Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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?

Short Answer

Expert verified
There are approximately \( 9.24 \times 10^{23} \) hydrogen atoms present in a 2.00-g sample of styrene.

Step by step solution

01

Determine the molecular formula of styrene

As given, the empirical formula of styrene is CH, and the molar mass is 104.14 g/mol. First, we need to figure out the molecular formula by finding the ratio of the molecular mass to the empirical formula mass. The molecular mass of CH is: Molar mass of C + Molar mass of H = 12.01 g/mol + 1.01 g/mol = 13.02 g/mol Now, to find the ratio: Molecular mass of Styrene / Molecular mass of CH = \( \frac{104.14}{13.02} \) ≈ 8 So, the molecular formula of styrene is C8H8 (multiply the empirical formula by the ratio).
02

Find the moles of styrene in the 2.00-g sample

We can find the moles of styrene using the formula: Moles of styrene = \( \frac{Mass of Styrene~(g)}{Molar mass of Styrene~(g/mol)} \) Moles of styrene = \( \frac{2.00}{104.14} \) = 0.0192 mol
03

Calculate the number of hydrogen atoms in the sample

In the molecular formula C8H8, there are 8 hydrogen atoms in one molecule of styrene. To find the total number of hydrogen atoms in the 2.00-g sample, we need to use Avogadro's number: Number of hydrogen atoms = Moles of styrene × (number of H atoms in 1 molecule of styrene) × Avogadro's number Number of hydrogen atoms = 0.0192 × 8 × \( 6.022 \times 10^{23} \) Number of hydrogen atoms = \( 9.24 \times 10^{23} \) There are approximately \( 9.24 \times 10^{23} \) hydrogen atoms present in a 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.

Molecular Mass Calculation
Molecular mass, also known as molecular weight, is the sum of the masses of all the atoms in a molecule. It's a fundamental concept in chemistry, especially when it comes to converting between moles and grams of a substance.

For instance, in the exercise, we calculate the molecular mass of styrene by first determining the mass of its empirical formula, CH. This involves adding the atomic mass of carbon (\(12.01 \text{g/mol}\)) to the atomic mass of hydrogen (\(1.01 \text{g/mol}\)), resulting in the empirical formula mass of \(13.02 \text{g/mol}\).

To determine how many times the empirical formula fits into the molecular mass of the compound, we divide the given molar mass of styrene (\(104.14 \text{g/mol}\)) by the mass of the empirical formula (\(13.02 \text{g/mol}\)). This ratio gives us the number of empirical formula units in the molecular formula, allowing us to deduce that the molecular formula of styrene is C8H8.
Moles and Molar Mass
In chemistry, the mole is a unit of measurement for the amount of substance. The molar mass is the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. It tells us how many grams are in one mole of that substance, essentially serving as a bridge between the macroscopic and microscopic worlds of chemistry.

Using the styrene example, we can calculate the number of moles present in a 2.00-gram sample by dividing the mass of the sample by the molar mass of styrene (\(104.14 \text{g/mol}\)). This step is crucial as it sets the stage for determining the quantity of atoms or molecules in a given sample. Through understanding moles and molar mass, we're able to measure and articulate the amount of a chemical substance in a way that can be practically and theoretically analyzed.
Avogadro's Number
One of the cornerstones of chemistry is Avogadro's number, which is \(6.022 \times 10^{23}\). This immense figure represents the number of particles, such as atoms or molecules, that are contained in one mole of a substance. It's named after the Italian scientist Amedeo Avogadro and is fundamental to understanding the composition of molecules and their interactions.

In our exercise with styrene, after determining the amount of moles in a given mass, we use Avogadro's number to calculate the number of individual hydrogen atoms. Multiplying the moles of styrene by the number of hydrogen atoms in one molecule of styrene, and then by Avogadro's number, provides us with the total count of hydrogen atoms in the sample. This application of Avogadro's number is the bridge connecting the molecular mass calculation and the concept of moles to actual, countable physical entities.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Silicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. \(\mathrm{SiO}_{2}(s)+\mathrm{C}(s) \frac{\text { Electic }}{\text { arc furmae }} \mathrm{Si}(s)+\mathrm{CO}(g)\) b. Liquid silicon tetrachloride is reacted with very pure solid magnesium, producing solid silicon and solid magnesium chloride. c. \(\mathrm{Na}_{2} \mathrm{SiF}_{6}(s)+\mathrm{Na}(s) \rightarrow \operatorname{Si}(s)+\operatorname{NaF}(s)\)

You take \(1.00 \mathrm{~g}\) of an aspirin tablet (a compound consisting solely of carbon, hydrogen, and oxygen), burn it in air, and collect \(2.20 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.400 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). You know that the molar mass of aspirin is between 170 and \(190 \mathrm{~g} / \mathrm{mol}\). Reacting 1 mole of salicylic acid with 1 mole of acetic anhydride \(\left(\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) gives you 1 mole of aspirin and 1 mole of acetic acid \(\left(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\right.\) ). Use this information to determine the molecular formula of salicylic acid.

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: $$ 2 \mathrm{C}_{6} \mathrm{H}_{3} \mathrm{Cl}+\mathrm{C}_{2} \mathrm{HOCl}_{3} \longrightarrow \mathrm{C}_{14} \mathrm{H}_{9} \mathrm{Cl}_{5}+\mathrm{H}_{2} \mathrm{O} $$ \(\begin{array}{lll}\text { chlorobenzene } & \text { chloral } & \text { DDT }\end{array}\) In a government lab, \(1142 \mathrm{~g}\) of chlorobenzene is reacted with \(485 \mathrm{~g}\) of chloral. a. What mass of DDT is formed, assuming \(100 \%\) yield? b. Which reactant is limiting? Which is in excess? c. What mass of the excess reactant is left over? d. If the actual yield of DDT is \(200.0 \mathrm{~g}\), what is the percent yield?

The molecular formula of acetylsalicylic acid (aspirin), one of the most commonly used pain relievers, is \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\). a. Calculate the molar mass of aspirin. b. A typical aspirin tablet contains \(500 . \mathrm{mg} \mathrm{C}_{9} \mathrm{H}_{\mathrm{g}} \mathrm{O}_{4} .\) What amount (moles) of \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) molecules and what number of molecules of acetylsalicylic acid are in a \(500 .-\mathrm{mg}\) tablet?

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of \(0.157 \mathrm{~g}\) of the compound produced \(0.213 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0310 \mathrm{~g} \mathrm{H}_{2} \mathrm{O} .\) In another experiment, it is found that \(0.103 \mathrm{~g}\) of the compound produces \(0.0230 \mathrm{~g} \mathrm{NH}_{3} .\) What is the empirical formula of the compound? Hint: Combustion involves reacting with excess \(\mathrm{O}_{2}\). Assume that all the carbon ends up in \(\mathrm{CO}_{2}\) and all the hydrogen ends up in \(\mathrm{H}_{2} \mathrm{O}\). Also assume that all the nitrogen ends up in the \(\mathrm{NH}_{3}\) in the second experiment.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free