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A LevelChemistryCambridge (CIE)Revision Notes2. Atoms, Molecules & StoichiometryReacting Masses & Volumes (of Solutions & Gases)Reacting Masses & Volumes

Reacting Masses & Volumes (Cambridge (CIE) A Level Chemistry)

Revision Note

Richard Boole

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Mole Calculations

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

number space of space moles equals fraction numerator mass space of space straight a space substance space open parentheses straight g close parentheses over denominator molar space mass space open parentheses straight g space mol to the power of negative 1 end exponent close parentheses end fraction

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

    • Eg. 1 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 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 is needed:

    • The mass of the reactants

    • The molar mass of the reactants

    • The balanced equation

Percentage yield

  • In a lot of reactions, not all reactants react to form products which can be due to several factors:

    • Other reactions take place simultaneously

    • The reaction does not go to completion

    • Reactants or products are lost to the atmosphere

  • The percentage yield shows how much of a particular product you get from the reactants compared to the maximum theoretical amount that you can get:

begin mathsize 16px style percentage space yield equals fraction numerator actual space yield over denominator predicted space or space theoretical space yield end fraction cross times 100 end style

  • Where actual yield is the number of moles or mass of product obtained experimentally

  • The predicted yield is the number of moles or mass obtained by calculation

Worked Example

Mass calculation using moles

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

magnesium + oxygen → magnesium oxide

Answer

  • Step 1: The symbol equation is:

    • 2Mg (s) + O2 (g) → 2MgO (s)

  • Step 2: The relative atomic and formula masses are:

    • Magnesium: 24.3

    • Oxygen: 32

    • Magnesium oxide: 40.3

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

    • Number of moles = fraction numerator 6.0 space straight g over denominator 24.3 space straight g space mol to the power of negative 1 end exponent end fraction = 0.2469 moles

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

 

Magnesium

Magnesium oxide

Moles

2

2

Ratio

1

1

Change in moles 

- 0.2469

+ 0.2469

  • Therefore, 0.2469 mol of MgO is formed

  • Step 5: Find the mass of magnesium oxide

    • mass = mol x Mr

    • mass = 0.2469 mol x 40.3 g mol-1

    • mass = 9.95 g

  • Therefore, the mass of magnesium oxide produced is 9.95 g

Worked Example

Calculate % yield using moles

In an experiment to displace copper from copper(II) sulfate, 6.54 g of zinc was added to an excess of copper(II) sulfate solution.

The copper was filtered off, washed and dried.

The mass of copper obtained was 4.80 g.

Calculate the percentage yield of copper.

Answer

  • Step 1: The symbol equation is:

    • Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

  •  Step 2: Calculate the amount of zinc reacted in moles

    • Number of moles = fraction numerator 6.54 space straight g over denominator 65.4 space straight g space mol to the power of negative 1 end exponent end fraction = 0.10 moles

  • Step 3: Calculate the maximum amount of copper that could be formed from the molar ratio:

    • Since the ratio of Zn(s) to Cu(s) is 1:1 a maximum of 0.10 moles can be produced

  • Step 4: Calculate the maximum mass of copper that could be formed (theoretical yield)

    • Mass = mol x Mr

    • Mass = 0.10 mol x 63.5 g mol-1

    • Mass = 6.35 g

  • Step 5: Calculate the percentage yield of copper

    • Percentage yield = fraction numerator 4.80 space straight g over denominator 6.35 space straight g end fraction x 100 = 75.6 %

Excess & limiting reagents

  • Sometimes, there is an excess of one or more of the reactants (excess reagent)

  • The reactant which is not in excess is called the limiting reagent

  • To determine which reactant is limiting:

    • The number of moles of the reactants should be calculated

    • The ratio of the reactants shown in the equation should be taken into account eg:

    C  + 2H2      →     CH4

There are 10 mol of Carbon reacting with 3 mol of Hydrogen

  • Hydrogen is the limiting reagent and since the ratio of C : H2 is 1:2 only 1.5 mol of C will react with 3 mol of H2

Worked Example

Excess & limiting reagent

9.2 g of sodium is reacted with 8.0 g of sulfur to produce sodium sulfide, Na2S.

