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In most of its ionic compounds, cobalt is either \(\mathrm{Co}(\mathrm{II})\) or \(\mathrm{Co}(\mathrm{III})\). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A \(0.256-\mathrm{g}\) sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of \(0.308 \mathrm{~g}\). A second sample of \(0.416 \mathrm{~g}\) of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was \(0.145 \mathrm{~g}\). a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

Short Answer

Expert verified
The ionic compound is CoCl鈧冣⒙糎鈧侽, with a percent composition by mass of 49.49% cobalt, 44.96% chloride, and 1.90% water. The balanced equations for the reactions are: a) CoCl鈧冣⒙糎鈧侽 (s) -> CoCl鈧 (aq) + \( \frac{1}{4} \) H鈧侽 (l) b) CoCl鈧 (aq) + 3AgNO鈧 (aq) -> 3AgCl (s) + Co(NO鈧)鈧 (aq) c) CoCl鈧 (aq) + 3NaOH (aq) -> Co(OH)鈧 (s) + 3 NaCl (aq) d) 2Co(OH)鈧 (s) -> Co鈧侽鈧 (s) + 3H鈧侽 (l)

Step by step solution

01

Calculate moles of silver chloride and cobalt(III) oxide

We can begin by calculating the moles of silver chloride and cobalt(III) oxide obtained from the given masses. Using their molar masses: Molar mass of AgCl = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol Molar mass of Co鈧侽鈧 = 2 * 58.93 (Co) + 3 * 16 (O) = 165.86 g/mol Moles of AgCl = Mass of AgCl / Molar mass of AgCl Moles of AgCl = \( \frac{0.308 g}{143.32 g/mol} \) = 0.00215 mol Moles of Co鈧侽鈧 = Mass of Co鈧侽鈧 / Molar mass of Co鈧侽鈧 Moles of Co鈧侽鈧 = \( \frac{0.145 g}{165.86 g/mol} \) = 0.000873 mol
02

Calculate moles of cobalt(II/III) and chloride ions

Using the stoichiometry of silver chloride and cobalt(III) oxide, we can determine the moles of cobalt(II/III) and chloride ions in the compound: AgCl -> Ag鈦 + Cl鈦 1 mol of AgCl gives 1 mol of Cl鈦 ions 0.00215 mol of AgCl gives 0.00215 mol of Cl鈦 ions Co鈧侽鈧 -> 2Co(III) + 3O虏鈦 1 mol of Co鈧侽鈧 gives 2 mol of Co(III) ions 0.000873 mol of Co鈧侽鈧 gives 0.000873 * 2 = 0.00175 mol of Co(III) ions
03

Determine the formula of the compound

We have that the compound contains 0.00215 mol of Cl鈦 ions and 0.00175 mol of Co(III) ions. Since the compound contains one cobalt ion per formula unit, we can calculate the number of water molecules and chloride ions in the formula: Divide moles of Cl鈦 ions by moles of Co(III) ions. \( \frac{0.00215}{0.00175} \) = 1.23 鈮 1 Therefore, there is approximately one chloride ion for each Co(III) ion indicating the compound is CoCl鈧. Now, we'll find the number of water molecules in the compound. Moles of CoCl鈧 = 0.00175 mol Mass of CoCl鈧 = Moles * Molar mass = 0.00175 mol * 233.45 g/mol = 0.408 g Mass of the hydrate - Mass of CoCl鈧 = Mass of water 0.416 g - 0.408 g = 0.008 g Divide the mass of water by its molar mass (18.015 g/mol) to find the moles of water: Moles of water = \( \frac{0.008 g}{18.015 g/mol} \) = 0.000444 mol Ratio of moles of water to moles of CoCl鈧 = \( \frac{0.000444}{0.00175} \) = 0.254 鈮 1/4 Since the ratio of water molecules to CoCl鈧 is approximately 1:4, the formula of the compound is CoCl鈧冣⒙糎鈧侽.
04

