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Mole Concept

  1. Feb 12, 2006 #1
    would someone be kind enough to explain througly the concept of moles....i have missed classes on that part of the unit and need to get a better grasp of it...the text is not helping...if anyone can do it in a step by step type of way...i would greatly appreciate it....thank you ahead of time.:smile:
  2. jcsd
  3. Feb 12, 2006 #2


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    Do you know what a dozen is?

    That's all you need to know.

    1 dozen = 12 eggs/cans/whatever
    1 mole ~ 6.022*10^23 molecules/atoms/particles (and you could even say eggs or cans! except 1 mole of eggs is probably more eggs then mankind has ever or ever will eat for all time)

    Don't be intimidated by the large number. If there are 1 mole of NaCl (otherwise known as table salt), that means there are 6.022*10^23 molecules of table salt just as if someone were to say there is 1 dozen NaCl molecules.
    Last edited: Feb 12, 2006
  4. Feb 12, 2006 #3
    i understand....now can you explain to me how to apply this concept with the whole moler mass thing...as in CaCO3....i dont understand that part
  5. Feb 12, 2006 #4


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    Ok, well, the period table has a # value normally right under an element's symbol. For hydrogen, it is 1.0079, Oxygen is 15.999, etc etc. What these numbers mean is that 1 mole of hydrogen atoms is equal to 1.0079 grams. 1 mole of oxygen is 15.999 grams. The 15.999 grams is considered the molar mass of an atom.

    Calcium carbonate is CaCO3. So what you would try to find out is how many grams does 1 mole of calcium carbonate weigh? Simple! Just add up 1 calcium, 1 carbon, and 3 oxygen. 40.08 + 12.01 + 3(15.999) = 100.087 grams/ 1 mole of CaCO3.

    This basically means if you have 10 grams of pure calcium carbonate, you have 1/10th of a mole of calcium carbonate or roughly 6.022*10^22 molecules of CaCO3
  6. Feb 12, 2006 #5
    thanks..now i understand....
  7. Feb 12, 2006 #6
    now from what you told me..if i apply that in this question...

    *find the mass of 0.250 moles of AgNO3*
    to slove it...i just take the total molar mass of AgNO3 and multiply it by 0.250(since the total mass is 1 mole...) that would give me the mass of 0.25 worth of moles.....do i have the right idea???
  8. Feb 12, 2006 #7


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    Yes, that is correct.

    Dimensional analysis seems to be a very important safeguard for people getting use to the idea of molar masses so you might want to remember to use that.
  9. Feb 14, 2006 #8
    here are some more problems that i am trying to do for practise....some of them have new stuff i need help with....
    would it be possiable for you to explain to me how to find out how many of an elements are in a compound.
    **What is the number of hydrogen atoms in 20 Kg of propane (C*3H*8)**

    i am not sure how to even start this....i saw it in the textbook and was confused.......
    Last edited: Feb 14, 2006
  10. Feb 14, 2006 #9


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    The first thing you need to do is figure out how many Hydrogen atoms there are in just one propant molecule. You can do this just by looking at the formula, there are 8 Hydrogens. So now you know for each propant molecule, you have 8 Hydrogens.
    Next, how many propane molecules do you have? You know that there are 6.022142 E23 molecules of propane per mole, so how many moles do you have? You can calculate the molar mass of propane by adding each of the atomic masses that make it up. (3*12.01 + 8*1.01 = 44.11 g/mol).
    To determine the number of moles, diveide the mass of the total propane by the molar mass.
    Once you know how many moles you have, you can calculate how many molecules you have, and from that, you can figure out how many Hydrogen atoms you have.
  11. Feb 14, 2006 #10
    so i take 20 kg turn it into 20 000 g and divide it by 44.11 g which is the molar mass...that gives me 453.4 moles??....
    then i multiply that by 6.02 x 10^23 to give me the number of molecules?...what do i do from there if have done everything correctly so far? do i divide that by 8? to give me the number of molecules of Hydrogen
  12. Feb 14, 2006 #11


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    I would multiply by 8, not divide.

    Multiplying the number of moles of propane by Avagodros number will give you the number of molecules of propane, right?
    In each molecule of propane, there are 8 Hydrogen atoms.
  13. Feb 14, 2006 #12
    OHHH....i see...thanks...that was a bit of tough to understand all togather but i see how clear the explanation was...thanks....i understand that part now...eww....if you have some time would you explain this other type of question to me???

