Mole concept & volumetric calc.

  1. Exactly 0.350 grams of divalent metal carbonate, RCO3 were dissolved completely in 100cm3 of 0.10M hydrochloric acid (HCl) solution. The excess acid required 20.0cm3 of 0.15M sodium hydroxide (NaOH) solution for complete neutralization. Find the relative molecular mass of the metal carbonate. Identify metal R.
    2. Suggested Answers
    i. RCO3 + 2HCl → RCl2 + H2 + CO2 ..equation (i) 
    0.35g  100cm3
       0.1mol/dm3

    The excess acid:
    ii. HCl + NaOH → NaCl + H2 ......... equation (ii)
           20cm3
       0.15mol/dm3

 From acid-base mole ratio:
	   Va = MbVbna
	         Manb
	Va = 0.15mol/dm3 x 20cm3 x 1	 
		 0.1mol/dm3 x 1
	Va = 30cm3
    
	Therefore, the acid reacted with RCO3 can be obtained as follows:
	100cm3 - 30cm3 = 70cm3
    Number of moles of acid reacted
    0.1mol →1000cm3
 ? mol → 70cm3
 = 0.007mol of HCl
    From the equation (i) 
    1 mole of RCO3 → 2 moles of HCl
? moles of RCO3 → 0.007 moles of HCl
 
 This gives:
 = 0.0035 moles of RCO3

 Number of moles = mass of the substance 
                   molar mass 
Therefore, 
Molar mass of RCO3 = mass of RCO3
                     number of moles of RCO3                        
                 = 0.35g	
                  0.0035mol		
               = 100g/mol

    To identify the element R
    RCO3 = 100g/mol
    (R + 12 + 16 X 3)g/mol = 100g/mol
    R + 60 = 100
    R = 100 - 60
    R = 40
    The relative atomic mass of the element R is 40.
    The element R therefore is calcium

    Ref:Volumetric analysis and Mole calculations
  2. A gaseous hydrocarbon X contains 20% hydrogen by mass. 7.5g of X occupy 5.6dm3 at S.T.P.
    1. Calculate
      1. the empirical formula of X.
      2. the molar mass of X.
      3. the molecular formula of X.
    2. (b) write
      1. the name of the hydrocarbon.
      2. the structural formula of X.

    3. Suggested answers

    Elements C H
    Percentage composition 80 20
    Relative atomic mass 12 1
    % composition divide by R.A.M 80/12 = 6.67 20/1 = 20
    Divide by smallest number 6.67/6.67= 1 20/6.67 =3
    Whole number ratios 1 3
    The empirical formula is CH3
    Case 1:   In terms of volume of a gas;
 Number of moles of a gas = volume of a gas 
                             molar volume
    
                     = 5.6dm3
                       22.4dm3/mol
                     = 0.25mol
Case 2: In terms of mass of a gas;  
  Number of moles = mass of the gas
                      molar mass
                      
The molar mass of the gas is not known
                      
Therefore;
Number of moles = 7.5g
                  xg/mol
 Because both are the ways of expressing the number of moles of the same gas:
 7.5g = 0.25mol
 xg/mol
 
 =7.5/0.25
 =30
 Molecular formula =(Empirical formula)n
    
    where:
    
relative molecular mass of empirical formula ( CH3 )= 12 + (1x3) = 15
 
 the value of n can be obtained by using this relation:   
     n = Mr of the compound
         Mr of the empirical formula
    n = 30
        15
   n=2
          
To get molecular formula we insert the value of n to (CH3)2
The molecular formula is C3H6
iii. The name of this compound is propene.
    Ref: empirical formula and molecular formula, Mole calculations and Organic chemistry