Suppose that a fictitious element, x, has two isotopes: 59x (59.015 amu) and 62x (62.011 amu). the lighter isotope has an abundance of 81.7%. calculate the atomic mass of the element x.

Atomic Mass of Element = 59.57 amu

Given Data:

Element X Isotope 1 atomic weight = 59.015 amu  

Element X Isotope 2 atomic weight = 62.011 amu

The abundance of lighter isotope = 81.7%

Atomic Mass of Element X = ?

Solution:

81.7% is the abundance of element that is lighter so it is the of the isotope having weight 59.015 amu

Then the remaining abundance = 100 -81.7 = 18.3 %

18.3 % abundance of the isotope weight = 62.011 amu

Formula:

Mass of the element =  % of abundance (atomic mass of Isotope 1) + % of abundance (atomic mass of Isotope 2)

Put the values

Atomic mass of Element X = 81.7/100 (59.015) + 18.3 /100 (62.011)

Atomic mass of Element = 0.817 (59.015) + 0.183 (62.011)

Atomic mass of Element = 48.22 + 11.35

Atomic Mass of Element = 59.57 amu

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g