Enduring Understanding 1.A.3: Chemical Analysis

Because atoms and molecules interact with each other on the atomic level, and because they cannot be created or destroyed, chemical processes can be described as balanced chemical equations, in which the number and type of atoms on each side of the equation must be the same.

Example: The combustion of ethanol, C₂H₅OH, with oxygen:

C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O

In the above equation, note that the total number of carbon, hydrogen and oxygen atoms to the left of the arrow is equal to the total number to the right.

Chemical Analysis is a technique that can provide information about the mass percent, and therefore help determine the ratio of atoms of different elements in a compound, and/or determine the purity of a sample.

To calculate the molar ratio of a compound from given masses, the following equations are useful:

# of moles in sample = mass of element in sample / atomic mass of element

and
molar ratio of A:B = # of moles of A in sample : # of moles of B in sample

Example 1: A sample of a compound of nitrogen and oxygen is found by analysis to contain 7 g nitrogen and 20 g oxygen. To determine its empirical formula, divide by the atomic masses:

N: 7 g/14.001 g/mol = 0.5 mol N

O: 20 g/15.999 g/mol = 1.25 mol O

The mole ratio of N to O is 0.5:1.25, which in whole numbers is 2:5 (remember, you can never have fractions of atoms in a molecule!)

The empirical formula of the molecule is therefore N₂O₅.

A more advanced chemical analysis problem involves using a balanced chemical equation as well as masses.

Example 2: Find the empirical formula of a 4.41 g sample of hydrocarbon that produces 13.2 g CO₂ and 7.21 g H₂O on combustion in O₂.

This is a combustion, each C produces one CO₂ and each H produces 1/2 H₂O, so the balanced reaction must be:

C_XH_Y + ZO₂ → X CO2 + (Y/2) H₂O

Moles of CO₂ = 13.2 g / 44.01 g/mol = 0.300 mol CO₂

Moles of H₂O = 7.21 g / 18.02 g/mol = 0.400 mol H₂O

From the balanced equation, the number of moles of C originally present must be 0.300 mol, and the number of moles of H originally present must be (0.400 x 2) or 0.800, so the ratio of C:H is 0.3:0.8

In whole numbers, the ratio of C:H must be 3:8, so the empirical formula of the hydrocarbon must be C₃H₈.

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Enduring Understanding 1.A.3: Chemical Analysis

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