Which reactant is in excess and which is the limiting reagent?

Answer

  • Step 1: Calculate the moles of each reactant

    • Number of moles (Na) = fraction numerator 9.2 space straight g over denominator 23.0 space straight g space mol to the power of negative 1 end exponent end fraction = 0.40 mol

    • Number of moles (S) = fraction numerator 8.0 space straight g over denominator 32.1 space straight g space mol to the power of negative 1 end exponent end fraction = 0.25 mol

  • Step 2: Write the balanced equation and determine the molar ratio

    • 2Na + S → Na2S

    • The molar ratio of Na: Na2S is 2:1

  • Step 3: Compare the moles and determine the limiting reagent

    • To completely react 0.40 moles of Na requires 0.20 moles of S

    • Since there are 0.25 moles of S, then S is in excess

    • Therefore, Na is the limiting reactant

Volumes of gases

  • Avogadro suggested that ‘equal volumes of gases contain the same number of molecules’

    • This is also called Avogadro’s hypothesis

  • At room temperature and pressure one mole of any gas has a volume of 24.0 dm3 

    • Room temperature is 20 oC

    • Room pressure is 1 atmosphere 

  • This molar gas volume of 24.0 dm3 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 24.0

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

Error converting from MathML to accessible text.

Worked Example

Calculating the volume of gas using excess & limiting reagents

Calculate the volume that 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

Gas

Amount of gas (mol)

Volume of gas (dm3)

Hydrogen

3.0

3.0 x 24.0 = 72.0

Carbon dioxide

0.25

0.25 x 24.0 = 6.0

Oxygen

5.4

5.4 x 24.0 = 129.6

Ammonia

0.02

0.02 x 24.0 = 0.48

Methane

fraction numerator bold 225 bold. bold 6 over denominator bold 24 bold. bold 0 end fraction = 9.4

225.6

Carbon monoxide

fraction numerator bold 7 bold. bold 2 over denominator bold 24 bold. bold 0 end fraction = 0.30

7.2

Sulfur dioxide

fraction numerator bold 960 over denominator bold 24 bold. bold 0 end fraction = 40

960

Volumes & concentrations of solutions

  • 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

concentration space open parentheses mol space dm to the power of negative 3 end exponent close parentheses equals fraction numerator number space of space moles space of space solute open parentheses mol close parentheses over denominator volume space of space solution space open parentheses dm cubed close parentheses 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 1.0 mol dm-3 hydrochloric acid required to completely react 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:

    • Number of moles (CaCO3) = fraction numerator 2.5 space straight g over denominator 100 space straight g space mol to the power of negative 1 end exponent end fraction = 0.025 mol

  • 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:

    • Volume (HCl) = fraction numerator amount space open parentheses mol close parentheses over denominator concentration space open parentheses mol space dm to the power of negative 3 end exponent close parentheses end fraction = fraction numerator 0.05 space mol over denominator 1.0 space mol space dm to the power of negative 3 end exponent end fraction = 0.05 dm3

    • So, the volume of hydrochloric acid required is 0.05 dm3 

Worked Example

Neutralisation calculation

25.0 cm3 of 0.050 mol dm-3 sodium carbonate solution 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  →  Na2Cl2  +  H2O  +  CO2

  • Step 2: Calculate the amount, in moles, of sodium carbonate reacted

    • Rearrange the equation for the amount of substance (mol) and divide the volume by 1000 to convert cm3 to dm3

    • Number of moles (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:

    • Concentration (HCl) = begin mathsize 16px style fraction numerator amount space open parentheses mol close parentheses over denominator volume space open parentheses dm cubed close parentheses end fraction end style = fraction numerator 0.00250 over denominator 0.0200 end fraction = 0.125 mol dm-3

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:

number space of space moles equals fraction numerator mass space of space straight a space substance space stretchy left parenthesis straight g stretchy right parenthesis over denominator molar space mass space stretchy left parenthesis straight g space mol to the power of negative 1 end exponent stretchy right parenthesis end fraction

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

    • Eg. 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) + H2O (l)

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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.