Python 3, 566 bytes
n="HᱨἬᨊBCNOFỆᶎỻ᭬∋PS᱄ᳲKᥣ’≴VᷖNoᮞᴍ ̔ẟ⚬ᰃG₿uVIIὤ⓶Y⠔ᷜIV=╺1⁄7ὀᨧᨬ⎪ῆZI⊸ḙᤂᩯ⎠Ṹ∐⍗ᾉᮼ
.ἸẺ⏄−⊼ᲰW⁚⍽2⑀ƫ᳸⍰Ạᬒ⊰ᵴ⌼Ἤἒᤣ∠ṐU∠9.ᮭᲇᮖ᪲ỿ᷎"
m="!$')+,.03478;<?@CHGHMPSTWX[[`afikoptuxy{} £\©«¬°±¶ ̧1⁄2¿ÃÅÇÉÍÏÒÕØÚÞàãåéìïññòþÿĂăĈćĎčĔēėėěĜġ"
i=lambda w:f.reduce(lambda l,c:l*ord(c),list(w),1)
print(str(ord(m[n.find(chr(i(input())))])-32))
Python 3 , 560 bytes
import functools as f
n="HᱨἬᨊBCNOFỆᶎỻ᭬∋PS᱄ᳲKᥣ’≴VᷖNoᮞᴍ ̔ẟ⚬ᰃG₿uVIIὤ⓶Y⠔ᷜIV=╺1⁄7ὀᨧᨬ⎪ῆZI⊸ḙᤂᩯ⎠Ṹ∐⍗ᾉᮼ
.ἸẺ⏄−⊼ᲰW⁚⍽2⑀ƫ᳸⍰Ạᬒ⊰ᵴ⌼Ἤἒᤣ∠ṐU∠9.ᮭᲇᮖ᪲ỿ᷎"
m="!$')+,.03478;<?@CHGHMPSTWX[[`afikoptuxy{}€‚…‡ŠŒ“—š Ÿ£\©«¬°±¶ ̧1⁄2¿ÃÅÇÉÍÏÒÕØÚÞàãåéìïññòþÿĂăĈćĎčĔēėėěĜġ"
i=lambda w:f.reduce(lambda l,c:l*ord(c),list(w),1)
print(str(ord(m[n.find(chr(i(input())))])-32))
Explanation: So 2 variables, n and m are used to hold the abbreviations (more described below) and the elements atomic mass as strings. The input is then encrypted into a character to match to some character in n, using the string.find, an index of where the character is can be used to find the respective atomic mass in m. Due to shifting the mass values by 32 previously, they must be unshifted, then this number is output.
Each character of n is the ascii of the product of the ascii values of the characters in the name
Each character of m is the ascii of mass + 32
Below is the setup code for m and n
obfNames=""
for name in names:
prod=1
for c in name:
prod*=ord(c)
obfNames+=chr(prod)
obfMass=""
for m in mass:
obfMass+=chr(int(m)+32)
for c in obfNames:
print(c,end="")
print()
for c in obfMass:
print(c,end="")
obfNames=""
for name in names:
prod=1
for c in name:
prod*=ord(c)
obfNames+=chr(prod)
obfMass=""
for m in mass:
obfMass+=chr(int(m)+32)
for c in obfNames:
print(c,end="")
print()
for c in obfMass:
print(c,end="")
Python 3, 566 bytes
n="HᱨἬᨊBCNOFỆᶎỻ᭬∋PS᱄ᳲKᥣ’≴VᷖNoᮞᴍ ̔ẟ⚬ᰃG₿uVIIὤ⓶Y⠔ᷜIV=╺1⁄7ὀᨧᨬ⎪ῆZI⊸ḙᤂᩯ⎠Ṹ∐⍗ᾉᮼ
.ἸẺ⏄−⊼ᲰW⁚⍽2⑀ƫ᳸⍰Ạᬒ⊰ᵴ⌼Ἤἒᤣ∠ṐU∠9.ᮭᲇᮖ᪲ỿ᷎"
m="!$')+,.03478;<?@CHGHMPSTWX[[`afikoptuxy{} £\©«¬°±¶ ̧1⁄2¿ÃÅÇÉÍÏÒÕØÚÞàãåéìïññòþÿĂăĈćĎčĔēėėěĜġ"
i=lambda w:f.reduce(lambda l,c:l*ord(c),list(w),1)
print(str(ord(m[n.find(chr(i(input())))])-32))
Explanation: So 2 variables, n and m are used to hold the abbreviations (more described below) and the elements atomic mass as strings. The input is then encrypted into a character to match to some character in n, using the string.find, an index of where the character is can be used to find the respective atomic mass in m. Due to shifting the mass values by 32 previously, they must be unshifted, then this number is output.
