1) Proton, neutron and electron are sub-atomic particles in an atom.What is the relative atomic mass of these sub-atomic particles? Giving that the mass of a proton, neutron and electron is 1.67 x 10^-24g, 1.67 x 10^-24g and 9.11 x 10^-28g respectively.( Hint. By referring to the mass given, the mass of proton and neutron is almost similar. However, it is found that the proton is heavier than electron. How heavy proton is when comparing with electron?)
(r.a.m=relative atomic mass)
formula : mass of atom = r.a.m / avogadro constant
Ans: r.a.m of proton = 1.67 x 10^-24(6.02 x 10^23)
= 1.0053
r.a.m of neutron = 1.67 x 10^-24(6.02 x 10^23)
= 1.0053
r.a.m of electron = 9.11 x 10^-28(6.02 x 10^23)
= 5.48 x 10^-4
proton - electron = 1.0053-5.48 x 10^-4
= 1.0048 2) Based on the following isotope notations, state the number of protons, neutrons and electrons for each isotope.
244 56 78
Pu Fe3+ Se2-
94 26 34
Ans: Pu, no.of protons: 94
no.of neutrons: 244-94 = 150
no.of electrons: 94
Fe3+, no.of protons: 26
no.of neutrons: 56-26 = 30
no.of electron: 23
Se2-, no.of proton: 34
no.of neutron: 78-34 = 44
no.of electron: 36
3) An isotope M consists of 33 protons and 42 neutrons. Write its notation.
75
M
33
4) Write the isotope notation for krypton-78. Given that the proton number of K is 36.
78
Kr
36
5) What is the average atomic of element X if x consists of 78.7% of atoms with a mass of 24.0 a.m.u,10.1% of atoms with mass of 25.0 a.m.u and 11.2% of atoms of 26.0 a.m.u?
Formula: average atomic mass = total percentage of abundance x total relative atomic mass / total percentage abundance
Ans: average atomic mass = 78.7(24.0)+10.1(25.0)+11.2(26.0) / 100
= 24.325 a.m.u
6) Based on the above mass spectrum of neon,calculate the average atomic mass of neon.
Ans: average atomic mass = (90.5 x 20)+(0.3 x 21)+(9.2 x 22) / 100
= 20.19
7) Based on the above mass spectrum of chlorine, calculate the
(a) Percentage composition for each isotope of chlorine.
(b) Average atomic mass of chlorine.
Ans: (a) percentage composition = 3(35)+1(37) / 3+1
= 142 / 4
= 35.5
35
Cl = 3 / 4 x 100
= 75%
37
Cl = 1 / 4 x 100
= 25%
(b) average atomic mass of chlorine = 3(35)+1(37) / 3+1
= 142 / 4
= 35.5
8) Gallium consists of two naturally occurring isotopes, Ga-69 and Ga-71. The atomic mass of Ga-69 and Ga-71 is 68.9256 a.m.u and 70.9247 a.m.u respectively. If the average atomic mass of Ga is 69.723 a.m.u, what is the abundance for each isotopes of Ga?
Ans: assume that relative abundance of Ga-69 = Y, Ga-71 = 100-Y
69.723 = (68.9256 a.m.u x Y)+(70.9247 a.m.u x 100-Y) / Y+100-Y
69.723 = 68.9256Y+7092.47-70.9247Y / 100
6972.3 = -1.9991Y+7092.47
-120.17 = -1991Y
Y = 0.06
Ga-69 = 0.06%
Ga-70 = 100-0.06
= 99.94%
9) The naturally occurring Boron has only two isotopes, B-10 and B-11. If B-10 has an atomic mass of 10.0129 a.m.u and abundance of 19.9%, what is the atomic mass of B-11? Given the average atomic mass of Boron is 10.811 a.m.u.
Ans: 10.0129 a.m.u = 19.9%
10.811 = (10.0129 x 19.9)+(80.1 x Y) / 100
= 199.2567+(80.1 x Y) / 100
1081.1-199.2567 = 80.1Y
Y = 881.8433 / 80.1
= 11.0092
69.723 = 68.9256Y+7092.47-70.9247Y / 100
6972.3 = -1.9991Y+7092.47
-120.17 = -1991Y
Y = 0.06
Ga-69 = 0.06%
Ga-70 = 100-0.06
= 99.94%
9) The naturally occurring Boron has only two isotopes, B-10 and B-11. If B-10 has an atomic mass of 10.0129 a.m.u and abundance of 19.9%, what is the atomic mass of B-11? Given the average atomic mass of Boron is 10.811 a.m.u.
Ans: 10.0129 a.m.u = 19.9%
10.811 = (10.0129 x 19.9)+(80.1 x Y) / 100
= 199.2567+(80.1 x Y) / 100
1081.1-199.2567 = 80.1Y
Y = 881.8433 / 80.1
= 11.0092
