/*! 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 107 Give the number of atoms of the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 107

Give the number of atoms of the specified element in a formula unit of each of the following compounds, and calculate the molecular (formula) mass: (a) Oxygen in aluminum sulfate, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (b) Hydrogen in ammonium hydrogen phosphate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) (c) Oxygen in the mineral azurite, \(\mathrm{Cu}_{3}(\mathrm{OH})_{2}\left(\mathrm{CO}_{3}\right)_{2}\)

Short Answer

Expert verified
a) 12 atoms, 342.17 u; b) 9 atoms, 133.062 u; c) 8 atoms, 344.686 u

Step by step solution

01

- Identify the number of atoms of the specified element in the compound

(a) For \(\text{Al}_2(\text{SO}_4)_3\): \ \Each \(\text{SO}_4\) molecule contains 4 oxygen atoms. Since there are 3 \(\text{SO}_4\) molecules, the total number of oxygen atoms is \(4 \times 3 = 12\).
02

- Identify the number of atoms of the specified element in the compound

(b) For \( (\text{NH}_4)_2\text{HPO}_4 \): \ \Each \( \text{NH}_4 \) contains 1 hydrogen atom. Since there are 2 \( \text{NH}_4 \) molecules, there are \(2 \times 4 = 8 \) hydrogen atoms from \( \text{NH}_4 \). Additionally, there is 1 more hydrogen atom in \(\text{HPO}_4\). Therefore, the total number of hydrogen atoms is \(8 + 1 = 9\).
03

- Identify the number of atoms of the specified element in the compound

(c) For \( \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 \): \ \Each \( \text{OH} \) contains 1 oxygen atom. Since there are 2 \( \text{OH} \) molecules, there are \(2 \) oxygen atoms from \( \text{OH} \). Similarly, each \( \text{CO}_3 \) molecule contains 3 oxygen atoms. Since there are 2 \( \text{CO}_3 \) molecules, there are \(2 \times 3 = 6 \) oxygen atoms from \( \text{CO}_3 \). Therefore, the total number of oxygen atoms is \(2 + 6 = 8\).
04

- Calculate molecular (formula) mass

(a) For \(\text{Al}_2(\text{SO}_4)_3\): \ \Molecular mass = 2(Mass of Al) + 3[ \( \text{Mass of S + 4(Mass of O)} \)] \ \= 2(26.98) + 3[ \( 32.07 + 4(16.00) \)] \ \= 2(26.98) + 3[32.07 + 64.00] = 2(26.98) + 3[96.07] \ \= 53.96 + 288.21 = 342.17 \ \Thus, the molecular mass is 342.17 u.
05

- Calculate molecular (formula) mass

(b) For \( (\text{NH}_4)_2\text{HPO}_4 \): \ \Molecular mass = 2[ \( 14.01 + 4(1.008) \)] + [ \( 1.008 + 31.97 + 4(16.00) \)] \ \= 2[14.01 + 4.032] + [1.008 + 31.97 + 64] = 2[18.042] + 96.978 \ \= 36.084 + 96.978 = 133.062 \ \Thus, the molecular mass is 133.062 u.
06

- Calculate molecular (formula) mass

(c) For \( \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 \): \ \Molecular mass = 3(Mass of Cu) + 2[ \( 1.008 + 16.00 \)] + 2[ \( 12.01 + 3(16.00) \)] \ \= 3(63.55) + 2[17.008] + 2[60.01] \ \= 190.65 + 34.016 + 120.02 = 344.686 \ \Thus, the molecular mass is 344.686 u.

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.

