Problem 29
Question
Pheromones are a special type of compound secreted by the females of many insect species to attract the males for mating. One pheromone has the molecular formula \(\mathrm{C}_{19} \mathrm{H}_{38} \mathrm{O}\). Normally, the amount of this pheromone secreted by a female insect is about \(1.0 \times 10^{-12} \mathrm{~g} .\) How many molecules are there in this quantity?
Step-by-Step Solution
Verified Answer
There are approximately \(2.13 \times 10^9\) molecules of this pheromone in \(1.0 \times 10^{-12}\) g.
1Step 1: Determine the molar mass of the compound
First, the molar mass of the compound, C19H38O is calculated. One carbon atom has a molar mass of about 12.01 g/mol, a hydrogen atom has a molar mass of about 1.008 g/mol, and an oxygen atom has a molar mass of about 16.00 g/mol. So, multiplying these molar masses with the respective number of atoms in the compound and summing them up gives: molar mass = (19 x 12.01)+(38 x 1.008)+(16.00) = 282.522 g/mol.
2Step 2: Convert the quantity from grams to moles
Now, there is a formula that can be used to convert the amount of substance from grams to moles as follows: number of moles = mass of substance in g / molar mass of substance in g/mol. Substitute the values into the formula: number of moles = \(1.0 \times 10^{-12} \mathrm{~g}\) / 282.522 g/mol = 3.54 x 10^-15 mol.
3Step 3: Calculate the number of molecules
Finally, to find out the number of molecules, you multiply the number of moles by Avogadro’s number (\(6.022 \times 10^{23} \) molecules/mol). Therefore, the number of molecules = (3.54 x 10^-15 mol) x (\(6.022 \times 10^{23} \) molecules/mol) = \(2.13 \times 10^9\) molecules.
Key Concepts
Molar Mass CalculationMole-to-Gram ConversionAvogadro's Number
Molar Mass Calculation
Understanding the molar mass of a substance is crucial when it comes to chemical calculations. The molar mass is defined as the mass of one mole of a given substance and is expressed in grams per mole (g/mol). It is a collective property for all the atoms present in a molecule.
To calculate the molar mass, we need to know the molecular formula of the substance. In our pheromone example, \( \mathrm{C}_{19} \$H_{38} \mathrm{O}\), we have 19 carbon atoms (C), 38 hydrogen atoms (H), and one oxygen atom (O). The molar mass of each type of atom is a standard value: approximately 12.01 g/mol for carbon, 1.008 g/mol for hydrogen, and 16.00 g/mol for oxygen.
This step is the foundation for all subsequent calculations involving the substance, as molar mass ties together the mass of a sample to the amount of substance in moles.
To calculate the molar mass, we need to know the molecular formula of the substance. In our pheromone example, \( \mathrm{C}_{19} \$H_{38} \mathrm{O}\), we have 19 carbon atoms (C), 38 hydrogen atoms (H), and one oxygen atom (O). The molar mass of each type of atom is a standard value: approximately 12.01 g/mol for carbon, 1.008 g/mol for hydrogen, and 16.00 g/mol for oxygen.
Step-by-Step Calculation:
First, we multiply the molar mass of each atom by the number of atoms of that element present in the molecule. Then, we sum these values to get the total molar mass of the molecule: \[ \text{molar mass} = (19 \times 12.01) + (38 \times 1.008) + (16.00) = 282.522 \text{ g/mol}\].This step is the foundation for all subsequent calculations involving the substance, as molar mass ties together the mass of a sample to the amount of substance in moles.
Mole-to-Gram Conversion
Once the molar mass is known, converting between mass and moles becomes a straightforward task. The conversion uses the formula: \[ \text{number of moles} = \frac{\text{mass of substance in g}}{\text{molar mass of substance in g/mol}}\].
In practical terms, this formula helps us understand the number of moles of a substance present in a given mass. For the pheromone \( \mathrm{C}_{19} \$H_{38} \mathrm{O}\), with a molar mass of 282.522 g/mol, if we have a tiny amount - say, \(1.0 \times 10^{-12} \text{ g}\) - we can find how many moles this mass corresponds to:
Having determined the number of moles in the sample, we can further connect it to the number of individual molecules, illustrating the mass-to-particle relationship in chemistry.
In practical terms, this formula helps us understand the number of moles of a substance present in a given mass. For the pheromone \( \mathrm{C}_{19} \$H_{38} \mathrm{O}\), with a molar mass of 282.522 g/mol, if we have a tiny amount - say, \(1.0 \times 10^{-12} \text{ g}\) - we can find how many moles this mass corresponds to:
Applying the Conversion:
By inserting the values into the formula, we get: \[ \text{number of moles} = \frac{1.0 \times 10^{-12} \text{ g}}{282.522 \text{ g/mol}} = 3.54 \times 10^{-15} \text{ mol}\].Having determined the number of moles in the sample, we can further connect it to the number of individual molecules, illustrating the mass-to-particle relationship in chemistry.
Avogadro's Number
Avogadro's number is a key concept in chemistry, linking moles to actual numbers of particles, be they atoms, molecules, ions, or others. It states that one mole of any substance contains exactly \(6.022 \times 10^{23}\) particles. This allows chemists to work with manageable amounts of substances, even though the number of particles involved in chemical reactions is often very large.
For our pheromone example, we already determined that the sample contains \(3.54 \times 10^{-15} \text{ mol}\). Thus, to find the number of molecules in the sample, we calculate: \[ \text{number of molecules} = (3.54 \times 10^{-15} \text{ mol}) \times (6.022 \times 10^{23} \text{ molecules/mol}) = 2.13 \times 10^{9} \text{ molecules}\].
This impressive number shows how even a minuscule amount of substance contains a vast number of particles, demonstrating the utility of Avogadro's number in bridging the macroscopic and microscopic worlds.
From Moles to Molecules:
To get from moles to the number of molecules, simply multiply the amount of the substance in moles by Avogadro's number: \[ \text{number of molecules} = (\text{number of moles}) \times (6.022 \times 10^{23} \text{ molecules/mol})\].For our pheromone example, we already determined that the sample contains \(3.54 \times 10^{-15} \text{ mol}\). Thus, to find the number of molecules in the sample, we calculate: \[ \text{number of molecules} = (3.54 \times 10^{-15} \text{ mol}) \times (6.022 \times 10^{23} \text{ molecules/mol}) = 2.13 \times 10^{9} \text{ molecules}\].
This impressive number shows how even a minuscule amount of substance contains a vast number of particles, demonstrating the utility of Avogadro's number in bridging the macroscopic and microscopic worlds.
Other exercises in this chapter
Problem 26
How many molecules of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) are present in \(0.334 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\)
View solution Problem 27
Calculate the number of \(\mathrm{C}, \mathrm{H},\) and \(\mathrm{O}\) atoms in \(1.50 \mathrm{~g}\) of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O
View solution Problem 31
Describe the operation of a mass spectrometer.
View solution Problem 32
Describe how you would determine the isotopic abundance of an element from its mass spectrum.
View solution