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Conservation of Mass and Energy


Conversion of mass to energy in chemical/nuclear reactionsCan non-free forces change the rest mass?Why does mass change in to energy during a nuclear change?Energy & Mass of a PhotonWhat is the argument for detailed balance in chemistry?Would impact angle matter on relativistic impactor?Hypothetical special relativity with mass conservationDoes the mass of object really increase?Some calculations on the energy consumption of a relativistic rocketIf mass and energy are same what will be the equivalent of a homogeneous ball in terms of energy and information?













1












$begingroup$


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:
$$CH_4 + 2O_2 >>> 2H_2O + CO_2 + Energy$$



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











$endgroup$
















    1












    $begingroup$


    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:
    $$CH_4 + 2O_2 >>> 2H_2O + CO_2 + Energy$$



    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











    $endgroup$














      1












      1








      1





      $begingroup$


      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:
      $$CH_4 + 2O_2 >>> 2H_2O + CO_2 + Energy$$



      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:
      $$CH_4 + 2O_2 >>> 2H_2O + CO_2 + Energy$$



      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago









      Aaron Stevens

      12.8k42248




      12.8k42248










      asked 1 hour ago









      Dude156Dude156

      1307




      1307




















          3 Answers
          3






          active

          oldest

          votes


















          2












          $begingroup$

          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











          $endgroup$




















            1












            $begingroup$

            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
              55 mins ago










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









            $endgroup$












              Your Answer





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              3 Answers
              3






              active

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              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

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              2












              $begingroup$

              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











              $endgroup$

















                2












                $begingroup$

                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











                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  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 54 mins ago

























                  answered 1 hour ago









                  F16FalconF16Falcon

                  3007




                  3007





















                      1












                      $begingroup$

                      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
                        55 mins ago










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















                      1












                      $begingroup$

                      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
                        55 mins ago










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













                      1












                      1








                      1





                      $begingroup$

                      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 18 mins ago

























                      answered 1 hour ago









                      TechDroidTechDroid

                      60912




                      60912











                      • $begingroup$
                        scientificamerican.com/article/…
                        $endgroup$
                        – safesphere
                        55 mins ago










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
















                      • $begingroup$
                        scientificamerican.com/article/…
                        $endgroup$
                        – safesphere
                        55 mins ago










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















                      $begingroup$
                      scientificamerican.com/article/…
                      $endgroup$
                      – safesphere
                      55 mins ago




                      $begingroup$
                      scientificamerican.com/article/…
                      $endgroup$
                      – safesphere
                      55 mins ago












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




                      $begingroup$
                      Well, Einstein has taken it all. But thanks for the link, it added something.
                      $endgroup$
                      – TechDroid
                      17 mins 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 24 mins ago









                          PhysicsDavePhysicsDave

                          94547




                          94547



























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