Hey there! Sign in to join this conversationNew here? Join for free

Struggling with Chemistry revision: working out molecular formula of a Hydrocarbon watch

    • Thread Starter
    Offline

    17
    The question I am finding difficult gives me no empirical formula to work with, and every answer I come up with seems incorrect. Could someone please help?

    (Q): Work out the molecular formula of a hydrocarbon whose molecular formula has a RMM (Relative Molecular Mass) of 30, formed when 1.44g of carbon reacts with 0.36g of hydrogen.

    I thought that the empirical formula is maybe CH4 as 1.44/0.36 = 4, therefore 4 Hydrogens to every Carbon.

    But then if I divide the empirical mass by the RMM I get 1.875, so I'm a bit confused.
    Offline

    9
    ReputationRep:
    To work out the empirical formula you need to divide the masses of both C and H by their own individual atomic mass (i.e 12 for C and 1 for H). Then divide both answers by the smallest number to get a ratio. This gives you the empirical formula.

    Work out the RMM of the empirical formula and just scale up to get to the required RMM of 30.
    Offline

    1
    ReputationRep:
    Work out the Moles of each first of all
    Moles of C = 1.44/12 = 0.12
    Moles of H = 0.36/1 = 0.36
    Work out the ratio between these by dividing each by the smallest value (moles of C)
    0.12/0.12 = 1 0.36/0.12 = 3
    This shows that the empirical formula is three H's for every C (CH3)
    You can then work out the empirical mass of CH3 which is 15
    Then for the molecular formula, you diving the RMM by the empirical mass 30/15 = 2
    So there are two empirical units, so the answer is C2H6,
    Hope this helps
    • Thread Starter
    Offline

    17
    (Original post by Infamous*)
    To work out the empirical formula you need to divide the masses of both C and H by their own individual atomic mass (i.e 12 for C and 1 for H). Then divide both answers by the smallest number to get a ratio. This gives you the empirical formula.

    Work out the RMM of the empirical formula and just scale up to get to the required RMM of 30.
    I've followed your advice, therefore 1.44/12 = 0.12 and 0.36/1 =0.36

    This gives a 1:3 ratio

    Therefore the empirical formula is CH3? So the empirical mass is 15g

    So the answer is C2H6?

    ok it's correct thanks
    • Thread Starter
    Offline

    17
    (Original post by Infamous*)
    To work out the empirical formula you need to divide the masses of both C and H by their own individual atomic mass (i.e 12 for C and 1 for H). Then divide both answers by the smallest number to get a ratio. This gives you the empirical formula.

    Work out the RMM of the empirical formula and just scale up to get to the required RMM of 30.
    Could you please help me with this revision question: When Sulphur reacts with Chlorine, the compound formed contains 47.4% sulphur; work out the molecular formula if the relative molecular mass is 135.

    My answer seems incorrect but so far I have converted % into grams and then divided the amounts by the Relative atomic masses, therefore 47.4/31.1 = 1.477 and 52.6/35.5 = 1.482, therefore near enough a 1:1 ratio.

    Also I know chlorine is a diatomic element therefore would I divide the 1.482 by 2 first?

    But how do I scale up this is confusing me?
    Offline

    9
    ReputationRep:
    (Original post by Sam00)
    Could you please help me with this revision question: When Sulphur reacts with Chlorine, the compound formed contains 47.4% sulphur; work out the molecular formula if the relative molecular mass is 135.

    My answer seems incorrect but so far I have converted % into grams and then divided the amounts by the Relative atomic masses, therefore 47.4/31.1 = 1.477 and 52.6/35.5 = 1.482, therefore near enough a 1:1 ratio.

    Also I know chlorine is a diatomic element therefore would I divide the 1.482 by 2 first?

    But how do I scale up this is confusing me?
    ok so you have got the empirical formular correct : SCl

    Empirical mass= 67.6

    The RMM we need is 135. So we do 135/67.6 = 2

    So the factor by which we have to multiply the empirical formula is 2
    = S2Cl2
    • Thread Starter
    Offline

    17
    (Original post by Infamous*)
    To work out the empirical formula you need to divide the masses of both C and H by their own individual atomic mass (i.e 12 for C and 1 for H). Then divide both answers by the smallest number to get a ratio. This gives you the empirical formula.

    Work out the RMM of the empirical formula and just scale up to get to the required RMM of 30.
    Any chance of a hint please?
 
 
 
Poll
Who is your favourite TV detective?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.