Chemical Formula Composition
Understanding the chemical formula of a compound is essential in chemistry. Every chemical formula, like \( \text{Al}_2(\text{SO}_4)_3 \), \( (\text{NH}_4)_2\text{HPO}_4 \), and \( \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 \), represents the specific ratio of atoms of each element in the compound.
For example:
  • \( \text{Al}_2(\text{SO}_4)_3 \) tells us there are 2 aluminum (Al) atoms and 3 sulfate (SO鈧�) units in one formula unit.
  • \( (\text{NH}_4)_2\text{HPO}_4 \) indicates 2 ammonium (NH鈧�) groups and 1 hydrogen phosphate (HPO鈧�).
  • \( \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 \) shows 3 copper (Cu), 2 hydroxide (OH), and 2 carbonate (CO鈧�) groups.
This understanding helps in accurately determining the number of each type of atom in a molecule and in further calculations such as determining molecular mass.
Atom Count in Compounds
Counting the atoms of each element in a compound is a fundamental step in chemistry. Let's take some examples to understand it better:
In \( \text{Al}_2(\text{SO}_4)_3 \), each sulfate (SO鈧�) group contains 4 oxygen atoms. With 3 sulfate groups in the compound, the total number of oxygen atoms is \( 4 \times 3 = 12 \).
  • For \( (\text{NH}_4)_2\text{HPO}_4 \), each ammonium (NH鈧�) group has 4 hydrogen atoms. Since there are 2 ammonium groups, they contribute \( 2 \times 4 = 8 \) hydrogen atoms. Adding the single hydrogen in HPO鈧�, the total hydrogen count becomes \( 8 + 1 = 9 \).
  • In \( \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 \), each hydroxide (OH) group has 1 oxygen atom. With 2 such groups, there are \( 2 \) oxygen atoms. Each carbonate (CO鈧�) group has 3 oxygen atoms, and with 2 such groups, this contributes \( 2 \times 3 = 6 \) oxygen atoms. The total number of oxygen atoms is \( 2 + 6 = 8 \).
Practicing these calculations helps you develop a strong foundation for understanding and working with chemical formulas.
Molecular Mass Determination
Calculating the molecular mass of a compound involves summing the atomic masses of all atoms in the compound. Here's how you can do it step by step:
  • For \( \text{Al}_2(\text{SO}_4)_3 \):
    The atomic mass of Al (aluminum) is 26.98 u. So, with 2 Al atoms, the total mass contribution from Al is \( 2 \times 26.98 = 53.96 \) u.
    Each SO鈧� group has 1 sulfur (S) and 4 oxygens (O). The atomic masses are 32.07 u for S and 16.00 u for O. So, the mass of one SO鈧� is \( 32.07 + 4 \times 16.00 = 96.07 \) u.
    With 3 SO鈧� groups, the mass is \( 3 \times 96.07 = 288.21 \) u.
    Therefore, the total molecular mass is \( 53.96 + 288.21 = 342.17 \) u.
  • For \( (\text{NH}_4)_2\text{HPO}_4 \):
    The mass contribution from 2 NH鈧� groups is \( 2 \times [14.01 + 4 \times 1.008] = 2 \times 18.042 = 36.084 \) u.
    The mass from HPO鈧� is \( 1.008 + 31.97 + 4 \times 16.00 = 96.978 \) u.
    Therefore, the total molecular mass is \( 36.084 + 96.978 = 133.062 \) u.
  • For \( \text{Cu}_3(\text{OH})_2(\text{CO}_3)_2 \):
    The mass contribution from 3 Cu atoms is \( 3 \times 63.55 = 190.65 \) u.
    From 2 OH groups, it is \( 2 \times (1.008 + 16.00) = 34.016 \) u.
    From 2 CO鈧� groups, it is \( 2 \times (12.01 + 3 \times 16.00) = 120.02 \) u.
    Therefore, the total molecular mass is \( 190.65 + 34.016 + 120.02 = 344.686 \) u.
Mastering these calculations will help you understand the composition and quantitative aspects of chemical compounds better.

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

Nitrogen forms more oxides than any other element. The percents by mass of \(\mathrm{N}\) in three different nitrogen oxides are (I) \(46.69 \%\), (II) \(36.85 \%\), and (III) \(25.94 \% .\) For each compound, determine (a) the simplest whole-number ratio of \(\mathrm{N}\) to \(\mathrm{O}\) and (b) the number of grams of oxygen per \(1.00 \mathrm{~g}\) of nitrogen.

Can the relative amounts of the components of a mixture vary? Can the relative amounts of the components of a compound vary? Explain.

Describe Thomson's model of the atom. How might it account for the production of cathode rays?

Correct each of the following formulas: (a) Barium oxide is \(\mathrm{BaO}_{2}\). (b) Iron(II) nitrate is \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}\) (c) Magnesium sulfide is \(\mathrm{MnSO}_{3}\). (d) Potassium iodide is \(\mathrm{P}_{2} \mathrm{I}_{3}\).

Describe the type and nature of the bonding that often occurs between two nonmetals.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Chemical Analysis

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.