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Chapter 8: Problem 125

When \(1.00 \mathrm{g}\) of metallic chromium is heated with elemental chlorine gas, \(3.045 \mathrm{g}\) of a chromium chloride salt results. Calculate the empirical formula of the compound.

Short Answer

Expert verified
The empirical formula of the chromium chloride compound is CrCl3.

Step by step solution

01

Convert mass to moles

First, we need to convert the mass of each element to moles using their respective molar masses. The molar mass of chromium (Cr) is 51.996 g/mol and that of chlorine (Cl) is 35.453 g/mol. Given mass of chromium = 1.00 g Moles of chromium = Given mass / Molar mass = 1.00 g / 51.996 g/mol = 0.0192 mol Given mass of the compound = 3.045 g Mass of chlorine = Mass of the compound - Mass of chromium = 3.045 g - 1.00 g = 2.045 g Moles of chlorine = Given mass / Molar mass = 2.045 g / 35.453 g/mol = 0.0577 mol
02

Determine the mole ratio

Now, we need to divide the moles of each element by the smallest number of moles for any element in the compound to get their mole ratios. Smallest number of moles = 0.0192 mol Mole ratio of chromium = Moles of chromium / Smallest number of moles = 0.0192 mol / 0.0192 mol = 1 Mole ratio of chlorine = Moles of chlorine / Smallest number of moles = 0.0577 mol / 0.0192 mol = 3
03

Determine the empirical formula

As we have determined the mole ratio, the empirical formula can now be written as Cr_1Cl_3, or more simply, Empirical formula = CrCl3

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is a branch of chemistry that deals with the quantitative relationships of the elements and compounds as they undergo chemical reactions. By understanding stoichiometry, chemists can predict the amounts of substances consumed and produced in a given reaction.

For instance, if we know the amounts of chromium and chlorine that react, stoichiometry can help us determine how much of a chromium chloride compound will form. In the exercise, stoichiometry was used to relate the mass of the elements to the amount in moles, which then allowed for the determination of the empirical formula of chromium chloride.
Molar Mass
The molar mass of an element or compound is the mass of 1 mole of that substance. It's a bridge between the mass of a sample and the amount of substance in moles. The molar mass of elements is found on the periodic table, while the molar mass of compounds is the sum of the molar masses of its constituent elements.

In the exercise, molar mass was used to convert the given mass of chromium and chlorine into moles, which is a crucial step in finding the empirical formula. The molar mass of chromium (Cr) is 51.996 grams per mole, and chlorine (Cl) is 35.453 grams per mole. Knowing these values is essential for accurate stoichiometric calculations.
Mole Ratio
The mole ratio expresses the relative quantities of reactants and products in a chemical reaction. It's derived from the coefficients of the balanced chemical equation and shows the proportion of moles of one substance to the moles of another substance.

In the solution provided, once the moles of chromium and chlorine were calculated, the next step was to find the simplest whole number ratio (mole ratio) between them. By dividing the moles of each element by the smallest number of moles calculated, the mole ratios for chromium and chlorine were obtained. This ratio dictated the subscripts used in the empirical formula, leading to the final answer of CrCl3.
Chemical Compound
A chemical compound is a substance consisting of two or more different types of atoms bonded together in a fixed proportion. Compounds have a unique set of physical and chemical properties, and they are often represented by empirical or molecular formulas that indicate the types and numbers of atoms present.

The empirical formula is the simplest whole-number ratio of elements in a compound. For example, the exercise provided leads to the determination of the empirical formula CrCl3 for a chromium chloride compound, which signifies that for every chromium atom, there are three chlorine atoms in the compound.

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Most popular questions from this chapter

A particular small laboratory cork weighs \(1.63 \mathrm{g}\) whereas a rubber lab stopper of the same size weighs 4.31 g. How many corks would there be in 500. g of such corks? How many rubber stoppers would there be in \(500 .\) g of similar stoppers? How many grams of rubber stoppers would be needed to contain the same number of stoppers as there are corks in \(1.00 \mathrm{kg}\) of corks?

The mass fraction of an element present in a compound can be obtained by comparing the mass of the particular element present in 1 mol of the compound to the ______ mass of the compound.

Calculate the number of moles of the indicated substance present in each of the following samples. a. \(1.28 \mathrm{g}\) of iron(II) sulfate b. \(5.14 \mathrm{mg}\) of mercury(II) iodide c. \(9.21 \mu g\) of tin(IV) oxide d. 1.26 lb of cobalt(II) chloride e. \(4.25 \mathrm{g}\) of copper(II) nitrate

Calculate the percent by mass of the element listed first in the formulas for each of the following compounds. a. copper(II) bromide, \(\mathrm{CuBr}_{2}\) b. copper(I) bromide, CuBr c. iron(II) chloride, \(\mathrm{FeCl}_{2}\) d. iron(III) chloride, \(\mathrm{FeCl}_{3}\) e. cobalt(II) iodide, \(\mathrm{Col}_{2}\) f. cobalt(III) iodide, \(\mathrm{CoI}_{3}\) g. tin(II) oxide, \(\operatorname{SnO}\) h. tin(IV) oxide, \(\operatorname{SnO}_{2}\)

Calculate the mass of carbon in grams, the percent carbon by mass, and the number of individual carbon atoms present in each of the following samples. a. 7.819 g of carbon suboxide, \(\mathrm{C}_{3} \mathrm{O}_{2}\) b. \(1.53 \times 10^{21}\) molecules of carbon monoxide c. 0.200 mol of phenol, \(C_{6} \mathrm{H}_{6} \mathrm{O}\)
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