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O Level Chemistry & IP Chemistry Notes by 10 Year Series Author – Chemistry Specialist

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O Level Chemistry: Molecular & Empirical Formulae

For all students learning O Levels Chemistry (and other equivalents), it is important that you understand the difference between the 3 types of formula commonly used – namely Molecular Formula, Empirical Formula and Structural Formula.

Previously, we have discussed on this topic before. You can find them HERE.

Let’s define them now, for better understanding and application:

Molecular Formula

– shows the exact number of atoms of each element in a molecule

Empirical Formula

– shows the types of elements (as well as the simplest ration of different types of atoms) in the compound

– commonly known as the Simplest Formula

Structural Formula

– shows the covalent bonds involved (single, double or triple covalent bonds) between each atom in the compound

– more for Covalent Molecules

Let’s look at some examples to look at the difference between Molecular Formula and Empirical Formula:

Example 1: Hydrogen Peroxide

Molecular Formula is H2O2

Empirical Formula is HO

Example 2: Ethane

Molecular Formula is C2H6

Empirical Formula is CH3

Example 3: Ammonia

Molecular Formula is NH3

Empirical Formula is NH3

Enjoying your self? Now, try out a tricky question that was emailed to me by my Sec 4 O Level Chemistry student.

Question 1:

A student can only recall the formula of a mineral as K2Al6(SO4)x(OH)y, but remembers clearly that its empirical formula is KAl3S2O14H6. What are the values of x and y in the formula?

Try it out and leave your suggested answer below. Feel free to challenge each other in your answer. That’s the way how learning should be.

PS: Feel free to send this blogpost link to your classmates, friends or students (if you are an educator) if you think it will help them.

Related Articles:

  • O Level Chemistry: Types of Formula
  • O Level Chemistry: Mole Concepts / Chemical Calculations
  • More Error in Marshall Cavendish GCE O-Level Chemistry Matters Textbook – Part 2
  • O Level Chemistry: Mole Concepts / Chemical Calculations
  • O Level Chemistry: Elements, Compounds & Mixture

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Comments

  1. sean says

    October 28, 2010 at 2:48 am

    You got it! =)

    Sean Chua

  2. sean says

    October 28, 2010 at 2:47 am

    No problem. In exam they will not ask you to show them the working…If your method gets you the final answer, thats good enough.

    And you know what? Your answer is correct! =)

    Cheers,

    Sean Chua

  3. sean says

    October 28, 2010 at 2:46 am

    Hi Max,

    Good attempt but answer is not correct.

    See my comments for Alex Lee.

    Cheers,

    Sean Chua

  4. sean says

    October 28, 2010 at 2:45 am

    Yup! =)

    Sean Chua

  5. sean says

    October 28, 2010 at 2:45 am

    Hi Alex Lee,

    Your method of expanding the empirical formula and then re-balancing it is a great way to attempt such questions.

    Answer is correct! Well done!

    Sean Chua

  6. sean says

    October 28, 2010 at 2:42 am

    Hi Andrew.

    Good attempt but the answer is incorrect.

    Check my suggested solution and answer below.

    Regards,

    Sean Chua

  7. sean says

    October 28, 2010 at 2:36 am

    The bond energy for C-H and C-Cl bonds are very strong. Not so easy to break them down using bacteria and chemical reagents.

    Hope this helps,

    Sean Chua

  8. Nur Raihan Binte Mohamad Ridzal says

    September 4, 2010 at 3:12 pm

    x=2
    y=6

  9. Kam Qiang Wei says

    September 3, 2010 at 6:40 am

    oh yah mr sean chua, do u mind telling me why the bonds between chloroethene cannot be broken down via bacteria and chemicals(acid and alkali)

  10. Kam Qiang Wei says

    September 3, 2010 at 6:33 am

    im not very sure. in fact i think im wrong…but anw

    K2Al6(SO4)x(OH)y

    K Al3 S2 O14 H6 as empirical formula

    via so called expanding.

    K2 AL6 Sx O4x Oy Hy

    compare it with the empirical formula, use Potassium and aluminium, since their values are known. hence the scalar factor from empirical
    formula and the real formula is 2 .
    thus use 014 multiply by 2 to get 28.
    and Oy + O4x = 28

    H6 times 2 to obtain 12 subscript
    y= 12 (sub to the oxygen ‘eqn’ )
    O12 +O 4x = 28
    4x= 16
    x=4

    thus x=4
    y=12

    please give me the correct answer, thnx

  11. JQ says

    September 2, 2010 at 9:38 pm

    Hi Sean!

    Heres my solution, i’m not sure if my presentation is correct and i hope you’ll take a look and correct me if theres something wrong

    2(KAl3S2O14H6) = K2Al6(SO4)x(OH)y
    K2Al6S4O28H12 = K2Al6(SO4)x(OH)y
    K2Al6(SO4)4O12H12= K2Al6(SO4)x(OH)y
    K2Al6(SO4)4(OH)12= K2Al6(SO4)x(OH)y

    Thus, x= 4, y=12

    Thanks!

  12. shawn says

    September 1, 2010 at 8:07 pm

    x is 4 y is 12?

  13. max says

    September 1, 2010 at 4:17 pm

    x=2, y=12

  14. shawn says

    September 1, 2010 at 3:42 pm

    x is 4 and y is 12?

  15. Alicia says

    September 1, 2010 at 3:14 pm

    x:4
    y:12

  16. Derrick says

    September 1, 2010 at 12:35 pm

    x=4
    y=12

  17. alex lee says

    August 31, 2010 at 9:13 pm

    A bit careless there Andrew.

    You have to start by multiplying everything by 2

    K2 Al6 S4 O28 H12
    K2 Al6 (SO4)4 (OH)y

    Your explanation is correct, due to 4 sulfur present

    Balance should be : 0K 0Al 0S 12O 12H

    Thus,

    K2 Al6 (SO4)4 (OH)12

    x = 4 , y = 12

  18. andrew yan says

    August 31, 2010 at 9:10 pm

    A student can only recall the formula of a mineral as K2Al6(SO4)x(OH)y, but remembers clearly that its empirical formula is KAl3S2O14H6. What are the values of x and y in the formula?

    Start isolating everything
    K2 Al6 (SO4)x (OH)y
    —>K Al3 S2 O14 H6
    K Al3 (SO4)2 (OH)y

    We leave SO4 as x2 because there is only 2 sulfur, so we use 8 oxygen along with it in (SO4)2

    Balance : 0K 0Al3 0S 6O 6H

    So it’s fairly obvious it will be(OH)6

    Hence

    K Al3 (SO4)2 (OH)6

    Where x = 2 , y = 6

    =)

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