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Two unknwon factors in this stoichiometry problem

  1. Nov 26, 2004 #1
    Can someone help me with this?

    What volume, cm³, would 36.25g of
    [tex] C_4H_1_0 [/tex]
    occupy at 85.20 kPa and 108 celsius?

    So im assuming i should use the ideal gas equation, PV = nRT, and isolate V so i get:

    [tex] V=\frac{nRT}{P} [/tex]

    heres where im stuck, how do i solve for "n" when only the mass is given? how do i find the molecular weight?

    and its my first time here .. so can some1 explain to me how these latex things work =X
    Last edited: Nov 26, 2004
  2. jcsd
  3. Nov 26, 2004 #2


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    Science Advisor

    you can find the number of moles (n) by dividing the number af grams you know you have by the molar weight of the substance.

    yuo can find the molar weight of the substance(grams/mole) by adding up each elements molar weight and totallying them together,

    for example:
    C4H10 is, 4 carbons (12.01 g/mol each) + 10 hydrogens (1.01) g/mol each) = 48.04 + 10.1 = 58.14 g/mol

    and since you already know you have 36.25 g of it, that means you have,
    36.25/58.14 = .6234 moles of it.

    remember, the ideal gas law only works for gasses, so make sure that your butane (C4H10) is a gas under those conditions, otherwise it wont work, but I am pretty sure that at that low of a pressure and that high of a temperature, it would be.
  4. Nov 26, 2004 #3
    alright tahnks i understand now
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