Conservation of Mass and Energy












2












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I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!



For example, consider combustion:
$$rm CH_4 + 2O_2 to 2H_2O + CO_2 + {Energy}$$



However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?










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  • $begingroup$
    Possible duplicate of physics.stackexchange.com/questions/11449/… ?
    $endgroup$
    – Shufflepants
    1 hour ago










  • $begingroup$
    Possible duplicate of Conversion of mass to energy in chemical/nuclear reactions
    $endgroup$
    – BowlOfRed
    32 mins ago
















2












$begingroup$


I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!



For example, consider combustion:
$$rm CH_4 + 2O_2 to 2H_2O + CO_2 + {Energy}$$



However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?










share|cite|improve this question











$endgroup$












  • $begingroup$
    Possible duplicate of physics.stackexchange.com/questions/11449/… ?
    $endgroup$
    – Shufflepants
    1 hour ago










  • $begingroup$
    Possible duplicate of Conversion of mass to energy in chemical/nuclear reactions
    $endgroup$
    – BowlOfRed
    32 mins ago














2












2








2





$begingroup$


I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!



For example, consider combustion:
$$rm CH_4 + 2O_2 to 2H_2O + CO_2 + {Energy}$$



However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?










share|cite|improve this question











$endgroup$




I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!



For example, consider combustion:
$$rm CH_4 + 2O_2 to 2H_2O + CO_2 + {Energy}$$



However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?







special-relativity conservation-laws mass-energy physical-chemistry






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share|cite|improve this question













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share|cite|improve this question








edited 1 hour ago









rob

41.2k974169




41.2k974169










asked 3 hours ago









Dude156Dude156

1357




1357












  • $begingroup$
    Possible duplicate of physics.stackexchange.com/questions/11449/… ?
    $endgroup$
    – Shufflepants
    1 hour ago










  • $begingroup$
    Possible duplicate of Conversion of mass to energy in chemical/nuclear reactions
    $endgroup$
    – BowlOfRed
    32 mins ago


















  • $begingroup$
    Possible duplicate of physics.stackexchange.com/questions/11449/… ?
    $endgroup$
    – Shufflepants
    1 hour ago










  • $begingroup$
    Possible duplicate of Conversion of mass to energy in chemical/nuclear reactions
    $endgroup$
    – BowlOfRed
    32 mins ago
















$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
1 hour ago




$begingroup$
Possible duplicate of physics.stackexchange.com/questions/11449/… ?
$endgroup$
– Shufflepants
1 hour ago












$begingroup$
Possible duplicate of Conversion of mass to energy in chemical/nuclear reactions
$endgroup$
– BowlOfRed
32 mins ago




$begingroup$
Possible duplicate of Conversion of mass to energy in chemical/nuclear reactions
$endgroup$
– BowlOfRed
32 mins ago










4 Answers
4






active

oldest

votes


















3












$begingroup$

Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.



So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Let's do an analysis and see how much of a difference this makes.
    The relevant enthalpies of formation are




    • Methane: −74.87 kJ/mol

    • Oxygen: 0

    • Water(vapor): −241.818 kJ/mol

    • Carbon dioxide: −393.509 kJ/mol


    Therefore:
    $$rm CH_4 + 2O_2 to 2H_2O + CO_2 + 802.3 text{kJ}$$
    The mass of the products and reactants not worrying about the energy would be:
    $$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
    Now checking the energy released:
    $$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{(3.0times 10^8 text{m/s})^2} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.



    So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
      $endgroup$
      – rob
      1 hour ago










    • $begingroup$
      BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
      $endgroup$
      – David White
      42 mins ago










    • $begingroup$
      "confusing" = "wrong". Feel free to edit when there's a mistake.
      $endgroup$
      – BowlOfRed
      35 mins ago



















    2












    $begingroup$

    It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.




    In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      scientificamerican.com/article/…
      $endgroup$
      – safesphere
      2 hours ago










    • $begingroup$
      Well, Einstein has taken it all. But thanks for the link, it added something.
      $endgroup$
      – TechDroid
      2 hours ago



















    0












    $begingroup$

    All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.



    In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.






    share|cite|improve this answer









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      4 Answers
      4






      active

      oldest

      votes








      4 Answers
      4






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.



      So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.



        So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.



          So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.






          share|cite|improve this answer











          $endgroup$



          Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.



          So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 hours ago

























          answered 2 hours ago









          F16FalconF16Falcon

          3107




          3107























              3












              $begingroup$

              Let's do an analysis and see how much of a difference this makes.
              The relevant enthalpies of formation are




              • Methane: −74.87 kJ/mol

              • Oxygen: 0

              • Water(vapor): −241.818 kJ/mol

              • Carbon dioxide: −393.509 kJ/mol


              Therefore:
              $$rm CH_4 + 2O_2 to 2H_2O + CO_2 + 802.3 text{kJ}$$
              The mass of the products and reactants not worrying about the energy would be:
              $$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
              Now checking the energy released:
              $$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{(3.0times 10^8 text{m/s})^2} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.



              So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
                $endgroup$
                – rob
                1 hour ago










              • $begingroup$
                BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
                $endgroup$
                – David White
                42 mins ago










              • $begingroup$
                "confusing" = "wrong". Feel free to edit when there's a mistake.
                $endgroup$
                – BowlOfRed
                35 mins ago
















              3












              $begingroup$

              Let's do an analysis and see how much of a difference this makes.
              The relevant enthalpies of formation are




              • Methane: −74.87 kJ/mol

              • Oxygen: 0

              • Water(vapor): −241.818 kJ/mol

              • Carbon dioxide: −393.509 kJ/mol


              Therefore:
              $$rm CH_4 + 2O_2 to 2H_2O + CO_2 + 802.3 text{kJ}$$
              The mass of the products and reactants not worrying about the energy would be:
              $$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
              Now checking the energy released:
              $$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{(3.0times 10^8 text{m/s})^2} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.



