Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 113

A mixture of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) is heated until all the water is lost. If \(5.020 \mathrm{~g}\) of the mixture gives \(2.988 \mathrm{~g}\) of the anhydrous salts, what is the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture?

Short Answer

Expert verified
The solution requires solving a system of equations. Please refer to the steps above for the details of the calculations needed. The final answer will be in the form of a percent which represents the percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the mixture.

Step by step solution

01

Calculate the Mass of Water Lost

First, calculate the mass of water lost by heating the mixture. This is done by subtracting the final mass of anhydrous salts from the initial mass of the hydrated salts. In this case, \(5.020 \mathrm{~g} - 2.988 \mathrm{~g} = 2.032 \mathrm{~g}\) of water is lost.
02

Calculate the Amount of each Salt

The water lost comes from two sources: the 5 molecules of water in \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and the 7 molecules of water in \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\). Therefore, the total mass of water lost can be written as the sum of the parts. Therefore, we can set up the following equation where x is the mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) and y is the mass of \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\): \(x\cdot \frac{5\cdot 2}{molar~mass~of~\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}} + y \cdot \frac{7\cdot 2}{molar~mass~of~\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}} = 2.032 \mathrm{~g}\)
03

Use the Fact that x + y equals the Total Mass of the Sample

Using the fact that \(x + y = 5.020 \mathrm{~g}\), we can solve the system of equations obtained in step 1 & 2 for x and y.
04

Calculate the Percent by mass of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) in the Mixture

Finally, once x and y are determined, the percent by mass can be calculated using the equation \(\%~by~mass = \frac{text{Mass of the part you are interested in}}{text{Total mass of the sample}} \cdot 100\%\). In this case, it would be \(\%~by~mass = \frac{x}{5.020 \mathrm{~g}} \cdot 100\%\)

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

How many moles of cobalt (Co) atoms are there in \(6.00 \times 10^{9}(6\) billion) Co atoms?

Carbohydrates are compounds containing carbon, hydrogen, and oxygen in which the hydrogen to oxygen ratio is 2: 1 . A certain carbohydrate contains 40.0 percent carbon by mass. Calculate the empirical and molecular formulas of the compound if the approximate molar mass is \(178 \mathrm{~g}\).

Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound gives the following percent composition by mass: C: 44.4 percent; \(\mathrm{H}: 6.21\) percent; \(\mathrm{S}: 39.5\) percent; \(\mathrm{O}\) 9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is about \(162 \mathrm{~g} ?\)

If we know the empirical formula of a compound, what additional information do we need to determine its molecular formula?

Calculate the molar mass of a compound if 0.372 mole of it has a mass of \(152 \mathrm{~g}\).
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

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.

Sign-up for free