Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 48

Tin(II) fluoride \(\left(\mathrm{SnF}_{2}\right)\) is often added to toothpaste as an ingredient to prevent tooth decay. What is the mass of \(\mathrm{F}\) in grams in \(24.6 \mathrm{~g}\) of the compound?

Short Answer

Expert verified
The mass of fluorine in 24.6 g of the compound is approximately 5.97 grams.

Step by step solution

01

Calculate the molar mass of Tin(II) fluoride

The molar mass of Tin(II) Fluoride (\(SnF_2\)) is calculated by adding the molar masses of Sn and 2F. The molar mass of Sn is 118.7 g/mol and that of F is 19.0 g/mol. So, the molar mass of \(SnF_2\) = 118.7 g/mol + 2 * 19.0 g/mol = 156.7 g/mol.
02

Calculate the moles of Tin(II) fluoride

Using the mass and molar mass of \(SnF_2\), to calculate the number of moles use the formula: moles = mass/molar mass. Plug in the values and get moles of \(SnF_2\) = \(24.6 g / 156.7 g/mol\) = 0.157 mol.
03

Find the mass of fluorine

There are 2 moles of F in each mole of \(SnF_2\). Therefore, the moles of F in 0.157 moles of \(SnF_2\) = \(2 * 0.157 mol\) = 0.314 mol. To find the mass of F, multiply the moles of F by its molar mass. Mass of F = \(0.314 mol * 19.0 g/mol\) = 5.966 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.

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

Which of the following has more atoms: \(1.10 \mathrm{~g}\) of hydrogen atoms or \(14.7 \mathrm{~g}\) of chromium atoms?

The atomic masses of \({ }_{17}^{35} \mathrm{Cl}(75.53\) percent \()\) and \({ }_{17}^{37} \mathrm{Cl}\) (24.47 percent) are 34.968 amu and 36.956 amu, respectively. Calculate the average atomic mass of chlorine. The percentages in parentheses denote the relative abundances.

One of the reactions that occurs in a blast furnace, where iron ore is converted to cast iron, is $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+3 \mathrm{CO} \longrightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_{2} $$ Suppose that \(1.64 \times 10^{3} \mathrm{~kg}\) of Fe are obtained from a \(2.62 \times 10^{3}-\mathrm{kg}\) sample of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). Assuming that the reaction goes to completion, what is the percent purity of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the original sample?

How many molecules of ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) are present in \(0.334 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{6} ?\)

Potash is any potassium mineral that is used for its potassium content. Most of the potash produced in the United States goes into fertilizer. The major sources of potash are potassium chloride \((\mathrm{KCl})\) and potassium sulfate \(\left(\mathrm{K}_{2} \mathrm{SO}_{4}\right)\). Potash production is often reported as the potassium oxide \(\left(\mathrm{K}_{2} \mathrm{O}\right)\) equivalent or the amount of \(\mathrm{K}_{2} \mathrm{O}\) that could be made from a given mineral. (a) If \(\mathrm{KCl}\) costs \(\$ 0.055\) per \(\mathrm{kg}\), for what price (dollar per \(\mathrm{kg}\) ) must \(\mathrm{K}_{2} \mathrm{SO}_{4}\) be sold in order to supply the same amount of potassium on a per dollar basis? (b) What mass (in \(\mathrm{kg}\) ) of \(\mathrm{K}_{2} \mathrm{O}\) contains the same number of moles of \(\mathrm{K}\) atoms as \(1.00 \mathrm{~kg}\) of \(\mathrm{KCl} ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Nuclear Chemistry

Read Explanation

The Earths Atmosphere

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