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    This is a topic I have found really had to get to grips with and with my exam literally just around the corner (Ie, tomorrow morning), I can understand the basics but some questions I have no clue about!! College has been shut all last week due to weather and I was gutted about it as we missed out on vital study lessons, and it's also been shut all this week so far.

    Anyway, can anyone please help me with this question? It's from a past exam paper (OCR)

    Epsom salts can be used as bath salts to help relieve aches and pains.

    Epsom salts are crystals of hydrated magnesium sulfate, mgSO4.xH2O.

    A sample of Epsom salts was heated to remove the water. 1.57g of water was removed. 1.51g of anhydrous mgSO4 was left behind.

    i) Calculate the amount, in mol, of anhydrous mgSO4 formed.

    ii) Calculated the amount, in mol, of H2O removed.

    iii) Calculate the value of x in mgSO4.xH2O.
    PS. I know it's a capital m in magnesium but my 'm' button doesn't work and I need to copy paste and was too lazy to find a capital for it. lol

    If anyone can help, I'd really, really appreciate it. Thanks!!
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    Moles= mass/molar mass
    the Mr of MgS04 is 24.3+32.1+(16x4)= 120.4 g/mol
    the mass of mgs04 s 1.51 (given in the question)
    so using the formula you do 1.51/120.4= 0.0125

    for the next part you do the same thing: the Mr of h20 is 18
    and 1.57/18=0.087222

    ok, so to work out the value of x you simply divide 0.08722/ 0.0125 which gives you 7 when rounded up. Do you get it?
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    i) Just use the formula Moles = Mass/ Molar Mass
    (having found the value of the Mr of the MgSO4 by adding together the Ar of all the elements in their proportions)
    ii) Same as above
    iii) You've now got the two values for the numbers of moles of each in the substance, which is the first step in finding the empirical formula. The next is to find the simplest ratio between the two: if it's not obvious, then just divide both by the smallest value. Given the phrasing of the question, that'll then give you a ratio of moles of MgSO4: moles of H20 in the form 1:x, with x being the value you're trying to find.
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    Thanks guys, now it looks so easy. Ha! I'll remember that for tomorrow.

    I have another question, sorry, should have put it in first post I guess:

    Calcium carbonate, CaCO3, reacts with hydrochloric acid.

    7.50 x 10^3 mol CaCO3 reacts with 0.200 mol dm-3 HCl

    i) Calculate the volume, in cm^3, of 0.200 mol dm-3 HCl required to react with 7.50 x 10^3 mol CaCO3.

    ii) Calculate the volume, in cm^3, of CO2 formed at room temperate and pressure.
    I know it's probably basic stuff but like I said I have struggled with it this term. Thanks again.
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    What paper is that CaCO3 question from cuz I don't get why it would be 10^3 .. it's a bit much isn't it? Ironically I did a very similar question just now!
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    (Original post by SpottySocks)
    Thanks guys, now it looks so easy. Ha! I'll remember that for tomorrow.

    I have another question, sorry, should have put it in first post I guess:



    I know it's probably basic stuff but like I said I have struggled with it this term. Thanks again.
    i) Firstly, you'll need a balanced equation, which will show that a mole of Calcium Carbonate reacts with 2 moles of HCL to give a mole of CaCl2, a mole of CO2, and a mole of water.
    So, if there's 7500 moles of Calcium Carbonate, then that means there'll be twice the number of moles of HCl, so 15,000 moles of that. Then use the equation Volume = Moles/ Concentration to find the volume of HCl (remembering to convert from decimetres into cm3 after you've done that)
    i) The ratio is 1:1 of Calcium Carbonate forming Carbon Dioxide, so 7500 moles of the former will form 7500 moles of the latter. One mole is 24dm^3 (=24,000 cm^3) at room temperature and pressure, so just multiply the number of moles by this molar volume, to give you the total volume.
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    (Original post by TwentyNine)
    What paper is that CaCO3 question from cuz I don't get why it would be 10^3 .. it's a bit much isn't it? Ironically I did a very similar question just now!
    I don't get what you mean; there's no reason for it being that value, it's just one they've randomly chosen to use in the question.
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    (Original post by Revolution is my Name)
    I don't get what you mean; there's no reason for it being that value, it's just one they've randomly chosen to use in the question.
    Haha yeah, but the question I just did, it was more like 10^-3 so I was getting confused
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    (Original post by TwentyNine)
    Haha yeah, but the question I just did, it was more like 10^-3 so I was getting confused
    Yeah, you're right, the values are usually quite small. Them being bigger is probably a good thing though, as I suppose you're considering them as actual numbers and you'll be more likely to notice if something goes amiss, whereas if they're small numbers in standard form there's more bunging them into the calculator and blindly following what it says.
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    (Original post by Revolution is my Name)
    Yeah, you're right, the values are usually quite small. Them being bigger is probably a good thing though, as I suppose you're considering them as actual numbers and you'll be more likely to notice if something goes amiss, whereas if they're small numbers in standard form there's more bunging them into the calculator and blindly following what it says.
    Yep there is :]
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    For part 1 I got 3cm^3, that can't be right? Do I multiply the 0.200 by 1000? The number would be too big then wouldn't it?

