Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 103

A \(1.013 \mathrm{g}\) sample of \(\mathrm{ZnSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) is dissolved in water and the sulfate ion precipitated as \(\mathrm{BaSO}_{4}\). The mass of pure, dry \(\mathrm{BaSO}_{4}\) obtained is \(0.8223 \mathrm{g}\). What is the formula of the zinc sulfate hydrate?

Short Answer

Expert verified
The formula of the zinc sulfate hydrate is \( \mathrm{ZnSO}_4 \cdot 7 \mathrm{H}_2 \mathrm{O} \).

Step by step solution

01

Calculate moles of BaSO4

Use the given mass of barium sulfate (BaSO4) and its molar mass to calculate the number of moles of BaSO4. The molar mass of BaSO4 is \(233.39 \, \mathrm{g/mol}\). Use the formula \(\frac{\text{given mass}}{\text{molar mass}}\) to get the moles of BaSO4; as such: \(\frac{0.8223 \, \mathrm{g}}{233.39 \, \mathrm{g/mol}} = 0.00352 \, \mathrm{mol}\). You now know that the precipitate formed contains 0.00352 moles of BaSO4.
02

Deduce moles of ZnSO4

Because the reaction between the sulfate ion in ZnSO4 and Ba to form BaSO4 is a 1:1 reaction, we can conclude that the moles of ZnSO4 in the original sample is also 0.00352 moles. This is because each SO4^2- ion in the ZnSO4 ended up in the BaSO4.
03

Calculate Molecular Mass of ZnSO4

After finding the moles of ZnSO4, we can now find the molecular mass of the hydrated compound. We do this by using the formula for molecular mass, which states that it's equal to the mass of the sample divided by the moles of the compound. So, \( \frac{1.013 \, \mathrm{g}}{0.00352 \, \mathrm{mol}} = 287.78 \, \mathrm{g/mol} \). This is the molar mass of ZnSO4 together with the water of hydration, i.e., ZnSO4. xH2O.
04

Determine Number of Water Molecules

Now, subtract the molar mass of ZnSO4(without the water of hydration) from the molar mass of the hydrated compound to find the molar mass of the water of hydration alone. The molar mass of ZnSO4 is \(161.47 \, \mathrm{g/mol}\). Therefore, the molar mass of the water of hydration is \(287.78 - 161.47 = 126.31 \, \mathrm{g/mol}\). Since the molar mass of water (H2O) is \(18.015 \, \mathrm{g/mol}\), dividing the molar mass of the water of hydration by the molar mass of water will give the number of water molecules in the compound, \(x\). So, \(x = \frac{126.31 \, \mathrm{g/mol}}{18.015 \, \mathrm{g/mol}} = 7\). Therefore, \(x = 7\).

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

The density of a mixture of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) and water is 1.78 g/mL. The percent composition of the mixture is to be determined by converting \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) If \(32.0 \mathrm{mL}\) of the mixture gives \(65.2 \mathrm{g}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) then what is the percent composition of the mixture?

Explain which of the following statement(s) is (are) correct concerning glucose (blood sugar), \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (a) The percentages, by mass, of \(\mathrm{C}\) and \(\mathrm{O}\) are the same as in CO. (b) The ratio of \(\mathrm{C}: \mathrm{H}: \mathrm{O}\) atoms is the same as in dihydroxyacetone, \(\left(\mathrm{CH}_{2} \mathrm{OH}\right)_{2} \mathrm{CO}\) (c) The proportions, by mass, of \(C\) and \(O\) are equal. (d) The highest percentage, by mass, is that of H.

Three of the following formulas might be either an empirical or a molecular formula. The formula that must be a molecular formula is (a) \(\mathrm{N}_{2} \mathrm{O} ;\) (b) \(\mathrm{N}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{NaCl} ;\) (d) \(\mathrm{NH}_{3}\)

Name the following compounds and specify which ones are best described as ionic: (a) \(\mathrm{OF}_{2} ;\) (b) \(\mathrm{XeF}_{2}\) (c) \(\operatorname{CuSO}_{3} ;\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\)

In many of its compounds, oxygen has an oxidation state of \(-2 .\) However, there are exceptions. What is the oxidation state of oxygen in each of the following compounds? (a) \(\mathrm{OF}_{2} ;\) (b) \(\mathrm{O}_{2} \mathrm{F}_{2} ;\) (c) \(\mathrm{CsO}_{2}\); (d) \(\mathrm{BaO}_{2}\)
See all solutions

Recommended explanations on Chemistry Textbooks

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Organic 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.

Sign-up for free