    **Determine the percentage composition of acrylamide C*3H*5ON***

    i think i have an understanding after your explanation.... i think after i find the molar mass of the compund i take each element and multiply by their molar mass and divide by the molar mass of the whole element?? it seemed like a logical way but i may have it compelety wrong.....
  14. Feb 14, 2006 #13
    \Also, multiply by the molar quantity of that element in a mole of that compound

    Acrylamide (i.e, [tex]{\text{C}}_3 {\text{H}}_5 {\text{NO}}[/tex] ) has-->a molar mass of 71.1g/mol.

    To find percent composition (by mass) of each element in acrylamide, simply calculate (total mass of element 'x' in one mole acrylamide)/(mass of one mole of acrylamide).

    Of course, below are many more steps than you'll really need-->but someone like you might understand the process/calculations more easily with some extra steps/>

    %Composition (<~by mass) of carbon:
    [tex]\left( {\frac{{12.0{\text{g}}\;{\text{C}}}}{{1\,{\text{mol}}\;{\text{C}}}}} \right)\left( {\frac{{3\,{\text{mol}}\;{\text{C}}}}{{{\text{1}}\,{\text{mol}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right)\left( \frac{{1\,{\text{mol}}\;{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{NO}}}}{{{\text{71.1g}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right) \approx 50.6\% \;\text{C} [/tex]

    %Composition (<~by mass) of hydrogen:
    [tex]\left( {\frac{{1.01{\text{g}}\;{\text{H}}}}{{1\,{\text{mol}}\;{\text{H}}}}} \right)\left( {\frac{{5\,{\text{mol}}\;{\text{H}}}}{{{\text{1}}\,{\text{mol}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right)\left( \frac{{1\,{\text{mol}}\;{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{NO}}}}{{{\text{71.1g}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right) \approx 7.10\% \;\text{H} [/tex]

    %Composition (<~by mass) of nitrogen:
    [tex]\left( {\frac{{14.0{\text{g}}\;{\text{N}}}}{{1\,{\text{mol}}\;{\text{N}}}}} \right)\left( {\frac{{1\,{\text{mol}}\;{\text{N}}}}{{{\text{1}}\,{\text{mol}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right)\left( \frac{{1\,{\text{mol}}\;{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{NO}}}}{{{\text{71.1g}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right) \approx 19.7\% \;\text{N} [/tex]

    %Composition (<~by mass) of oxygen:
    [tex]\left( {\frac{{16.0{\text{g}}\;{\text{O}}}}{{1\,{\text{mol}}\;{\text{O}}}}} \right)\left( {\frac{{1\,{\text{mol}}\;{\text{O}}}}{{{\text{1}}\,{\text{mol}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right)\left( \frac{{1\,{\text{mol}}\;{\text{C}}_{\text{3}}{\text{H}}_{\text{5}}{\text{NO}}}}{{{\text{71.1g}}\;{\text{C}}_{\text{3}} {\text{H}}_{\text{5}} {\text{NO}}}}} \right) \approx 22.5\% \;\text{O} [/tex]

    Hope this helps :smile:

    Though, this questions like these should be posted in the Homework section...
    Last edited: Feb 14, 2006
  15. Feb 14, 2006 #14
    thanks that helped.....would you help explain to me how to calculate the molar mass of two compunds??...like MgSO*4 (there is a dot here) 7*H*2O....

    i think that dot means adding??...so for this we just find the mass of Mg add it with S and Ox4..then we find the mass of oxygen and add it to the mass of H...then we multiply it by 7...then we just add the sum of MgSO*4 and add it to 7 x water...am i correct
  16. Feb 14, 2006 #15
    To calculate the molar mass of have magnesium sulfate heptahydrate (i.e., [tex]{\text{MgSO}}_4 \cdot 7{\text{H}}_2 {\text{O}}[/tex]),
    add the mass of one mole of magnesium sulfate to the mass of seven moles of water.

    MgSO4 has a molar mass of 120. g/mol, and
    H2O (water) has a molar mass of 18.0 g/mol.

    Since each molecule MgSO4 is hydrated with seven molecules H2O, the molar mass of your substance--magnesium sulfate heptahydrate--is 120g/mol + 7*18.0g/mol = 246g/mol.
    Last edited: Feb 14, 2006
  17. Feb 14, 2006 #16
    i now understand.....what is the empirical formula???

    this question is asking me for the emp formula of 48.1% of Ni 16.8% of P and 35% of O....

    so for this do i take 48.1% turn it into 48.1g and divide it by its molar mass?....i am a bit confused
    Last edited: Feb 15, 2006
  18. Feb 15, 2006 #17
    this one question has got me lost....can someone help me solve it?