              So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
                $endgroup$
                – rob
                1 hour ago










              • $begingroup$
                BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
                $endgroup$
                – David White
                42 mins ago










              • $begingroup$
                "confusing" = "wrong". Feel free to edit when there's a mistake.
                $endgroup$
                – BowlOfRed
                35 mins ago














              3












              3








              3





              $begingroup$

              Let's do an analysis and see how much of a difference this makes.
              The relevant enthalpies of formation are




              • Methane: −74.87 kJ/mol

              • Oxygen: 0

              • Water(vapor): −241.818 kJ/mol

              • Carbon dioxide: −393.509 kJ/mol


              Therefore:
              $$rm CH_4 + 2O_2 to 2H_2O + CO_2 + 802.3 text{kJ}$$
              The mass of the products and reactants not worrying about the energy would be:
              $$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
              Now checking the energy released:
              $$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{(3.0times 10^8 text{m/s})^2} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.



              So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.






              share|cite|improve this answer











              $endgroup$



              Let's do an analysis and see how much of a difference this makes.
              The relevant enthalpies of formation are




              • Methane: −74.87 kJ/mol

              • Oxygen: 0

              • Water(vapor): −241.818 kJ/mol

              • Carbon dioxide: −393.509 kJ/mol


              Therefore:
              $$rm CH_4 + 2O_2 to 2H_2O + CO_2 + 802.3 text{kJ}$$
              The mass of the products and reactants not worrying about the energy would be:
              $$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
              Now checking the energy released:
              $$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{(3.0times 10^8 text{m/s})^2} = 8.9 times 10^{-9}text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.



              So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 36 mins ago

























              answered 1 hour ago









              BowlOfRedBowlOfRed

              17.5k22743




              17.5k22743












              • $begingroup$
                This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
                $endgroup$
                – rob
                1 hour ago










              • $begingroup$
                BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
                $endgroup$
                – David White
                42 mins ago










              • $begingroup$
                "confusing" = "wrong". Feel free to edit when there's a mistake.
                $endgroup$
                – BowlOfRed
                35 mins ago


















              • $begingroup$
                This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
                $endgroup$
                – rob
                1 hour ago










              • $begingroup$
                BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
                $endgroup$
                – David White
                42 mins ago










              • $begingroup$
                "confusing" = "wrong". Feel free to edit when there's a mistake.
                $endgroup$
                – BowlOfRed
                35 mins ago
















              $begingroup$
              This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
              $endgroup$
              – rob
              1 hour ago




              $begingroup$
              This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $rm dalton = amu = GeV/mathit c^2$. A change of a few eV in a system like $rm CH_4 + 2O_2$ with total mass $sim 48,mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place.
              $endgroup$
              – rob
              1 hour ago












              $begingroup$
              BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
              $endgroup$
              – David White
              42 mins ago




              $begingroup$
              BowlOfRed, you didn't show "c" being squared in your equation. I'm sure that your final number accounted for this, but your answer looks a bit confusing as is.
              $endgroup$
              – David White
              42 mins ago












              $begingroup$
              "confusing" = "wrong". Feel free to edit when there's a mistake.
              $endgroup$
              – BowlOfRed
              35 mins ago




              $begingroup$
              "confusing" = "wrong". Feel free to edit when there's a mistake.
              $endgroup$
              – BowlOfRed
              35 mins ago











              2












              $begingroup$

              It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.




              In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.







              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                scientificamerican.com/article/…
                $endgroup$
                – safesphere
                2 hours ago










              • $begingroup$
                Well, Einstein has taken it all. But thanks for the link, it added something.
                $endgroup$
                – TechDroid
                2 hours ago
















              2












              $begingroup$

              It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.




              In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.







              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                scientificamerican.com/article/…
                $endgroup$
                – safesphere
                2 hours ago










              • $begingroup$
                Well, Einstein has taken it all. But thanks for the link, it added something.
                $endgroup$
                – TechDroid
                2 hours ago














              2












              2








              2





              $begingroup$

              It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.




              In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.







              share|cite|improve this answer











              $endgroup$



              It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few kilograms of mass can decay into.




              In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.








              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 hours ago

























              answered 2 hours ago









              TechDroidTechDroid

              61912




              61912












              • $begingroup$
                scientificamerican.com/article/…
                $endgroup$
                – safesphere
                2 hours ago










              • $begingroup$
                Well, Einstein has taken it all. But thanks for the link, it added something.
                $endgroup$
                – TechDroid
                2 hours ago


















              • $begingroup$
                scientificamerican.com/article/…
                $endgroup$
                – safesphere
                2 hours ago










              • $begingroup$
                Well, Einstein has taken it all. But thanks for the link, it added something.
                $endgroup$
                – TechDroid
                2 hours ago
















              $begingroup$
              scientificamerican.com/article/…
              $endgroup$
              – safesphere
              2 hours ago




              $begingroup$
              scientificamerican.com/article/…
              $endgroup$
              – safesphere
              2 hours ago












              $begingroup$
              Well, Einstein has taken it all. But thanks for the link, it added something.
              $endgroup$
              – TechDroid
              2 hours ago




              $begingroup$
              Well, Einstein has taken it all. But thanks for the link, it added something.
              $endgroup$
              – TechDroid
              2 hours ago











              0












              $begingroup$

              All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.



              In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.



                In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.



                  In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.






                  share|cite|improve this answer









                  $endgroup$



                  All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.



                  In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  PhysicsDavePhysicsDave

                  94547




                  94547






























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