A comparison analysis subjects a suspect specimen and a control specimen to the same tests and examinations for the ultimate purpose of determining: whether or not they have a common origin if they are identical in chemical composition if the same person handled them if they are alike in molecular structure All of these

whether or not they have a common origin

Explanation:

During the scientific investigation of a crime, a forensic scientist might be required to perform a comparison analysis. A comparison analysis is an examination where specific physical properties are obtained from a suspect specimen and a standard which are then compared to identify their common origin. This type of analysis is important for it helps in narrowing down the crime to a particular person. It also enables accurate delivery of justice.

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.