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ALevel

Chemistry-9701

Reacting Masses and Volumes

Paper-3

Reacting Masses and Volumes

1: Perform calculations including use of the mole concept, involving: 


(a) Reacting masses (from formulae and equations) including percentage yield calculations 


> We use the relative molecular mass or formula mass in grams (1000 g = 1 kg). The number of moles of a substance can be calculated by using the mass of the substance and the relative atomic mass (Ar) or relative molecular mass (Mr).  

          

           

       Cambridge International Examinations Lawrie Ryan, Roger Norris

> Finding the mass of a substance present in the provided number of moles requires simply rearranging the equation:

          
          
      Cambridge International Examinations Lawrie Ryan, Roger Norris

> There might be a probable need to know what mass of each provided reactant is required for the reaction so that they react exactly and there is no possibility of waste. To figure this out we need to know the chemical equation of the reaction. This exhibits the ratio of moles of the reactants and products in the reaction. 

> The balanced equation shows this stoichiometry. For example, in the reaction:

> 1 mole of iron (III) oxide reacts with 3 moles of carbon monoxide to form 2 moles of iron and 3 moles of carbon dioxide. The stoichiometry of the equation above is 1: 3: 2: 3. 

>The numbers that are present in the above equation (3, 2 and 3) are called stoichiometric numbers. For us to find the mass of products produced in the chemical reaction we use:

The mass of the reactants 
The molar mass of the reactants
The balanced equation.

  Cambridge International Examinations Lawrie Ryan, Roger Norris

> The stoichiometry of a reaction if we are aware of the amounts of each reactant that will exactly react together and the amounts of each product formed. For an illustration, if we react 4.0 g of hydrogen present with 32.0 g the of oxygen we get 36.0 g of water in the reaction. (Ar values will be: H = 1.0, O = 16.0)

Cambridge International Examinations Lawrie Ryan, Roger Norris

>This ratio is the ratio of stoichiometric numbers in the equation. So the equation is
>We can still calculate the stoichiometry of this reaction that is taking place even if we do not know the mass of oxygen that is present and reacts. The ratio observed of hydrogen to water is 1: 1. However, there is only one atom of oxygen in a molecule of water – half the amount in a present oxygen molecule. So the observed mole ratio of oxygen to water in the equation must be 1: 2.

Percentage Yield: 

>The formula of percentage yield is: 

       
                      chemrevise.org

>The following calculation pertains to percentage yield:

         
         chemrevise.org


(b) Volumes of gases (e.g. in the burning of hydrocarbons) 


> Equal (same) volumes of all the gases present contain the same number of molecules.  This idea is approximately accurate as long as we understand the conditions of pressure which should not be too high or the temperature too low.

> Hence, it is convenient to measure volumes of gases at room temperature (20 °C) and pressure (1 atmosphere). At room temperature and pressure (r.t.p.) one mole of any gas has a volume of 24.0 dm3

> Hence, 24.0dm3 of carbon dioxide and 24.0dm3 of hydrogen both constitute one mole of gas molecules. Also, we can use the molar gas volume of 24.0dm3 at r.t.p. to find:

  Volume of a given mass or number of moles of gas present 
  Mass or number of moles of a given volume of gas present

> Note: If we mix 20 cm3of hydrogen with 10 cm3 of oxygen and explode the mixture, we will find that the gases have exactly reacted together and no hydrogen or oxygen remains. 

> If equal volumes of gases contain equal numbers of molecules and hence equal numbers of moles of gases. Hence, the mole ratio present of hydrogen to oxygen is 2: 1.

> We can summarize this as:


Cambridge International Examinations Lawrie Ryan, Roger Norris

>We can extend this idea to experiments involving combustion data of hydrocarbons. The example provided below illustrates how the formula of propane and the stoichiometry of the equation can be calculated. Propane is a hydrocarbon – a compound of carbon and hydrogen only:


       Cambridge International Examinations Lawrie Ryan, Roger Norris


> Alternatively: The formula for finding the volume of gas can be used which is:

Gas Volume (dm3) = amount x 24

Note: This equation gives the volume of a gas at room pressure (1atm) and room temperature 25 ̊C.

Example: 

Calculate the volume in dm3 at room temperature and pressure of 50.0g of Carbon dioxide gas?



      chemrevise.org 


(c) Volumes and concentrations of solutions: 


> The concentration of a solution is the amount of solute dissolved in a solvent to make 1dm3 (one cubic decimeter) of solution. The solvent is usually water.

> There are 1000 cm3 in a cubic decimeter. When 1 mole of a compound is dissolved to make 1dm3 of solution the concentration is 1moldm–3

   

Cambridge International Examinations Lawrie Ryan, Roger Norris

> We use the terms ‘concentrated’ and ‘dilute’ to refer to the relative amount of solute in the solution. A solution with a low concentration of solute is a dilute solution. If there is a high concentration of solute, the solution is concentrated. 

> When performing calculations involving concentrations in moldm–3  
Remember: 

  • Change mass in grams to moles
  • Change cm3 to dm(by dividing the number of cm3 by 1000).

      

Cambridge International Examinations Lawrie Ryan, Roger Norris

>There is an occasional need to calculate the mass of a substance present in a solution of known concentration and volume. To do this we: 
Rearranging the concentration equation, we know, to: 

> Number of moles (moles) = concentration (mol dm–3 ) × vol(dm3

Multiply the moles of the solute present by its molar mass:

>       Mass of solute (g) = number of moles (moles) × molar mass (gmol–1)

       
    Cambridge International Examinations Lawrie Ryan, Roger Norris


>Note: Titration: Titration is known as a technique where a solution of known concentration is used to determine the concentration of an unknown solution in the reaction.

>Calculations of “titration” require the above knowledge, examples of which are seen through:

>Titration: A titration is known as a technique where a solution present of a known concentration is used to determine the concentration of an unknown solution in the reaction mixture. 

     
   Cambridge International Examinations Lawrie Ryan, Roger Norris


>We can use titration results to find the stoichiometry of a reaction. To do this, we need to know the concentrations and the volumes of both the reactants that are present. The example below shows how to we can deduce the stoichiometry of the reaction that takes place between a metal hydroxide and an acid that is present in the reaction. An example of which is given below:
       
       
  Cambridge International Examinations Lawrie Ryan, Roger Norris


(d) Limiting reagent and excess reagent:


>Here, we do not always require the molar mass of each of the reactants present. If one or more of the reactants is present in excess, we need only know the present mass in grams and the molar mass of the reactant that is not in excess (the limiting reactant).

Example:

>Deduce the maximum mass of titanium that is produced when it reacts with 100 g of TiCl4 with 80 g of sodium.

                  





(e) Deduce stoichiometric relationships from calculations such as those in 2.4.1 (a)–(d)