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Chapter 8: Problem 73

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.

Short Answer

Expert verified
The empirical formula of lithium fluoride is LiF since the ratio of Li and F in the compound is approximately 1:1.

Step by step solution

01

Recall the molar masses of lithium and fluorine

To calculate the moles of each element, we need the molar masses of lithium (Li) and fluorine (F). - Molar mass of Li = 6.94 g/mol - Molar mass of F = 19.00 g/mol
02

Calculate the mass of fluorine in the lithium fluoride

Since the mass of the lithium fluoride is 3.73 mg and the mass of lithium is 1.00 mg, we can determine the mass of fluorine in the compound: Mass of F = Total mass of lithium fluoride - Mass of Li Mass of F = 3.73 mg - 1.00 mg = 2.73 mg
03

Convert masses to moles

Now, we need to convert the masses of lithium and fluorine to moles using their respective molar masses: Moles of Li = (1.00 mg * (1 g/1000 mg)) / 6.94 g/mol = 1.44 x 10^(-4) mol Moles of F = (2.73 mg * (1 g/1000 mg)) / 19.00 g/mol = 1.43 x 10^(-4) mol
04

Find ratio between moles of Li and F

Divide the moles of Li and F by the smallest value to find the simplest whole number ratio: Li ratio = Moles of Li / smallest value = 1.44 x 10^(-4) mol / 1.43 x 10^(-4) mol 鈮� 1.01 F ratio = Moles of F / smallest value = 1.43 x 10^(-4) mol / 1.43 x 10^(-4) mol = 1
05

Determine the empirical formula

Since the ratio of Li and F in the compound is approximately 1:1, the empirical formula for lithium fluoride is: LiF

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Most popular questions from this chapter

When a \(2.118-g\) sample of copper is heated in an atmosphere in which the amount of oxygen present is restricted, the sample gains \(0.2666 \mathrm{g}\) of oxygen in forming a reddish-brown oxide. However, when 2.118 g of copper is heated in a stream of pure oxygen, the sample gains \(0.5332 \mathrm{g}\) of oxygen. Calculate the empirical formulas of the two oxides of copper.

Use the average atomic masses given inside the front cover of this book to calculate the number of moles of each element present in each of the following samples. a. 21.50 g of arsenic b. \(9.105 \mathrm{g}\) of phosphorus c. \(0.05152 \mathrm{g}\) of barium d. \(43.15 \mathrm{g}\) of carbon e. \(26.02 \mathrm{g}\) of chromium f. \(1.951 \mathrm{g}\) of platinum

Calculate the percent by mass of the element mentioned first in the formulas for each of the following compounds. a. sodium azide, \(\mathrm{NaN}_{3}\) b. copper(II) sulfate, \(\mathrm{CuSO}_{4}\) c. gold(III) chloride, \(\mathrm{AuCl}_{3}\) d. silver nitrate, \(\mathrm{AgNO}_{3}\) e. rubidium sulfate, \(\mathrm{Rb}_{2} \mathrm{SO}_{4}\) f. sodium chlorate, \(\mathrm{NaClO}_{3}\) g. nitrogen triiodide, \(\mathrm{NI}_{3}\) h. cesium bromide, \(\operatorname{CsBr}\)

A magnesium salt has the following elemental composition: \(16.39\% \) Mg, \(18.89\%\) N, \(64.72\%\) O. Determine the empirical formula of the salt.

When \(4.01 \mathrm{g}\) of mercury is strongly heated in air, the resulting oxide weighs \(4.33 \mathrm{g}\). Calculate the empirical formula of the oxide.
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