/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 147 The empirical formula of styrene... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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
The number of hydrogen atoms in a 2.00-g sample of styrene is approximately 9.23 x 10²² atoms.

Step by step solution

01

Find the molar mass of the empirical formula.

First, we need to find the molar mass of the empirical formula (CH). The molar mass can be found by adding the molar mass of carbon and hydrogen. Molar mass of Carbon (C): 12.01 g/mol Molar mass of Hydrogen (H): 1.01 g/mol Molar mass of CH = 12.01 g/mol +1.01g/mol = 13.02 g/mol
02

Find the number of CH units in one molecule of styrene.

Next, we need to find the number of CH units that are present in one molecule of styrene. For that, divide the molar mass of styrene by the molar mass of the empirical formula: Number of CH units in styrene = (molar mass of styrene) / (molar mass of CH) = 104.14 g/mol / 13.02 g/mol Number of CH units in styrene ≈ 8
03

Determine the number of moles in the 2.00-g sample.

Now that we know the molar mass of styrene, we can calculate the number of moles in the 2.00-g sample: Number of moles = (mass of sample) / (molar mass of styrene) = 2.00 g / 104.14 g/mol ≈ 0.0192 moles
04

Calculate the number of hydrogen atoms in the 2.00-g sample.

Since there are 8 hydrogen atoms for every molecule of styrene, we will multiply the number of moles by 8 to get the number of moles of hydrogen: Number of hydrogen atoms = (number of moles of styrene) × (number of hydrogen atoms per molecule of styrene) = 0.0192 moles × 8 ≈ 0.1536 moles Now we can convert the number of moles to the number of atoms using Avogadro's number (6.022 x 10²³ atoms/mol): Number of hydrogen atoms = 0.1536 moles × 6.022 x 10²³ atoms/mol ≈ 9.23 x 10²² atoms The number of hydrogen atoms in a 2.00-g sample of styrene is approximately 9.23 x 10²² atoms.

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 91Ó°ÊÓ!

Key Concepts

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

Molar Mass
The molar mass is a fundamental concept in chemistry that helps to determine the mass of one mole of a chemical substance. It is measured in grams per mole (\(\text{g/mol}\)). To compute the molar mass, simply sum up the atomic masses of all atoms present in the molecule.
For instance, in the case of styrene, we first need the empirical formula, which is \(\text{CH}\), to calculate its molar mass. This entails adding the molar masses of carbon and hydrogen:
  • Carbon (C) has a molar mass of 12.01 \(\text{g/mol}\).
  • Hydrogen (H) has a molar mass of 1.01 \(\text{g/mol}\).
So, the molar mass of \(\text{CH}\) is 13.02 \(\text{g/mol}\).
Understanding the molar mass is crucial as it serves as the bridge between the mass of a substance and the number of particles contained in that mass.
Moles Calculation
Calculating moles is a central task when dealing with chemical substances. The formula used is: \[\text{Number of moles} = \frac{\text{mass of the sample}}{\text{molar mass}}\]This calculation translates the macroscopic mass of a sample into a number that reflects how many microscopic particles, or moles, it contains.
Using styrene as an example, with a given molar mass of 104.14 \(\text{g/mol}\), if we have a 2.00 g sample, the number of moles is:
  • \(\frac{2.00 \text{ g}}{104.14 \text{ g/mol}} \approx 0.0192 \text{ moles}\)
This calculation, \(0.0192\) moles, indicates the amount of substance present in the sample, making it easier to determine further properties like the number of atoms.
Avogadro's Number
Avogadro's number is a large constant used in chemistry to relate moles to individual entities like atoms or molecules. It is given as: \(6.022 \times 10^{23}\) atoms/mol or molecules/mol.
This constant allows for the conversion from the macroscopic mole concept to a microscopic count of atoms. In our styrene sample, having determined the moles, we convert this to the actual number of hydrogen atoms:
  • First, find the moles of hydrogen in styrene by multiplying the moles of styrene by the number of hydrogen atoms per molecule.
  • Next, convert these moles to atoms using Avogadro's number.
  • In the example, \((0.1536 \text{ moles})\times(6.022 \times 10^{23} \text{ atoms/mol}) \approx 9.23 \times 10^{22} \text{ atoms}\)
This vast number gives a feel for the multitude of atoms present even in small chemical amounts.
Chemical Formula
Chemical formulas are the written symbol representation of the elements present in a compound, showcasing the proportions in which they combine. They can be empirical, molecular, or structural.
The chemical formula not only identifies the elements but also indicates the proportions. For styrene, the empirical formula is \(\text{CH}\), which represents the simplest whole-number ratio of carbon to hydrogen atoms in the compound.
This simplifies to signify that the compound contains eight times this ratio in a full molecular depiction, as seen in step 2 of our solution.
Understanding the chemical formula provides insights into chemical reactions and properties, serving as the fundamental language of chemistry. It allows chemists to convey a vast amount of information about the nature and composition of chemical substances efficiently.

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

A \(0.4230-\) g sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

Which of the following statements about chemical equations is(are) true? a. When balancing a chemical equation, you can never change the coefficient in front of any chemical formula. b. The coefficients in a balanced chemical equation refer to the number of grams of reactants and products. c. In a chemical equation, the reactants are on the right and the products are on the left. d. When balancing a chemical equation, you can never change the subscripts of any chemical formula. e. In chemical reactions, matter is neither created nor destroyed so a chemical equation must have the same number of atoms on both sides of the equation.

The space shuttle environmental control system handled excess \(\mathrm{CO}_{2}\) (which the astronauts breathe out; it is 4.0\(\%\) by mass of exhaled air) by reacting it with lithium hydroxide, LiOH, pellets to form lithium carbonate, Li \(_{2} \mathrm{CO}_{3},\) and water. If there were seven astronauts on board the shuttle, and each exhales \(20 .\) L of air per minute, how long could clean air be generated if there were \(25,000\) g of LiOH pellets available for each shuttle mission? Assume the density of air is 0.0010 \(\mathrm{g} / \mathrm{mL}\) .

Consider the following unbalanced equation: $$ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 \(\mathrm{kg}\) calcium phosphate with 1.0 \(\mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4} \text { by mass)? }\right.\)

Which (if any) of the following is(are) true regarding the limiting reactant in a chemical reaction? a. The limiting reactant has the lowest coefficient in a balanced equation. b. The limiting reactant is the reactant for which you have the fewest number of moles. c. The limiting reactant has the lowest ratio of moles available/coefficient in the balanced equation. d. The limiting reactant has the lowest ratio of coefficient in the balanced equation/moles available. Justify your choice. For those you did not choose, explain why they are incorrect.

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.