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Applying Chemistry to the real world

  1. Jan 20, 2005 #1
    Hi, I need to find the mass density and the number of atoms in a building that has a volume of 53306 cm^3. I know the building is made out of calcium carbonate, but I have no idea how to go about doing it. Do I need to assume the mass of the builing and find its mass density using D=M/V? What things can I assume? I know I need to use avogadro's number somewhere in there for the second part of the problem.

    thanks in advance,

    yours truely
     
  2. jcsd
  3. Jan 20, 2005 #2

    chem_tr

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    Hello, we all know that a structure with solid concrete (calcium carbonate) won't be much stable without any steel framework. However, we can even try to find its density, but keep in mind that this calculation will not reflect the reality. Otherwise, you may try to find the percentages of this building in terms of concrete, steel, glass, brick, etc.

    Well, the expression 53306 cc equals to 53.306 liters, so this is not as big as a building I first assumed. We are dealing with a very small copy of a big building, if I'm not wrong.

    Mass density is generally given in the form of grams/liter, but we don't know if the "building" is made up of solid (with no spaces in it) concrete. If there are some spaces, we'll have to subtract them in order to find the correct mass of the building. The unit density of calcium carbonate is, according to this website, 2.83 g/mL. You can do the calculation starting from this point, I think.
     
  4. Jan 21, 2005 #3
    clarify

    sorry, it's actually a pillar. And it's solely made out of calcium carbonate. Is there any way I can do the problem without having to use the unit density of calcium carbonate? The problem says we can assume many things...
     
  5. Jan 21, 2005 #4

    chem_tr

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    The weight of pillar must be given in order to calculate its density. This is because you were not given the weight, but volume instead. To use Avogadro's number, you'll need the weight. Frankly, I haven't found any way to calculate further without knowing the weight of the pillar.
     
  6. Jan 27, 2005 #5
    This is probably a stupid question, but i'll ak anyway...
    So after I calculated how many CaCO3 is is in the pillar, I need to multiply by 5 since there are five things that make up CaCO3, right?
     
  7. Jan 27, 2005 #6

    chem_tr

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    No, you are on a wrong lane to reach the correct answer. You must use mole amounts to learn the correct thing, in which one mole of CaCO3 makes 100 grams. I'll allow you to find the correct route by yourself.
     
  8. Jan 27, 2005 #7
    OK, say the mass is 50 grams. So 50 g CaCO3 X 1 mol CaCO3/100g CaCO3 X 6.022E23 CaCO3/1 mol CaCO3= number of molecules of CaCO3, right? Then I multiply it by 5 since there are five atoms that make up CaCO3, one Ca, one C, and 3 O's? Is this right?
     
  9. Jan 28, 2005 #8

    chem_tr

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    Well, it seems okay, since you are looking for the number of atoms. About density, what do you think?
     
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