Answer:
12.34 amu
Explanation:
Let the 1st isotope be A
Let the 2nd isotope be B
Let the 3rd isotope be C
From the question given above, the following data were obtained:
1st Isotope (A):
Mass of A = 12.32 amu
Abundance (A%) = 19.5%
2nd isotope (B):
Mass of B = 13.08 amu
Abundance (B%) = 26.23%
3rd isotope (C):
Mass of C = 11.99 amu
Abundance (C%) = 54.27%
Atomic mass of X =?
The atomic mass of the element X can be obtained as follow:
Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100] + [(Mass of C × C%)/100]
= [(12.32 × 19.5)/100] + [(13.08 × 26.23)/100] + [(11.99 × 54.27)/100]
= 2.402 + 3.431 + 6.507
= 12.34 amu
Thus, the atomic mass of the element X is 12.34 amu
Answer:
C₂H₃Cl₃O₂
Explanation:
Given parameters:
Percentage composition:
14.52% C
1.83% H
64.30% Cl
19.35% O
Molar mass = 165.4 g/mo
Unknown:
Empirical formula = ?
Solution:
The empirical formula of a compound is the simplest formula of a compound. To solve this problem, let us follow the process below:
C H Cl O
%composition 14.52 1.83 64.3 19.35
Molar mass 12 1 17 16
number of
moles 14.52/12 1.83/1 64.3/3.5 19.35/16
1.21 1.83 1.81 1.21
Divide by
the smallest 1.21/1.21 1.83/1.21 1.81/1.21 1.21/1.21
1 1.51 1.51 1
multiply
by 2 2 3 3 2
Empirical formula C₂H₃Cl₃O₂
I think the answer is Density