    I did (15,000/0.200)/1,000

    Gawd I sound so dumb lol

    and yep sorry missed out the -! It was 10^-3.
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    (Original post by SpottySocks)
    For part 1 I got 3cm^3, that can't be right? Do I multiply the 0.200 by 1000? The number would be too big then wouldn't it?

    I did (15,000/0.200)/1,000

    Gawd I sound so dumb lol

    and yep sorry missed out the -! It was 10^-3.
    15,000 moles and a concentration of 0.2 mol dm^-3 gives:
    Volume = Moles/ Concentration
    = 15,000/ 0.2
    = 75,000 dm^3
    1 dm^3= 1,000 cm^3, hence to convert from dm^3 to cm^3 we need to multiply by 1,000, giving:
    75,000,000 cm^3

    By the looks of your answer, I think you accidentally multiplied Moles and concentration, instead of dividing the first by the second.
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    (Original post by Revolution is my Name)
    15,000 moles and a concentration of 0.2 mol dm^-3 gives:
    Volume = Moles/ Concentration
    = 15,000/ 0.2
    = 75,000 dm^3
    1 dm^3= 1,000 cm^3, hence to convert from dm^3 to cm^3 we need to multiply by 1,000, giving:
    75,000,000 cm^3

    By the looks of your answer, I think you accidentally multiplied Moles and concentration, instead of dividing the first by the second.
    Ahh I see thank you. I got that number first but thought it didn't look right. Not much looks right when it comes to me and Chemistry. :rolleyes: Thanks for your help I am understanding it better now.
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    (Original post by SpottySocks)
    For part 1 I got 3cm^3, that can't be right? Do I multiply the 0.200 by 1000? The number would be too big then wouldn't it?

    I did (15,000/0.200)/1,000

    Gawd I sound so dumb lol

    and yep sorry missed out the -! It was 10^-3.
    Don't call yourself dumb!

    Okay moles=concentration X volume right? So if you want to know the volume you simply re-arrange your formula.. hence volume = moles/concentration
    So thats: 7.50 x 10^-3 / 0.200 = 0.0375 dm^3 if you want, multiply that by 1000 to get cm^3 which will = 37.5 cm^3
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    (Original post by TwentyNine)
    Don't call yourself dumb!

    Okay moles=concentration X volume right? So if you want to know the volume you simply re-arrange your formula.. hence volume = moles/concentration
    So thats: 7.50 x 10^-3 / 0.200 = 0.0375 dm^3 if you want, multiply that by 1000 to get cm^3 which will = 37.5 cm^3
    There's twice as many moles of HCl as there is of Calcium Carbonate, so that should be 1.5x10^-2
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    (Original post by Revolution is my Name)
    There's twice as many moles of HCl as there is of Calcium Carbonate, so that should be 1.5x10^-2
    Ahh yeahh I forgot the equation, well, I was hoping to outline the basic idea =]
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    Thank you so much Revolution and TwentyNine for your help last night. Definitely helped me for my exam this morning, it went very well surprisingly!!!
 
 
 
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