 250 cm³ sample of CO₂ is collected at 100 kPa and 298 K. Calculate the amount of CO₂.

Convert into ideal gas equation units:
$100000\ \text{Pa}, \quad V = 2.50 \times 10^{-4}\ \text{m}^3$
Rearrange the equation:
$\frac{pV}{RT}$
Substitute and calculate:
$\frac{100000 \times 2.50 \times 10^{-4}}{8.314 \times 298} = 0.0101\ \text{mol}$

Stoichiometry and titrations

Stoichiometry means using the mole ratios in a balanced chemical equation. The coefficients in the equation give the reacting ratio in moles.

For example:

HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

This is a 1:1 reaction, so the moles of HCl equal the moles of NaOH at the equivalence point.

In an acid-base titration, you usually:

use a pipette to measure a fixed volume into a conical flask
add a suitable indicator
use a burette to add the other solution
do a rough titration first
repeat until you obtain concordant titres, then calculate a mean titre

Tip

Using the rough titre

If your rough titre is about 24 cm³, you can add solution quickly to around 22 cm³ in accurate runs, then add dropwise near the endpoint.

Example

Calculating concentration from titration data

25.00 cm³ of NaOH is titrated with 0.1000 mol dm⁻³ HCl. The titres are: rough 24.20 cm³, then 23.60 cm³, 23.55 cm³ and 23.90 cm³. Calculate the NaOH concentration.

Select concordant titres. 23.60 cm³ and 23.55 cm³ are close; 23.90 cm³ is not used.
$cm3\text{mean titre} = \frac{23.60 + 23.55}{2} = 23.575\ \text{cm}^3$
Convert the mean titre into dm³ and calculate moles of HCl:
$moln(\text{HCl}) = 0.1000 \times 0.023575 = 0.0023575\ \text{mol}$
Use the 1:1 equation, so $moln(\text{NaOH}) = 0.0023575\ \text{mol}$ in 25.00 cm³.
Calculate the NaOH concentration:
$dm−3c(\text{NaOH}) = \frac{0.0023575}{0.02500} = 0.0943\ \text{mol dm}^{-3}$

Atom economy and percentage yield

Atom economy measures how much of the reactant atoms end up in the desired product.

products×100\text{atom economy} = \frac{M_r\ \text{of desired product}}{\text{sum of } M_r\ \text{values of all products}} \times 100

Remember to include balancing coefficients.

Percentage yield compares the actual mass made with the theoretical maximum mass.

yield×100\text{percentage yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100

Example

Calculating atom economy and percentage yield

Glucose ferments to ethanol:

C₆H₁₂O₆(aq) → 2C₂H₅OH(aq) + 2CO₂(g)

18.0 g of glucose produces 7.80 g of ethanol. Calculate the atom economy for ethanol and the percentage yield.

Calculate atom economy using product formula masses:
$economy=2×46.0(2×46.0)+(2×44.0)×100=51.1%\text{atom economy} = \frac{2 \times 46.0}{(2 \times 46.0) + (2 \times 44.0)} \times 100 = 51.1\%$
Find moles of glucose:
$\frac{18.0}{180} = 0.100\ \text{mol}$
Use the equation ratio: 1 mol glucose forms 2 mol ethanol, so 0.100 mol glucose forms 0.200 mol ethanol.
Calculate theoretical ethanol mass and percentage yield:
$yield=7.809.20×100=84.8%\begin{aligned} m &= 0.200 \times 46.0 = 9.20\ \text{g} \\ \text{percentage yield} &= \frac{7.80}{9.20} \times 100 = 84.8\% \end{aligned}$

Percentage error and significant figures

Every measurement has uncertainty. A burette reading might be ±0.05 cm³, but a titre uses two readings, so the titre uncertainty is often ±0.10 cm³.

value×100\text{percentage uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100

When combining measurements in a calculation, percentage uncertainties are often added to estimate the overall percentage uncertainty.

Example

Estimating percentage uncertainty in a titre

A titre is 23.58 cm³. Each burette reading has an uncertainty of ±0.05 cm³. Estimate the percentage uncertainty in the titre.

A titre is calculated from final reading minus initial reading, so two readings are involved:
$cm3\text{absolute uncertainty} = 0.05 + 0.05 = 0.10\ \text{cm}^3$
Calculate percentage uncertainty:
$0.1023.58×100=0.424%\frac{0.10}{23.58} \times 100 = 0.424\%$
Use this to judge precision: quoting many extra digits in a final concentration would not be meaningful.

Exam technique

In the exam

Start by converting the given data into moles using the correct equation and units.
Use the balanced equation ratio before converting into the final requested quantity.
For titrations, ignore rough titres and obvious outliers; average concordant titres only.
Check units carefully: dm³ for concentration, m³ for $p V = n R T$ , and K for gas temperature.

Self review

Check yourself

Why does chlorine gas show molecular ion peaks at 70, 72 and 74?
When should you use $n = c V$ rather than $\frac{m}{M}$ ?
How would you decide which titres to include in a mean titre?

What you'll learn

Why “amount of substance” matters

Relative mass terms

Mass spectrometry and isotopes

The mole, Avogadro constant and molar mass

Empirical and molecular formulae

Concentration and solubility

Gases: molar volume and the ideal gas equation

Stoichiometry and titrations

Atom economy and percentage yield

Percentage error and significant figures

Review Chemical calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical calculations

What you'll learn

Why “amount of substance” matters

Relative mass terms

Mass spectrometry and isotopes

The mole, Avogadro constant and molar mass

Empirical and molecular formulae

Concentration and solubility

Gases: molar volume and the ideal gas equation

Stoichiometry and titrations

Atom economy and percentage yield

Percentage error and significant figures

Review Chemical calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical calculations

What you'll learn

Why “amount of substance” matters

Relative mass terms

Mass spectrometry and isotopes

The mole, Avogadro constant and molar mass

Empirical and molecular formulae

Concentration and solubility

Gases: molar volume and the ideal gas equation

Stoichiometry and titrations

Atom economy and percentage yield

Percentage error and significant figures

Review Chemical calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical calculations Revision Guide

Chemical calculations

What you'll learn

Why “amount of substance” matters

Relative mass terms

Mass spectrometry and isotopes

The mole, Avogadro constant and molar mass

Empirical and molecular formulae

Concentration and solubility

Gases: molar volume and the ideal gas equation

Stoichiometry and titrations

Atom economy and percentage yield

Percentage error and significant figures

Review Chemical calculations by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Chemical calculations Revision Guide