Calculate percent composition

To find the percent composition, we'll first determine the molar mass of the compound CoCl鈧冣⒙糎鈧侽. Molar mass of CoCl鈧冣⒙糎鈧侽 = 233.45 (CoCl鈧) + 0.25 * 18.015 (H鈧侽) = 237.90 g/mol Percent composition, by mass: % Co = \( \frac{ 2*58.93 }{ 237.90 } \) * 100% = 49.49% % Cl = \( \frac{ 3*35.45 }{ 237.90 } \) * 100% = 44.96% % H鈧侽 = \( \frac{0.25*18.015}{237.90} \) * 100% = 1.90 %
05

Write balanced equations for the reactions

Now we'll write balanced equations for the three reactions described. a) Dissolving the compound in water: CoCl鈧冣⒙糎鈧侽 (s) -> CoCl鈧 (aq) + \( \frac{1}{4} \) H鈧侽 (l) b) Reaction with excess silver nitrate: CoCl鈧 (aq) + 3AgNO鈧 (aq) -> 3AgCl (s) + Co(NO鈧)鈧 (aq) c) Reaction with excess sodium hydroxide: CoCl鈧 (aq) + 3NaOH (aq) -> Co(OH)鈧 (s) + 3 NaCl (aq) d) Heating Co(OH)鈧 to form cobalt(III) oxide: 2Co(OH)鈧 (s) -> Co鈧侽鈧 (s) + 3H鈧侽 (l)

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Key Concepts

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

Percent Composition by Mass
Understanding the percent composition by mass of a compound is vital as it provides insights into the relative amount of each element within the substance. To determine the percent composition, you calculate the mass of each element in one mole of the compound and then divide that by the molar mass of the compound, finally multiplying by 100% to get a percentage.

For a compound like CoCl鈧冣⒙糎鈧侽, you first find the mass of cobalt, chlorine, and water that would be present in one mole of the compound. You'd calculate the molar mass of the compound as the sum of the masses of its constituent elements and water molecules. As calculated in the example, Co contributes to nearly half of the compound's mass, while Cl accounts for the majority of the remaining mass, and H鈧侽 鈥 the water of hydration 鈥 makes up a small percentage.

Understanding percent composition is crucial for chemistry students as it forms a foundation for further study of stoichiometry, empirical and molecular formulas, and chemical reactions.
Molar Mass Calculation
The molar mass of a substance is the weight of 6.022 x 10虏鲁 (Avogadro's number) of its particles, usually atoms or molecules, and is expressed in grams per mole (g/mol). It's akin to finding the collective weight of all participants in a massive identical twins convention where everyone has the same mass.

To calculate the molar mass, you'd sum the atomic masses of all the atoms in the molecule. This means multiplying the atomic mass of each element by the number of times it appears in the formula and adding these values together. For example, the molar mass of CoCl鈧冣⒙糎鈧侽 is computed by adding the molar masses of one cobalt atom, three chlorine atoms, and a quarter of a water molecule.

The accurate calculation of molar mass is a pillar of chemical stoichiometry, as it allows you to convert between grams and moles, facilitating the comparison and quantification of substances involved in chemical reactions.
Chemical Reaction Equations
Chemical reaction equations are the shorthand of chemistry; they tell the story of reactants transforming into products. A balanced chemical equation obeys the Law of Conservation of Mass, meaning the number of atoms of each element is the same before and after the reaction.

In our case, the equations elegantly describe the transformation of a hydrated cobalt(III) chloride compound when it reacts with silver nitrate and sodium hydroxide, as well as the subsequent heating to form cobalt(III) oxide. Every step is accounted for, with the reactants on the left side of the arrows and the products on the right.

Writing balanced chemical equations is a fundamental skill in chemistry that aids in predicting the quantities of substances consumed and produced in a reaction. It translates the qualitative description of a chemical reaction into a precise quantitative language that can be used to calculate reactant and product masses, volumes, and even energy changes.

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