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ASChemistryOCRRevision Notes2. Foundations in ChemistryAmount of SubstanceReaction Calculations

Reaction Calculations (OCR AS Chemistry A)

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Richard Boole

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Mass Calculations

  • The number of moles of a substance can be found by using the following equation:

n u m b e r space o f space m o l e s space left parenthesis m o l right parenthesis space equals space fraction numerator m a s s space o f space a space s u b s t a n c e space left parenthesis g right parenthesis over denominator m o l a r space m a s s space left parenthesis g space m o l to the power of negative 1 end exponent right parenthesis end fraction

  • It is important to be clear about the type of particle you are referring to when dealing with moles

    • E.g. one mole of CaF2 contains one mole of CaF2 formula units, but one mole of Ca2+ and two moles of F- ions

Reacting masses

  • The masses of reactants are useful to determine how much of the reactants exactly react with each other to prevent waste

  • To calculate the reacting masses, the balanced chemical equation is required

  • This equation shows the ratio of moles of all the reactants and products, also called the stoichiometry, of the equation

  • To find the mass of products formed in a reaction the following pieces of information are needed:

    • The mass of the reactants

    • The molar mass of the reactants

    • The balanced equation

Worked Example

Mass calculation using moles

Calculate the mass of magnesium oxide that can be made by completely burning 6 g of magnesium in oxygen.

magnesium (s)  +  oxygen (g)  → magnesium oxide (s)

Answer

Step 1: The balanced symbol equation is:

2Mg (s) + O2 (g) → 2MgO (s)

Step 2: The relative formula masses are:

Mg = 24.3, O2 = 32.0, MgO = 40.3

Step 3: Calculate the moles of magnesium used in the reaction:

n u m b e r space o f space m o l e s space equalsbegin mathsize 16px style space fraction numerator 6.0 space g over denominator 24.3 space g space m o l to the power of negative 1 end exponent end fraction end style = 0.25 moles

Step 4: Find the ratio of magnesium to magnesium oxide using the balanced chemical equation

Atoms, Molecules & Stoichiometry Table 1, downloadable AS & A Level Chemistry revision notes

Therefore, 0.25 mol of MgO is formed

Step 5: Find the mass of magnesium oxide

Mass = mol x Mr

Mass = 0.25 mol x 40 g mol-1 

Mass = 10 g

Therefore, the mass of magnesium oxide produced is 10 g

Stoichiometric relationships

  • The stoichiometry of a reaction can be found if the exact amounts of reactants and products formed are known

  • The amounts can be found by using the following equation:

n u m b e r space o f space m o l e s space left parenthesis m o l right parenthesis space equals space fraction numerator m a s s space o f space a space s u b s t a n c e space left parenthesis g right parenthesis over denominator m o l a r space m a s s space left parenthesis g space m o l to the power of negative 1 end exponent right parenthesis end fraction

  • The gas volumes can be used to deduce the stoichiometry of a reaction

    • E.g. in the combustion of 50 cm3 of propane reacting with 250 cm3 of oxygen, 150 cm3  of carbon dioxide is formed suggesting that the ratio of propane : oxygen : carbon dioxide is 1 : 5 : 3

C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (l)

Volume Calculations

  • The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of  solution

    • The solute is the substance that dissolves in a solvent to form a solution

    • The solvent is often water

c o n c e n t r a t i o n space left parenthesis m o l space d m to the power of negative 3 end exponent right parenthesis space equals space fraction numerator n u m b e r space o f space m o l e s space left parenthesis m o l right parenthesis over denominator v o l u m e space o f space s o l u t i o n space left parenthesis d m cubed right parenthesis end fraction

  • A concentrated solution is a solution that has a high concentration of solute

  • A dilute solution is a solution with a low concentration of solute

  • When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:

    • Change mass in grams to moles

    • Change cm3 to dm3 

  • To calculate the mass of a substance present in solution of known concentration and volume:

    • Rearrange the concentration equation

number of moles (mol) = concentration (mol dm-3) x volume (dm3)

  • Multiply the moles of solute by its molar mass

mass of solute (g) = number of moles (mol) x molar mass (g mol-1)

