Chemical Mass and Empirical Formulas
Molecular Mass (MM)
Definition: The molecular mass (MM) is the sum of all atomic masses of all atoms in the molecule.
For ionic compounds, the term formula mass (FM) is used instead of molecular mass.
Reason: Ionic compounds consist of ions, not molecules.
Thus, the formula mass is calculated as:
FM = ext{Sum of all atomic masses of all atoms in a formula unit}
In calculations, molecular mass (MM) and formula mass (FM) will be used interchangeably.
Unit: The unit of measure for MM and FM is atomic mass units (amu) or Dalton (Da).
Precision in Measurements:
Generally, 3 significant figures (SF) are considered sufficient for most purposes.
In contexts requiring higher precision, 2 decimal points may be used for larger measurements of MM.
Mole and Molar Mass
Definition of Mole:
A mole is defined as the quantity of substance that contains as many molecules or formula units as there are atoms in 12 grams of carbon-12 (C-12).
This quantity is known as Avogadro’s number (NA):
N_A = 6.0221367 \times 10^{23}
Therefore, one mole (mol) is equivalent to having $N_A$ of molecules (or formula units).
Definition of Molar Mass:
Molar mass is the mass of one mole of a substance.
The numerical value of molar mass is equal to the molecular mass (MM).
Unit of molar mass: grams per mole (g/mol).
Determining Chemical Formulas
Calculation of Percentage Composition:
The % composition is the mass percentage of each element in a compound.
Methods to determine % composition include:
Combustion analysis (e.g., CHN or CHNS analysis)
Gravimetric analysis
Optical atomic spectroscopy (both emission and absorption)
Neutron activation analysis
Formula for Mass Percentage:
The mass percentage of element A in a compound can be calculated using:
ext{Mass \% of element A} = \frac{\text{mass of element A in compound}}{\text{mass of compound}} \times 100\%
Combustion Analysis for Empirical Formula Determination
Procedure for Combustion Analysis:
An unknown compound is burned in oxygen to isolate and weigh the products (water and carbon dioxide).
C (carbon) and H (hydrogen) are the easiest elements to determine, as they produce CO₂ and H₂O, respectively.
Diagram Overview:
The unknown compound is combusted in a furnace, generating CO₂ and H₂O.
CO₂ is absorbed by a CO₂ absorber, and H₂O is captured by an H₂O absorber.
Key Steps in Determining Empirical Formula:
From the given % composition of all elements:
Assume a 100 g sample and convert percentages into corresponding masses.
Convert these masses into moles of each element.
Formulate a “pseudoformula” based on mole ratios.
If the subscripts are fractions after calculations:
Divide all subscripts by the smallest subscript.
If resulting subscripts remain fractional, multiply all by a small whole number to convert into whole numbers.
For specific fractional scenarios:
If the subscripts are 0.2, 0.4, or 0.8, multiply by 5.
If the subscripts are 0.25 or 0.75, multiply by 4.
If the subscripts are 0.33 or 0.67, multiply by 3.
If the subscripts are 0.50, multiply by 2.
Example Problem 1:
Given combustion data: 60.00% C, 4.48% H, 35.52% O, identify the empirical formula.
Further Examples of Determining Empirical Formula from Combustion Data
From Masses of Combustion Products:
Convert the masses of CO₂ and H₂O produced into moles.
Calculate moles of C and H based on stoichiometric relationships to CO₂ and H₂O:
For CO₂: 1 mole of CO₂ contains 1 mole of C.
For H₂O: 1 mole of H₂O contains 2 moles of H.
Compile a “pseudoformula” based on calculated moles.
Follow the previous steps for reducing fractions to smallest whole numbers.
Example Problem 2:
Find the empirical formula for a compound containing only C and H, if a sample produces 1.83 g CO₂ and 0.901 g H₂O upon combustion.
Example Problem 3:
Find the empirical formula for a compound containing C, H, and O if a sample weighing 0.8233 g produces 2.445 g CO₂ and 0.6003 g H₂O upon combustion.