Question

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

The empirical formula of the chromium chloride compound is CrCl3.

Step-by-step solution

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

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

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

Key concepts

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.