/*! 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 36 Calculate the mass in grams of e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 36

Calculate the mass in grams of each of the following samples. a. \(6.14 \times 10^{-4}\) moles of sulfur trioxide b. \(3.11 \times 10^{5}\) moles of lead(IV) oxide c. 0.495 mole of chloroform, \(\mathrm{CHCl}_{3}\) d. \(2.45 \times 10^{-8}\) moles of trichloroethane, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3}\) e. 0.167 mole of lithium hydroxide f. 5.26 moles of copper(I) chloride

Short Answer

Expert verified
The mass in grams for each of the samples are: a. 0.0492 g b. \(7.43 \times 10^7\) g c. 59.09 g d. \(3.28 \times 10^{-6}\) g e. 4.00 g f. 520.74 g

Step by step solution

01

a. Mass of 6.14 x 10^{-4} moles of sulfur trioxide (SO3)

1. Find the molar mass of SO3: \(M(SO3) = (1 \times M(S)) + (3 \times M(O))\) Using the periodic table, we find the atomic masses: M(S) = 32.06 g/mol, M(O) = 16.00 g/mol Calculate \(M(SO3)\) \(M(SO3) = (1\times 32.06) + (3 \times 16.00) = 80.06\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 6.14 \times 10^{-4} moles \times 80.06\ g/mol = 0.0492\ g\)
02

b. Mass of 3.11 x 10^{5} moles of lead(IV) oxide (PbO2)

1. Find the molar mass of PbO2: \(M(PbO2) = (1 \times M(Pb)) + (2 \times M(O))\) Using the periodic table, we find the atomic masses: M(Pb) = 207.2 g/mol, M(O) = 16.00 g/mol Calculate \(M(PbO2)\) \(M(PbO2) = (1 \times 207.2) + (2 \times 16.00) = 239.2\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 3.11 \times 10^5 moles \times 239.2\ g/mol = 7.43 \times 10^7\ g\)
03

c. Mass of 0.495 mole of chloroform (CHCl3)

1. Find the molar mass of CHCl3: \(M(CHCl3) = (1 \times M(C)) + (1 \times M(H)) + (3 \times M(Cl))\) Using the periodic table, we find the atomic masses: M(C) = 12.01 g/mol, M(H) = 1.01 g/mol, M(Cl) = 35.45 g/mol Calculate \(M(CHCl3)\) \( M(CHCl3) = (1\times 12.01) + (1\times 1.01) + (3 \times 35.45) = 119.37\ g/mol \) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 0.495 moles \times 119.37\ g/mol = 59.09\ g\)
04

d. Mass of 2.45 x 10^{-8} moles of trichloroethane (C2H3Cl3)

1. Find the molar mass of C2H3Cl3: \(M(C2H3Cl3) = (2 \times M(C)) + (3 \times M(H)) + (3 \times M(Cl))\) Using the periodic table, we find the atomic masses: M(C) = 12.01 g/mol, M(H) = 1.01 g/mol, M(Cl) = 35.45 g/mol Calculate \(M(C2H3Cl3)\) \(M(C2H3Cl3) = (2\times 12.01) + (3\times 1.01) + (3 \times 35.45) = 133.93\ g/mol \) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 2.45 \times 10^{-8} moles \times 133.93\ g/mol = 3.28 \times 10^{-6}\ g\)
05

e. Mass of 0.167 mole of lithium hydroxide (LiOH)

1. Find the molar mass of LiOH: \(M(LiOH) = (1 \times M(Li)) + (1 \times M(O)) + (1 \times M(H))\) Using the periodic table, we find the atomic masses: M(Li) = 6.94 g/mol, M(O) = 16.00 g/mol, M(H) = 1.01 g/mol Calculate \(M(LiOH)\) \(M(LiOH) = (1\times 6.94) + (1\times 16.00) + (1 \times 1.01) = 23.95\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 0.167 moles \times 23.95\ g/mol = 4.00\ g\)
06

f. Mass of 5.26 moles of copper(I) chloride (CuCl)

1. Find the molar mass of CuCl: \(M(CuCl) = (1 \times M(Cu)) + (1 \times M(Cl))\) Using the periodic table, we find the atomic masses: M(Cu) = 63.55 g/mol, M(Cl) = 35.45 g/mol Calculate \(M(CuCl)\) \(M(CuCl) = (1\times 63.55) + (1\times 35.45) = 99.00\ g/mol\) 2. Convert moles to grams: \(mass = moles \times molar\_mass\) \(mass = 5.26 moles \times 99.00\ g/mol = 520.74\ g\)

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 91影视!

