A solution is prepared by dissolving \(5.00 \mathrm{~g}\) of caffeine in \(100.0 \mathrm{~g}\) of carbon tetrachloride. The solution is cooled and the temperature plotted over time: (a) What is the molar mass of caffeine? (b) Combustion analysis reveals that the empirical formula of caffeine is \(\mathrm{C}_{4} \mathrm{H}_{5} \mathrm{~N}_{2} \mathrm{O}\). What is the molecular formula?

Understanding how to calculate the molar mass of an empirical formula is crucial in chemistry. It represents the mass of one mole of a substance, which is equivalent to Avogadro's number of molecules or atoms of that substance.
To calculate the molar mass of the empirical formula, one must sum the molar masses of each element within the formula, considering their respective quantities in the formula. For example, let's look at caffeine's empirical formula, C4H5N2O, where the atomic masses for carbon (C), hydrogen (H), nitrogen (N), and oxygen (O) are approximately 12.01, 1.01, 14.01, and 16.00 g/mol, respectively.

• For carbon: 4 atoms * 12.01 g/mol = 48.04 g/mol
• For hydrogen: 5 atoms * 1.01 g/mol = 5.05 g/mol
• For nitrogen: 2 atoms * 14.01 g/mol = 28.02 g/mol
• For oxygen: 1 atom * 16.00 g/mol = 16.00 g/mol

Adding these together gives us the molar mass of the empirical formula of caffeine, which is 97.11 g/mol. It's important to note that these calculations rely on precise atomic mass values which are critical for accurate results.

When it comes to identifying the empirical formula of a compound, combustion analysis is a revered laboratory method. It involves burning a known mass of the compound in an excess of oxygen to produce carbon dioxide and water. We can then measure the masses of these products to back-calculate the amount of carbon, hydrogen, and other elements in the original sample.
In this case, combustion analysis revealed that caffeine has the empirical formula C4H5N2O. To reach this conclusion, the amounts of CO2 and H2O produced during combustion are converted to the quantities of C and H in the original compound and simplified to find the simplest whole number ratio of elements. Nitrogen can also be determined by analyzing the leftover residue or by other quantitative methods, while the presence of oxygen is accounted for by difference – subtracting the sum of the masses of carbon, hydrogen, and nitrogen from the total mass of the compound.
It's pertinent for students to grasp that combustion analysis results in an empirical formula, which may not reflect the actual number of atoms in a molecule but gives the proportion of atoms to one another in its simplest form.

After establishing the empirical formula of a compound, we may step further to its molecular formula, which shows the actual number of atoms of each element in a molecule of the compound. It can be derived by comparing the molar mass of the empirical formula with the molar mass of the compound itself. Here’s how:
Find the ratio of the molar mass of the compound to the molar mass of the empirical formula. This ratio will be a simple whole number, which we will denote as 'x'. The molecular formula is then the empirical formula multiplied by this ratio, represented algebraically as (empirical formula)x.
For instance, if the molar mass of a compound was 194.22 g/mol, and you have already calculated its empirical formula’s molar mass as 97.11 g/mol, 'x' would be 2 because 194.22 / 97.11 = 2. Therefore, the molecular formula would be simply the empirical formula (C4H5N2O) doubled, resulting in C8H10N4O2.
It's important for students to realize that without an accurate molar mass of the compound itself, 'x' cannot be determined, and hence, the exact molecular formula remains unknown. Compounds with the same empirical formula can have different molecular formulas; hence, additional information through experimental data, such as the molar mass found from other measurements, is crucial for this determination.