Chapter 3: Problem 135
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 \(2.00-\mathrm{g}\) sample of styrene?
Short Answer
Expert verified
In a 2.00-g sample of styrene, approximately \(1.16 \times 10^{22}\) hydrogen atoms are present.
Step by step solution
01
Determine the number of moles of styrene in the sample
To find the number of moles in the sample, we will use the equation:
Number of moles = mass / molar mass
Where the mass is given as 2.00 g, and the molar mass is 104.14 g/mol. Plugging in the values, we get:
Number of moles = \(2.00~\text{g}\) / \(104.14\frac{\text{g}}{\text{mol}}\)
02
Calculate the number of moles of styrene
Dividing 2.00 g by 104.14 g/mol, we have:
Number of moles = 0.0192 mol
So, there are 0.0192 moles of styrene in the 2.00-g sample.
03
Determine the number of hydrogen atoms in one mole of styrene
Since the empirical formula is CH, there is one hydrogen atom for each styrene molecule. Therefore, there will be 1 mole of hydrogen atoms for each mole of styrene molecules.
04
Calculate the number of hydrogen atoms in the sample
We can now use Avogadro's number (6.022 x \(10^{23}\) atoms/mol) to calculate the number of hydrogen atoms in the 0.0192 moles of styrene.
Number of hydrogen atoms = moles of hydrogen x Avogadro's number
Number of hydrogen atoms = \(0.0192\text{ mol} \times 6.022 \times 10^{23}\text{ atoms/mol}\)
05
Determine the number of hydrogen atoms in the sample
Multiplying 0.0192 mol by 6.022 x \(10^{23}\) atoms/mol, we get:
Number of hydrogen atoms = \(1.16 \times 10^{22}\) atoms
So, there are approximately \(1.16 \times 10^{22}\) hydrogen atoms in the 2.00-g sample of styrene.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Molar Mass
When working with chemical substances, it's important to know the molar mass, which helps in determining how much a sample weighs in terms of moles. The molar mass is expressed in grams per mole (g/mol) and represents the mass of one mole of a substance.
For styrene, which has an empirical formula of \( \mathrm{CH} \), the molar mass is given as 104.14 g/mol.
Here's why molar mass is significant:
For styrene, which has an empirical formula of \( \mathrm{CH} \), the molar mass is given as 104.14 g/mol.
Here's why molar mass is significant:
- It allows for the conversion between the mass of a substance and the number of moles, which are two different ways of measuring the same quantity.
- Understanding the molar mass of a compound is crucial for stoichiometric calculations in chemical reactions.
Moles of Substance
Moles are a fundamental concept in chemistry, serving as a bridge between the atomic scale and the macroscopic scale that we can measure directly. A mole of any substance always contains \(6.022 \times 10^{23}\) entities (atoms, molecules, electrons, etc.), thanks to Avogadro's number.
For example, with a 2.00 g sample of styrene, using its molar mass (104.14 g/mol), you can calculate the number of moles: \[\text{Number of moles} = \frac{2.00 \text{ g}}{104.14 \text{ g/mol}} \approx 0.0192 \text{ moles}\] This means your sample contains 0.0192 moles of styrene molecules. Thus, this calculation helps convert weight into a more usable form for further chemical analysis or reactions.
For example, with a 2.00 g sample of styrene, using its molar mass (104.14 g/mol), you can calculate the number of moles: \[\text{Number of moles} = \frac{2.00 \text{ g}}{104.14 \text{ g/mol}} \approx 0.0192 \text{ moles}\] This means your sample contains 0.0192 moles of styrene molecules. Thus, this calculation helps convert weight into a more usable form for further chemical analysis or reactions.
Hydrogen Atoms
Hydrogen atoms are the most abundant and simplest of all atoms. In the context of our styrene example, it is vital to know how many hydrogen atoms are present in a given amount of substance.
The empirical formula of styrene, \( \mathrm{CH} \), tells us that for each molecule of styrene, there is exactly one hydrogen atom.
The empirical formula of styrene, \( \mathrm{CH} \), tells us that for each molecule of styrene, there is exactly one hydrogen atom.
- This means if you have 1 mole of styrene, it corresponds directly to 1 mole of hydrogen atoms.
- Therefore, to find the total hydrogen atoms in any sample, you simply determine the moles of styrene present.
Avogadro's Number
Avogadro's number is one of the cornerstones of chemical calculations, providing the link between the atomic world and our everyday measurements. This constant, \(6.022 \times 10^{23}\), tells us how many atoms or molecules are in a mole of a substance.
Using Avogadro's number:- You can convert moles of a substance into the actual number of molecules or atoms.- In our styrene example, once we have calculated there are 0.0192 moles of styrene, we multiply by Avogadro's number to find the number of hydrogen atoms.\[\text{Number of hydrogen atoms} = 0.0192 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 1.16 \times 10^{22} \text{ atoms}\]This calculation allows you to comprehend the massive scale of atoms even in a small sample, revealing the micro-scale intricacies of chemistry.
Using Avogadro's number:- You can convert moles of a substance into the actual number of molecules or atoms.- In our styrene example, once we have calculated there are 0.0192 moles of styrene, we multiply by Avogadro's number to find the number of hydrogen atoms.\[\text{Number of hydrogen atoms} = 0.0192 \text{ moles} \times 6.022 \times 10^{23} \text{ atoms/mole} \approx 1.16 \times 10^{22} \text{ atoms}\]This calculation allows you to comprehend the massive scale of atoms even in a small sample, revealing the micro-scale intricacies of chemistry.