Key Concepts

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

Molar Mass Calculation
The concept of molar mass is crucial in chemistry for converting moles to grams, and vice versa. Molar mass essentially represents the mass of one mole of a given substance. Here's how you calculate it:
  • Identify the chemical formula of the compound you're dealing with.
  • Refer to the periodic table to look up the atomic mass of each element in the compound.
  • Multiply the atomic mass of each element by the number of times it appears in the formula.
  • Add up all these values to get the total molar mass.
For example, sulfur trioxide (\(\text{SO}_3\)) requires you to find the atomic masses of sulfur and oxygen. Sulfur has an atomic mass of 32.06 g/mol, and oxygen has an atomic mass of 16.00 g/mol. Calculating molar mass would involve:
\(M(\text{SO}_3) = (1 \times 32.06) + (3 \times 16.00) = 80.06\text{ g/mol}\).
This simplicity is key to performing stoichiometric calculations and converting between moles and grams.
Chemical Compounds
Chemical compounds consist of two or more different elements that are chemically bonded together in fixed proportions. Understanding the composition of a compound helps you interpret its chemical formula, which is vital for further calculations.
Here are some things to keep in mind about chemical compounds:
  • Each compound has a unique formula reflecting its specific ratio of atoms.
  • An example is \(\text{PbO}_2\), lead(IV) oxide, which includes one lead (Pb) atom and two oxygen (O) atoms.
  • The type of bond鈥攊onic or covalent鈥攁ffects the properties of the compound.
  • Compounds can be broken down into simpler substances through chemical reactions.
Recognizing the differences between compounds, such as sulfur trioxide and chloroform, \(\text{CHCl}_3\), and their unique formulas, is fundamental for calculating molar masses and performing conversions in stoichiometric problems.
Stoichiometry
Stoichiometry is the area of chemistry that involves the calculation of reactants and products in chemical reactions. It is a handy tool for scientists to predict yields and measure reaction efficiency.
Here鈥檚 what makes stoichiometry vital:
  • It explores the quantitative relationships using balanced chemical equations.
  • Stoichiometry uses the mole concept to convert between mass and number of particles.
  • It allows for scaling lab reactions to industrial processes based on reactant or product usages and yields.
To convert moles to grams in stoichiometric ratios, such as with other samples of chloroform or trichloroethane (\(\text{C}_2\text{H}_3\text{Cl}_3\)), you need to follow these steps:
  • Determine the molar mass as described earlier.
  • Use the molar mass to convert from moles to grams using the formula: \(\text{mass} = \text{moles} \times \text{molar mass}\).
  • This conversion is often essential for experimental guidance and proportion calculation in reactions.
A firm grasp of stoichiometry provides a deeper understanding of how much of each compound is consumed or produced in a chemical reaction, aiding in precise chemical analysis and synthesis.

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

You have a sample of copper (Cu) and a sample of aluminum (Al) . You have an equal number of atoms in each sample. Which of the following statements concerning the masses of the samples is true? a. The mass of the copper sample is more than twice as great as the mass of the aluminum sample. b. The mass of the copper sample is more than the mass of the aluminum sample, but it is not more than twice as great. c. The mass of the aluminum sample is more than twice as great as the mass of the copper sample. d. The mass of the aluminum sample is more than the mass of the copper sample, but it is not more than twice as great. e. The masses of the samples are equal.

A certain transition metal ion \(\left(\mathrm{M}^{\mathrm{n}+}\right)\) forms a compound with oxygen \(\left(\mathrm{M}_{\mathrm{x}} \mathrm{O}_{y}\right)\). The molar mass of the compound is \(250.2 \mathrm{~g} / \mathrm{mol}\). If the charge on the transition metal ion is \(+3,\) what is the identity of the transition metal, \(\mathrm{M} ?\) a. \(\mathrm{Th}\) b. Ti c. Hg d. \(\mathrm{Ru}\) e. \(\mathrm{Ag}\)

When \(1.00 \mathrm{mg}\) of lithium metal is reacted with fluorine gas \(\left(\mathrm{F}_{2}\right),\) the resulting fluoride salt has a mass of \(3.73 \mathrm{mg}\). Calculate the empirical formula of lithium fluoride.

The mass fraction of an element present in a compound can be obtained by comparing the mass of the particular element present in 1 mole of the compound to the _____ mass of the compound.

Calculate the number of grams of lithium that contain the same number of atoms as \(1.00 \mathrm{~kg}\) of zirconium.
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Physical Chemistry

Read Explanation
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.