# 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:
• CXHY + 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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