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Chapter 3: Problem 86

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

Expert verified
The empirical formula of urea is NH_2CO.

Step by step solution

01

Determine the moles of each element in the sample

First, we need to calculate the number of moles of each element present in the sample. We can do this by dividing the mass of the element by its molar mass. The molar mass of N = 14.01 g/mol, H = 1.01 g/mol, C = 12.01 g/mol, and O = 16.00 g/mol. - For nitrogen (N): \(\frac{1.121 g}{14.01 g/mol} = 0.0800\) moles - For hydrogen (H): \(\frac{0.161 g}{1.01 g/mol} = 0.159\) moles - For carbon (C): \(\frac{0.480 g}{12.01 g/mol} = 0.0400\) moles - For oxygen (O): \(\frac{0.640 g}{16.00 g/mol} = 0.0400\) moles
02

Find the smallest mole value amongst the elements

Now, we need to find the smallest mole value amongst the elements: The smallest mole value is 0.0400 (C and O).
03

Determine the mole ratio of each element

Now, we need to find the ratio between the actual moles of each element and the smallest mole value. This will give us the mole ratio for each element: - For nitrogen (N): \(\frac{0.0800}{0.0400} = 2\) - For hydrogen (H): \(\frac{0.159}{0.0400} \approx 4\) - For carbon (C): \(\frac{0.0400}{0.0400} = 1\) - For oxygen (O): \(\frac{0.0400}{0.0400} = 1\)
04

Write the empirical formula

Finally, write the empirical formula using the mole ratios as subscripts: The empirical formula is NH_2CO.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moles Calculation
Understanding moles is key in chemistry. It's a way to count atoms and molecules, much like using a dozen to count eggs. To calculate moles, divide the mass of an element by its molar mass. The molar mass is the mass of one mole of an element, specific to each element. For example, nitrogen has a molar mass of 14.01 g/mol. In the given exercise, we have:
  • Nitrogen: Dividing 1.121 g by 14.01 g/mol gives 0.0800 moles.
  • Hydrogen: Dividing 0.161 g by 1.01 g/mol gives 0.159 moles.
  • Carbon: Dividing 0.480 g by 12.01 g/mol gives 0.0400 moles.
  • Oxygen: Dividing 0.640 g by 16.00 g/mol gives 0.0400 moles.
These calculations show how each gram divides into moles, providing a foundation for finding elemental ratios.
Molar Mass
Molar mass serves as a bridge between grams and moles. It's defined as the mass of one mole of a substance, expressed in grams per mole (g/mol). This property is crucial for converting a substance's mass into a measurable amount of atoms or molecules, known as moles. For instance, in the exercise:
  • Nitrogen (N) has a molar mass of 14.01 g/mol.
  • Hydrogen (H) is lighter with a molar mass of 1.01 g/mol.
  • Carbon (C) weighs 12.01 g/mol.
  • Oxygen (O) has a molar mass of 16.00 g/mol.
In chemistry, using molar mass simplifies reactions and helps achieve more accurate calculations.
Elemental Composition
Analyzing the elemental composition of a compound reveals the proportion of each element in a sample. This is vital to discovering the empirical formula, which represents the simplest whole number ratio of atoms in a compound. The exercise demonstrated calculating moles, a fundamental step in determining elemental composition. By comparing the moles of different elements, the smallest value (in this case, 0.0400 for carbon and oxygen) becomes a reference point. Finding the mole ratio involves dividing each element's moles by the smallest mole value discovered among them:
  • Nitrogen: 0.0800 moles divided by 0.0400, resulting in a mole ratio of 2.
  • Hydrogen: 0.159 moles divided by 0.0400, resulting in an approximate mole ratio of 4.
  • Carbon: 0.0400 moles yields a ratio of 1.
  • Oxygen: 0.0400 moles also yields a ratio of 1.
This ratio helps formulate the compound's empirical formula, NH_2CO, illustrating the composition in its simplest form.

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Most popular questions from this chapter

Adipic acid is an organic compound composed of \(49.31 \% \mathrm{C}\), \(43.79 \% \mathrm{O}\), and the rest hydrogen. If the molar mass of adipic acid is \(146.1 \mathrm{~g} / \mathrm{mol}\), what are the empirical and molecular formulas for adipic acid?

An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$ \mathrm{M}+\mathrm{X}_{2} \longrightarrow \mathrm{MX}_{3} $$ A \(0.105-\mathrm{g}\) sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of \(54.47 \% \mathrm{X}\) by mass. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\), and what is the correct name for \(\mathrm{MX}_{3}\) ? Starting with \(1.00 \mathrm{~g}\) each of \(\mathrm{M}\) and \(\mathrm{X}_{2}\), what mass of \(\mathrm{MX}_{3}\) can be prepared?

In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen \((77 \mathrm{~K})\) was discovered. The approximate formula of this substance is \(\mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}\). Calculate the percent composition by mass of this material.

Cumene is a compound containing only carbon and hydrogen that is used in the production of acetone and phenol in the chemical industry. Combustion of \(47.6 \mathrm{mg}\) cumene produces some \(\mathrm{CO}_{2}\) and \(42.8 \mathrm{mg}\) water. The molar mass of cumene is between 115 and \(125 \mathrm{~g} / \mathrm{mol}\). Determine the empirical and molecular formulas.

The empirical formula of styrene is \(\mathrm{CH} ;\) the molar mass of styrene is \(104.14 \mathrm{~g} / \mathrm{mol}\). What number of \(\mathrm{H}\) atoms are present in a \(2.00-g\) sample of styrene?
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