/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 119 A sample of \(\mathrm{CuSO}_{4} ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 119

A sample of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\) was heated to \(110^{\circ} \mathrm{C}\), where it lost water and gave another hydrate of copper(II) ion that contains \(32.50 \%\) Cu. A 98.77 -mg sample of this new hydrate gave \(116.66 \mathrm{mg}\) of barium sulfate precipitate when treated with a barium nitrate solution. What is the formula of the new hydrate?

Short Answer

Expert verified
The formula of the new hydrate is \( \mathrm{CuSO}_4 \cdot 0.5 \mathrm{H}_2\mathrm{O} \).

Step by step solution

01

Determine Moles of BaSO鈧�

First, calculate the number of moles of barium sulfate (BaSO鈧�) produced. The molar mass of BaSO鈧� is 233.39 g/mol.Convert the mass from mg to grams: \[ 116.66 \text{ mg} = 0.11666 \text{ g} \]Then calculate the moles:\[ \text{moles of BaSO鈧剗 = \frac{0.11666 \text{ g}}{233.39 \text{ g/mol}} = 0.0004995 \text{ mol} \]
02

Determine Moles of Sulfate Ion

Since each molecule of BaSO鈧� contains one sulfate ion (SO鈧劼测伝), the number of moles of sulfate ion is equal to the number of moles of BaSO鈧�.\[ \text{moles of SO鈧剗^{2-} = 0.0004995 \text{ mol} \]
03

Calculate Moles of Copper in Hydrate

Using the given percentage by mass of copper in the new hydrate (32.50% Cu), first calculate the mass of copper in the sample:\[ \text{mass of Cu} = 0.3250 \times 98.77 \text{ mg} = 32.122 \text{ mg} \]Convert to grams:\[ 32.122 \text{ mg} = 0.032122 \text{ g} \]Then find the number of moles of copper, using the molar mass of Cu (63.546 g/mol):\[ \text{moles of Cu} = \frac{0.032122 \text{ g}}{63.546 \text{ g/mol}} = 0.0005056 \text{ mol} \]
04

Find Moles Ratio of Cu to SO鈧�

The moles of Cu and SO鈧劼测伝 are approximately equal since they are both around 0.0005 mol. This suggests that in the hydrate formula, there is likely a 1:1 ratio, meaning \[ \mathrm{CuSO}_4 \cdot n \mathrm{H}_2\mathrm{O} \] where one copper ion corresponds to one sulfate ion.
05

Identify Hydrate Formula

Since you know the moles of Cu is equal to moles of SO鈧劼测伝, the formula of the hydrate should reflect that each Cu atom pairs with one SO鈧� to make CuSO鈧�. By further empirical evaluation, the remaining weight in the hydrate must account for the water. The molecular weight analysis and expected values point towards the formula \[ \mathrm{CuSO}_4 \cdot 0.5 \mathrm{H}_2\mathrm{O} \], suggesting exactly half a molecule of water per molecule.

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Key Concepts

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

Molar Mass Calculation
In order to carry out many chemical calculations, knowing how to determine the molar mass of compounds is crucial. The molar mass is essentially the weight of one mole of a substance. Take the compound barium sulfate (BaSO鈧�) as an example. Its molar mass is calculated by summing up the atomic masses of all the atoms it contains.
  • Barium (Ba): 137.33 g/mol
  • Sulfur (S): 32.07 g/mol
  • Oxygen (O) for four atoms: 16.00 g/mol 脳 4 = 64.00 g/mol
Add these together to get the molar mass of BaSO鈧�: 137.33 + 32.07 + 64.00 = 233.4 g/mol.
Such precise calculations help determine how many moles are present in a given mass of a substance. In calculations, we convert between mass and moles using the formula:\[\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]
This fundamental conversion allows you to explore further chemical relationships and reactions.
Sulfate Ion Determination
Understanding how to determine the presence and quantity of sulfate ions is a typical procedure in analytical chemistry. In the given exercise, the formation of BaSO鈧� precipitate indicates the presence of sulfate ions, because BaSO鈧� is highly insoluble in water. Each barium sulfate particle contains one sulfate ion ( SO鈧刕{2-} ), allowing us to directly relate the moles of BaSO鈧� to moles of sulfate ion.
For example, when calculating sulfate ion moles, if you find 0.0004995 moles of BaSO鈧� , then there are exactly 0.0004995 moles of SO鈧刕{2-} . This 1:1 relationship helps identify how many sulfate ions are present in a sample simply by determining the mass of BaSO鈧� formed.
Recognizing the formation of such precipitates is a crucial skill for identifying ionic components in solutions.
Mole Ratio Analysis
Mole ratio analysis is an essential tool for deducing chemical formulas from various analytical data. In hydrates such as the copper-based compound in our exercise, it involves comparing the moles of different components, like Cu and SO鈧刕{2-} , to determine the empirical formula of the compound.
In the exercise, the approximate equality of moles for copper and sulfate ions (both around 0.0005 mol) suggests a 1:1 mole ratio. This hints that for each copper atom, a sulfate ion is paired, suggesting the presence of CuSO鈧� as the fundamental unit.
Further, using mass percentage of elements can also involve considering the water in hydrates. For instance, if there is leftover mass unaccounted for after determining SO鈧刕{2-} and Cu, it is typically due to water ( H鈧侽 ) molecules.
Such mole ratio analysis is key for constructing accurate chemical formulas and understanding the structure of compounds.

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Most popular questions from this chapter

Calculate the concentrations of each ion present in a solution that results from mixing \(50.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M}\) \(\mathrm{NaClO}_{3}(a q)\) solution with \(25.0 \mathrm{~mL}\) of a \(0.20 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) solution. Assume that the volumes are additive.

An aqueous solution is made from \(0.798 \mathrm{~g}\) of potassium permanganate, \(\mathrm{KMnO}_{4}\). If the volume of solution is \(50.0 \mathrm{~mL},\) what is the molarity of \(\mathrm{KMnO}_{4}\) in the solution?

How many milliliters of \(0.250 \mathrm{M} \mathrm{KMnO}_{4}\) are needed to react with \(3.55 \mathrm{~g}\) of iron(II) sulfate, \(\mathrm{FeSO}_{4}\) ? The reaction is as follows: $$ \begin{array}{r} 10 \mathrm{FeSO}_{4}(a q)+2 \mathrm{KMnO}_{4}(a q)+8 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \\ 5 \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+2 \mathrm{MnSO}_{4}(a q)+\mathrm{K}_{2} \mathrm{SO}_{4}(a q)+ \\ 8 \mathrm{H}_{2} \mathrm{O}(l) \end{array} $$

What volume of a solution of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O},\) that is \(94.0 \%\) ethanol by mass contains \(0.200 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) ? The density of the solution is \(0.807 \mathrm{~g} / \mathrm{mL}\) 4.150 What volume of a solution of ethylene glycol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\), that is \(56.0 \%\) ethylene glycol by mass contains \(0.350 \mathrm{~mol}\) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2} ?\) The density of the solution is \(1.072 \mathrm{~g} / \mathrm{mL}\)

A 71.2-g sample of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), was dissolved in \(1.00 \mathrm{~L}\) of solution. How would you prepare \(2.50 \mathrm{~L}\) of \(0.150 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\) from this solution?
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