What can you learn about a compound from its chemical formula

From the chemical formula of a compound you can learn about its composition:

Also, with the ratios and the atomic masses of each kind of atom, you can obtain the mass composition (%) of the compound.

Explanation:

Using indexes (subscripts), the chemical formula of a compound shows  the elements (kind of atoms) in the compound and the relative proportions of those elements.

The following example will show you how you learn that information from the chemical formula.

A very familiar compound is water and its chemical formula (also very familiar) is H₂O.

From that, you learn that water (a chemical compound):

        * hydrogen, H: 2 × 1.008 g/ mol = 2.016 g/,ol

        * oxygen, O: 1 × 15.999 g/mol = 15.999 g/mol

        * Molar mass of water: 2.016 g/mol + 15.999 g/mol = 18.015 g/ mol

        * H: (2.016 g/mol / 18.015 g/mol) × 100 = 11.19%

        * O: (15.999 g/mol / 18.015 g/mol) × 100 = 88.81%

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years