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Percentage of ASA in tablet

  1. Jun 21, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate the number of moles and the mass of ASA in one tablet. Determine the percentage of the original mass of the tablet that was actually ASA.

    (this question was based off an experiment previously in the lesson)


    2. Relevant equations
    c = n/v
    n = m/M


    3. The attempt at a solution
    Given:
    [itex] HC_9H_7O_4 + NaOH --> C_9H-7O-4 + H_2O [/itex]
    I'm using the subscript b to represent the base, and subscript a to represent the acid.
    [itex] c_b = 0.100 mol/L [/itex]
    [itex] V_b = 18.0 mL = 0.018 L [/itex]
    [itex] n_b = c_bV_b = 0.100mol/L x 0.018L = 1.8x10^{-3} mol [/itex]

    The mol ratio is 1:1, so the number of moles of the acid is also [itex] 1.8x10^{-3} [/itex]
    [itex] n =m/M [/itex]
    [itex] M_a = 180.17 g/mol [/itex]
    [itex] m = (1.8x10^{-3})(180.17 g/mol) [/itex]
    [itex] = 0.324 g [/itex]
    [itex] % purity = 0.324g/0.500g x 100% = 64.8% [/itex]
    (The original tablet was 0.500 g)

    I was hoping someone could let me know if I'm doing this correctly.
    Thank you!
     
  2. jcsd
  3. Jun 21, 2013 #2
    * % purity = 0.324g/0.500g = 64.8%
     
  4. Jun 22, 2013 #3

    Borek

    User Avatar

    Staff: Mentor

    Looks OK to me.
     
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