Problem 8 - Chemical Principles: The Quest for Insight

Question

Urea is used as a commercial fertilizer because of its nitrogen content. An analysis of \(25.0 \mathrm{mg}\) of urea showed that it contained \(5.0 \mathrm{mg} \mathrm{C}, 11.68 \mathrm{mg} \mathrm{N}\), \(6.65 \mathrm{mg} \mathrm{O}\), and the remainder hydrogen. What is the empirical formula of urea?

Step-by-Step Solution

Answer

The empirical formula of urea is CH4N2O.

Step 1: Determine the mass of hydrogen

Subtract the total mass of carbon, nitrogen, and oxygen from the total mass of the urea sample to find the mass of hydrogen. Use the equation: Mass of hydrogen = Total mass of urea - (Mass of carbon + Mass of nitrogen + Mass of oxygen).

Step 2: Calculate moles of each element

Calculate the number of moles of each element by dividing the mass of each by its atomic mass (C: 12.01 g/mol, N: 14.01 g/mol, O: 16.00 g/mol, H: 1.008 g/mol). Use the formula moles = mass (mg) / atomic mass (g/mol).

Step 3: Determine the mole ratio

Find the mole ratio of the elements by dividing each by the smallest number of moles obtained in step 2. This gives the simplest whole number ratio.

Step 4: Write the empirical formula

Use the whole number ratios as subscripts to get the empirical formula. If any subscript is not a whole number, multiply all subscripts by the smallest number to obtain whole numbers.

Key Concepts

Stoichiometry

Stoichiometry is a section of chemistry that involves the calculation of the quantities of reactants and products in chemical reactions. It's directly connected to the conservation of mass where the total mass of the reactants equals the total mass of the products. Understanding stoichiometry is essential for determining the correct proportions of elements that combine to form compounds.

In the context of our urea problem, stoichiometry guides us to determine the exact amount of each element in a given sample. We utilize the masses to ultimately establish the empirical formula, which represents the simplest whole-number ratio of atoms in a compound. It provides a fundamental understanding of how elements combine and react with one another in fixed proportions to form substances like urea.

Mole Concept

The mole concept is a fundamental principle in chemistry that allows chemists to count atoms, molecules, and ions based on their amount of substance rather than their mass. One mole contains exactly 6.022 x 10²³ (Avogadro's number) of particles, be it atoms, ions, or molecules.

For instance, when calculating the moles of carbon in urea, we divide the mass of carbon by its atomic mass unit in g/mol. This conversion from mass to moles lets us compare the number of particles of each element in a compound directly. In our exercise, by converting the mass of C, N, O, and H to moles, we lay the groundwork for finding the mole ratio, which is vital to determine the empirical formula.

Chemical Composition Analysis

Chemical composition analysis involves identifying and quantifying the elements or compounds within a chemical substance. This analytical technique is crucial for determining the purity and formulation of a compound.

In the exercise provided, we conducted a composition analysis by breaking down urea into its elemental components—carbon, nitrogen, oxygen, and hydrogen—and measuring their masses. From this analysis, we can infer how much of each element is present in a given sample of urea. The determination of relative amounts by mass leads to the calculation of moles, which in turn is used to ascertain the empirical formula.

Molecular Formula

The molecular formula of a substance provides information on the actual number of atoms of each element present in a molecule of that substance. It is derived from the empirical formula, which represents the simplest ratio of elements in a compound. The molecular formula can be the same as the empirical formula or a multiple thereof.

In our exercise, after calculating the empirical formula, we may or may not need to determine the molecular formula. If further information, such as the molar mass of the compound, is provided, one can compare it with the mass of the empirical formula to find a ratio. This ratio, if greater than one, is multiplied by the empirical subscripts to determine the molecular formula. However, in the given problem, we are only asked for the empirical formula, which is the stepping stone towards finding the molecular formula should we need it.