Problem 139 - Chemistry An Atoms First Approach

Question

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains \(2.59 \times 10^{23}\) atoms of hydrogen and is \(17.3 \%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and \(65 \mathrm{g} / \mathrm{mol}\), what amount (moles) of compound is present, and what is the mass of the sample?

Step-by-Step Solution

Answer

The amount of the hydrocarbon present is approximately 2.485 moles, and the mass of the sample is approximately 179.26 g. The most likely molecular formula of the hydrocarbon is C5H12.

Step 1: Calculate the moles of hydrogen

First, we will calculate the moles of hydrogen atoms. We can do this by dividing the number of hydrogen atoms by Avogadro's number (approximately \(6.022 \times 10^{23}\) atoms/mol).\ Moles of hydrogen = \(\frac{2.59 \times 10^{23}}{6.022 \times 10^{23}} = 0.43 \textrm{ mol}\)

Step 2: Calculate the moles of the hydrocarbon

Next, we will calculate the moles of the hydrocarbon. Since the sample is 17.3% hydrogen by mass, the moles of hydrogen to the moles of hydrocarbon can be represented as:\ \(0.173 \times n_{hydrocarbon} = 0.43mol\) \(\Rightarrow n_{hydrocarbon} = \frac{0.43}{0.173} = 2.485 \textrm{ mol}\)

Step 3: Find the empirical formula of the hydrocarbon

Since the empirical formula of a hydrocarbon is CxHy, where x and y are integers representing the number of carbon and hydrogen atoms, respectively,\ Carbon moles = Hydrocarbon moles - Hydrogen moles\ Carbon moles = \(2.485 - 0.43 = 2.055\) Since the carbon to hydrogen ratio must be integers, we can assume x and y are integers that satisfy the whole number ratio of carbon and hydrogen moles. The simplest possible empirical formula is C5H12 (for a ratio of 2.055:0.43, which is close to 5:1).

Step 4: Calculate the molar mass of the empirical formula and determine the most likely molecular formula

The molar mass of the empirical formula C5H12 is,\ Molar mass = 5(12.01) + 12(1.01) = 72.15 g/mol Given that the molar mass of the hydrocarbon is between 55 and 65 g/mol, the empirical formula of C5H12 is the most likely molecular formula of our hydrocarbon since its molar mass (approximately 72.15 g/mol) is closest to the given range.

Step 5: Calculate the mass of the sample

Now that we have found the moles of the hydrocarbon, and we have the most likely molecular formula (C5H12), we can calculate the mass of the sample.\ Mass of sample = Moles of hydrocarbon × Molar mass of hydrocarbon\ Mass of sample = 2.485 mol × 72.15 g/mol = 179.26 g Therefore, the amount of the hydrocarbon present is approximately 2.485 moles, and the mass of the sample is approximately 179.26 g.

Key Concepts

Mole Concept

Understanding the mole concept is foundational in chemistry. It serves as the bridge between the microscopic world of atoms and the macroscopic world we experience daily. Simply put, a mole represents a massive collection of particles, usually atoms or molecules, just like a dozen represents a collection of 12 items. The importance of the mole concept lies in its ability to quantify substances in chemical reactions.

The number of particles in one mole is defined by Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles per mole. This number is not arbitrary but is based on the number of atoms in exactly 12 grams of carbon-12, which is a standard reference. When we refer to one mole of a substance, we mean it contains Avogadro's number of particles, be it atoms in an element or molecules in a compound.

In the given problem, we used the mole concept to convert the number of hydrogen atoms into moles by dividing by Avogadro's number. This step is crucial in stoichiometry, allowing us to relate the microscopic particles to measurable, macroscopic quantities like mass.

Molar Mass

Molar mass, expressed in grams per mole (g/mol), is another key concept closely linked to the mole. It represents the weight of one mole of a chemical element or compound. For instance, the molar mass of water (H2O) is approximately 18 g/mol because one mole of water, which contains Avogadro's number of water molecules, weighs about 18 grams.

The molar mass of a compound like our example hydrocarbon (CxHy) is calculated by adding together the molar masses of its individual elements, taking into account their respective quantities in the compound. Carbon has a molar mass of approximately 12.01 g/mol, and hydrogen has a molar mass of approximately 1.01 g/mol. Therefore, to find the molar mass of a hydrocarbon, we multiply the number of each type of atom by its atomic molar mass and sum the results.

In our exercise, we used molar mass to determine the molecular formula of the compound and to calculate the mass of the hydrocarbon sample.

Avogadro's Number

Avogadro's number is immensely significant in chemistry and is crucial to the mole concept. Not to be confused with molar mass, Avogadro's number is a constant representing the number of particles found in one mole of any substance. Its approximation, \(6.022 \times 10^{23}\), allows us to handle chemical equations and perform calculations involving atoms and molecules on a macroscopic scale.

In practice, when we have a particular number of particles, we can determine the corresponding number of moles by dividing the number of particles by Avogadro's number. This was demonstrated in the problem's step-by-step solution where we calculated the moles of hydrogen atoms in the hydrocarbon sample.

Percent Composition

The percent composition of a compound is a reflection of the relative amounts of each element in that compound, presented as a percentage of the compound's total mass. For a hydrocarbon, the percent composition tells us the fractions of carbon and hydrogen by mass.

By knowing the percent composition, we can deduce the empirical formula of the substance, which is the simplest whole number ratio of elements in a compound. In our exercise, the hydrocarbon was 17.3% hydrogen by mass. Using this information, we confirmed the mole relationship between hydrogen and the overall hydrocarbon, which ultimately aided in determining the compound's empirical formula and the mass of the sample.

It is essential to understand the percent composition as it sets the foundation to calculate empirical and molecular formulas, which describe the proportions of each element in a compound and are used in stoichiometry for various calculations in chemistry.