Question

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O} .\) What is the empirical formula of urea?

Short answer

The empirical formula of urea is \(\mathrm{CH_4N_2O}\).

Step-by-step solution

Determine the moles of each element in the sample

First, we must convert the grams of each element to moles, using their molar masses: - Nitrogen (N): \(1.121 \mathrm{~g \space N} \times \frac{1 \mathrm{~mol \space N}}{14.01 \mathrm{~g \space N}} = 0.0800 \mathrm{~mol \space N}\) - Hydrogen (H): \(0.161 \mathrm{~g \space H} \times \frac{1 \mathrm{~mol \space H}}{1.008 \mathrm{~g \space H}} = 0.160 \mathrm{~mol \space H}\) - Carbon (C): \(0.480 \mathrm{~g \space C} \times \frac{1 \mathrm{~mol \space C}}{12.01 \mathrm{~g \space C}} = 0.0400 \mathrm{~mol \space C}\) - Oxygen (O): \(0.640 \mathrm{~g \space O} \times \frac{1 \mathrm{~mol \space O}}{16.00 \mathrm{~g \space O}} = 0.0400 \mathrm{~mol \space O}\)

Find the mole ratio of each element in the sample

To find the mole ratio, divide the moles of each element by the smallest number of moles: - Nitrogen (N): \(\frac{0.0800 \mathrm{~mol \space N}}{0.0400} = 2\) - Hydrogen (H): \(\frac{0.160 \mathrm{~mol \space H}}{0.0400} = 4\) - Carbon (C): \(\frac{0.0400 \mathrm{~mol \space C}}{0.0400} = 1\) - Oxygen (O): \(\frac{0.0400 \mathrm{~mol \space O}}{0.0400} = 1\)

Determine the empirical formula using the mole ratio

The empirical formula is obtained by taking the whole number ratio of the moles of each element and using them as subscripts: - Nitrogen (N): 2 - Hydrogen (H): 4 - Carbon (C): 1 - Oxygen (O): 1 The empirical formula of urea is therefore \(\mathrm{CH_4N_2O}\).

Key concepts

Molar Mass Calculation

The molar mass calculation is fundamental in chemistry, especially when dealing with the composition of substances. It represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). In simpler terms, it tells us the weight of one mole, or 6.022 x 10²³ particles, of a given substance.

For instance, the molar mass of hydrogen (H) is 1.008 g/mol, which means that one mole of hydrogen atoms weighs 1.008 grams. Calculating the molar mass of a compound involves summing up the molar mass of each element in the compound according to its subscript in the molecular formula.

For example, water (H₂O) has a molar mass of approximately 18.02 g/mol, calculated from (2 x 1.008) + 16.00. Understanding molar mass is vital to convert grams to moles, which is a step often used in stoichiometry.

Mole Ratio

The mole ratio bridges the gap between the microscopic world of atoms and molecules and the macroscopic world we can measure. The mole ratio is used to compare the number of moles of each element in a compound. It's calculated by dividing the number of moles of each component by the smallest number of moles present in the mixture.

This comparison allows us to determine the simplest whole number ratio of the elements, which is crucial for writing empirical formulas. The empirical formula represents the simplest proportion of the elements in a compound. For example, in the given urea sample, the mole ratio helped to simplify the ratio of nitrogen, hydrogen, carbon, and oxygen into the whole number ratio 2:4:1:1, which directly translates to the empirical formula CH₄N₂O.

Stoichiometry

Stoichiometry is like the recipe for chemistry. It involves the calculation of reactants and products in chemical reactions. This branch of chemistry is built on the conservation of mass and the concept of moles, which allows chemists to predict quantitative outcomes of chemical reactions. Effective stoichiometry should take into account the mole ratios of each reactant and product involved in a reaction.

For example, balancing chemical equations is an application of stoichiometry that ensures the number of atoms for each element is the same on both sides of the reaction. Stoichiometry is not only about balancing equations; it's also about understanding the relationships between different substances in a reaction, which allows for the calculation of unknown quantities given the known quantities. Grasping molar mass calculation and mole ratio is essential for mastering stoichiometry, as these concepts allow you to quantify and relate the substances participating in a chemical reaction.