Worked Example

Calculating volume from concentration

Calculate the volume of hydrochloric acid of concentration 1.0 mol dm-3 that is required to react completely with 2.5 g of calcium carbonate

Answer

Step 1: Write the balanced symbol equation

CaCO3  +  2HCl  →  CaCl2  +  H2O  +  CO2

Step 2: Calculate the amount, in moles, of calcium carbonate that reacts

n u m b e r space o f space m o l e s space left parenthesis C a C O subscript 3 right parenthesis space equals space fraction numerator 2.5 space g over denominator 100 space g space m o l to the power of negative 1 end exponent end fraction space equals space 0.025 space m o l

Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of CaCO3 requires 2 mol of HCl

So 0.025 mol of CaCO3 requires 0.05 mol of HCl

Step 4: Calculate the volume of HCl required

v o l u m e space o f space H C l space left parenthesis d m cubed right parenthesis space equals space fraction numerator a m o u n t space left parenthesis m o l right parenthesis over denominator c o n c e n t r a t i o n space left parenthesis m o l space d m to the power of negative 3 end exponent right parenthesis end fraction space equals space fraction numerator 0.05 space m o l over denominator 1.0 space m o l space d m to the power of negative 3 end exponent end fraction space equals space 0.05 space d m cubed

Volume of hydrochloric acid = 0.05 dm3

Worked Example

Neutralisation calculation

25.0 cm3 of 0.050 mol dm–3 sodium carbonate was completely neutralised by 20.0 cm3  of dilute hydrochloric acid. 

Calculate the concentration in mol dm–3 of the hydrochloric acid.  

Answer

Step 1: Write the balanced symbol equation

Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

Step 2: Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing the volume by 1000 to convert cm3 to dm3

amount (Na2CO3) = 0.025 dm3 x 0.050 mol dm-3 = 0.00125 mol

Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2

Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

Step 4: Calculate the concentration, in mol dm-3, of hydrochloric acid

c o n c e n t r a t i o n space o f space H C l space left parenthesis m o l space d m to the power of negative 3 end exponent right parenthesis space equals space fraction numerator a m o u n t space left parenthesis m o l right parenthesis over denominator v o l u m e space left parenthesis d m cubed right parenthesis end fraction space equals space fraction numerator 0.00250 over denominator 0.0200 end fraction space equals space 0.125 space m o l space d m to the power of negative 3 end exponent

Volumes of gases

  • Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’ (also called Avogadro’s hypothesis)

  • At room temperature (20 degrees Celsius) and pressure (1 atm) one mole of any gas has a volume of 24.0 dm3

  • This molar gas volume of 24.0 dm3 mol-1 can be used  to find:

    • The volume of a given mass or number of moles of gas:

volume of gas (dm3) = amount of gas (mol) x 24 dm3 mol-1

  • The mass or number of moles of a given volume of gas:

a m o u n t space o f space g a s space left parenthesis m o l right parenthesis space equals space fraction numerator v o l u m e space o f space g a s space left parenthesis d m cubed right parenthesis over denominator 24.0 space left parenthesis d m cubed space m o l to the power of negative 1 end exponent right parenthesis end fraction

Worked Example

Calculation volume of gas using excess & limiting reagents

Calculate the volume the following gases occupy:

  1. Hydrogen (3 mol)

  2. Carbon dioxide (0.25 mol)

  3. Oxygen (5.4 mol)

  4. Ammonia (0.02 mol)

Calculate the moles in the following volumes of gases:

  1. Methane (225.6 dm3)

  2. Carbon monoxide (7.2 dm3)

  3. Sulfur dioxide (960 dm3)

Answer:

Atoms, Molecules & Stoichiometry Table 2, downloadable AS & A Level Chemistry revision notes

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    Richard Boole

    Author: Richard Boole

    Expertise: Chemistry

    Richard has taught Chemistry for over 15 years as well as working as a science tutor, examiner, content creator and author. He wasn’t the greatest at exams and only discovered how to revise in his final year at university. That knowledge made him want to help students learn how to revise, challenge them to think about what they actually know and hopefully succeed; so here he is, happily, at SME.