    ***the hydrate of barium hydroxide is 45.6% water. Determine the molecular formula of hydrate***
  19. Feb 15, 2006 #18
    Right, you convert to a mole ratio as:

    For nickel,
    [tex]\left( {48.1{\text{g}}\;{\text{Ni}}} \right)\left( {\frac{{1\,{\text{mol}}\;{\text{Ni}}}}{{58.7{\text{g}}\;{\text{Ni}}}}} \right) = 0.819\,{\text{mol}}[/tex]

    For phosphorus,
    [tex]\left( {16.8{\text{g}}\;{\text{P}}} \right)\left( \frac{{1\,{\text{mol}}\;{\text{P}}}}{{31.0{\text{g}}\;{\text{P}}}}} \right) = 0.542\,{\text{mol}}[/tex]

    Finally, for oxygen
    [tex]\left( {35{\text{g}}\;{\text{O}}} \right)\left( {\frac{{1\,{\text{mol}}\;{\text{O}}}}{{16.0{\text{g}}\;{\text{O}}}}} \right) = 2.2\,{\text{mol}}[/tex]

    Next, your textbook might ask that you "divide by the smallest molar quantity," which in this case, will be our 0.542mol P.

    As such, the ratio {0.819mol Ni, 0.542mol P, 2.2mol O} 'reduces' to {1.51mol Ni, 1mol P, 4.06mol O}. Multiplying each quantity by 2 yields {3.02mol Ni, 2mol P, 8.12mol O}, which we can round to {3mol Ni, 2mol P, 8mol O}.

    *And thus, the empirical formula of your substance is
    [tex]{\text{Ni}}_3 {\text{P}}_2 {\text{O}}_8[/tex]

    -Which is sensibly:
    [tex]{\text{Ni}}_{\text{3}} \left( {{\text{PO}}_{\text{4}} } \right)_2[/tex]
    As 100% - 45.6% = 55.4%, you can solve as
    [tex]\left( {\frac{{{\text{1}}\,{\text{mol}}\;{\text{H}}_2 {\text{O}}}}
    {{18.0{\text{g}}\;{\text{H}}_2 {\text{O}}}}} \right)\left( {\frac{{0.456\,{\text{g}}\;{\text{H}}_{\text{2}} {\text{O}}}}{{1.000{\text{g Ba}}\left( {{\text{OH}}} \right)_2 \cdot n{\text{H}}_2 {\text{O}}}}} \right)\left( {\frac{{1.000{\text{g Ba}}\left( {{\text{OH}}} \right)_2 \cdot n{\text{H}}_2 {\text{O}}}}{{0.554{\text{g}}\;{\text{Ba}}\left( {{\text{OH}}}\right)_2 }}} \right)\left( {\frac{{{\text{171g}}\;{\text{Ba}}\left( {{\text{OH}}} \right)_2 }}
    {{{\text{1}}\,{\text{mol}}\;{\text{Ba}}\left( {{\text{OH}}} \right)_2 \cdot n{\text{H}}_2 {\text{O}}}}} \right) = \frac{{7.82\,{\text{mol}}\;{\text{H}}_{\text{2}} {\text{O}}}}{{1\,{\text{mole}}\;{\text{Ba}}\left( {{\text{OH}}} \right)_2 \cdot n{\text{H}}_2 {\text{O}}\;}}[/tex]

    [tex] \approx \frac{{8\,{\text{mol}}\;{\text{H}}_{\text{2}} {\text{O}}}}
    {{1\,{\text{mole}}\;{\text{Ba}}\left( {{\text{OH}}} \right)_2 \cdot n{\text{H}}_2 {\text{O}}}}[/tex]

    And so, the molecular formula of your hydrate is:
    [tex]{\text{Ba}}\left( {{\text{OH}}} \right)_2 \cdot 8{\text{H}}_2 {\text{O}}[/tex]
    Last edited: Feb 15, 2006
  20. Feb 15, 2006 #19
    thanks i got that...
    Last edited: Feb 15, 2006
  21. Feb 15, 2006 #20
    ohh...my text book does another step after finding the moles....it divides it by the lowest.....like 0.82/0.54.. 0.54/0.54 and 2.2/0.54.....wat is this about?
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