Each character of n is the ascii of the product of the ascii values of the characters in the name
Each character of m is the ascii of mass + 32
Below is the setup code for m and n
obfNames=""
for name in names:
prod=1
for c in name:
prod*=ord(c)
obfNames+=chr(prod)
obfMass=""
for m in mass:
obfMass+=chr(int(m)+32)
for c in obfNames:
print(c,end="")
print()
for c in obfMass:
print(c,end="")
Python 3 , 560 bytes
import functools as f
n="HᱨἬᨊBCNOFỆᶎỻ᭬∋PS᱄ᳲKᥣ’≴VᷖNoᮞᴍ ̔ẟ⚬ᰃG₿uVIIὤ⓶Y⠔ᷜIV=╺1⁄7ὀᨧᨬ⎪ῆZI⊸ḙᤂᩯ⎠Ṹ∐⍗ᾉᮼ
.ἸẺ⏄−⊼ᲰW⁚⍽2⑀ƫ᳸⍰Ạᬒ⊰ᵴ⌼Ἤἒᤣ∠ṐU∠9.ᮭᲇᮖ᪲ỿ᷎"
m="!$')+,.03478;<?@CHGHMPSTWX[[`afikoptuxy{}€‚…‡ŠŒ“—š Ÿ£\©«¬°±¶ ̧1⁄2¿ÃÅÇÉÍÏÒÕØÚÞàãåéìïññòþÿĂăĈćĎčĔēėėěĜġ"
i=lambda w:f.reduce(lambda l,c:l*ord(c),list(w),1)
print(str(ord(m[n.find(chr(i(input())))])-32))
Explanation: So 2 variables, n and m are used to hold the abbreviations (more described below) and the elements atomic mass as strings. The input is then encrypted into a character to match to some character in n, using the string.find, an index of where the character is can be used to find the respective atomic mass in m. Due to shifting the mass values by 32 previously, they must be unshifted, then this number is output.
Each character of n is the ascii of the product of the ascii values of the characters in the name
Each character of m is the ascii of mass + 32
Below is the setup code for m and n
obfNames=""
for name in names:
prod=1
for c in name:
prod*=ord(c)
obfNames+=chr(prod)
obfMass=""
for m in mass:
obfMass+=chr(int(m)+32)
for c in obfNames:
print(c,end="")
print()
for c in obfMass:
print(c,end="")
Python 3, 566 bytes
n="HᱨἬᨊBCNOFỆᶎỻ᭬∋PS᱄ᳲKᥣ’≴VᷖNoᮞᴍ ̔ẟ⚬ᰃG₿uVIIὤ⓶Y⠔ᷜIV=╺1⁄7ὀᨧᨬ⎪ῆZI⊸ḙᤂᩯ⎠Ṹ∐⍗ᾉᮼ
.ἸẺ⏄−⊼ᲰW⁚⍽2⑀ƫ᳸⍰Ạᬒ⊰ᵴ⌼Ἤἒᤣ∠ṐU∠9.ᮭᲇᮖ᪲ỿ᷎"
m="!$')+,.03478;<?@CHGHMPSTWX[[`afikoptuxy{} £\©«¬°±¶ ̧1⁄2¿ÃÅÇÉÍÏÒÕØÚÞàãåéìïññòþÿĂăĈćĎčĔēėėěĜġ"
i=lambda w:f.reduce(lambda l,c:l*ord(c),list(w),1)
print(str(ord(m[n.find(chr(i(input())))])-32))
Explanation: So 2 variables, n and m are used to hold the abbreviations (more described below) and the elements atomic mass as strings. The input is then encrypted into a character to match to some character in n, using the string.find, an index of where the character is can be used to find the respective atomic mass in m. Due to shifting the mass values by 32 previously, they must be unshifted, then this number is output.
Each character of n is the ascii of the product of the ascii values of the characters in the name
Each character of m is the ascii of mass + 32
Below is the setup code for m and n
obfNames=""
for name in names:
prod=1
for c in name:
prod*=ord(c)
obfNames+=chr(prod)
obfMass=""
for m in mass:
obfMass+=chr(int(m)+32)
for c in obfNames:
print(c,end="")
print()
for c in obfMass:
print(c,end="")