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The "alum" used in cooking is potassium aluminum sulfate hydrate, \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} \cdot x \mathrm{H}_{2} \mathrm{O} .\) To find the value of \(x,\) you can heat a sample of the compound to drive off all of the water and leave only \(\mathrm{KAl}\left(\mathrm{SO}_{4}\right)_{2} .\) Assume you heat \(4.74 \mathrm{g}\) of the hydrated compound and that the sample loses \(2.16 \mathrm{g}\) of water. What is the value of \(x ?\)

Short Answer

Expert verified
The value of \(x\) is 12.

Step by step solution

01

Determine Mass of Anhydrate

First, find the mass of the anhydrous compound after the water has been driven off. This can be done by subtracting the mass of water lost from the initial mass of the hydrated compound. \[\text{Mass of anhydrate} = 4.74\, \text{g} - 2.16\, \text{g} = 2.58\, \text{g}\]
02

Calculate Moles of Anhydrate

Next, calculate the moles of the anhydrous compound \(\text{KAl(SO}_4)_2\). Use the molar mass provided in the periodic table:\[\text{Molar Mass of } \text{KAl(SO}_4)_2 = K(39.10) + Al(26.98) + 2\times(S(32.07) + 4\times O(16.00))\]\[= 39.10 + 26.98 + 2 \times (32.07 + 64.00) = 258.21 \, \text{g/mol}\]Now:\[\text{Moles of anhydrate} = \frac{2.58\, \text{g}}{258.21\, \text{g/mol}} = 0.00999 \, \text{mol}\]
03

Calculate Moles of Water Lost

Calculate the moles of water that were lost. Use the molar mass of water (\(18.02\, \text{g/mol}\)):\[\text{Moles of } \text{H}_2\text{O} = \frac{2.16\, \text{g}}{18.02\, \text{g/mol}} = 0.1199 \, \text{mol}\]
04

Determine Value of x

Now find the value of \(x\) by dividing the moles of water lost by the moles of the anhydrate:\[x = \frac{0.1199 \, \text{mol}}{0.00999 \, \text{mol}} = 12.00\]Therefore, the value of \(x\) is 12. This indicates there are 12 water molecules per formula unit of the compound.

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Key Concepts

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

Molar Mass Calculation
Molar mass calculation is a fundamental concept in chemistry, crucial for converting between grams and moles of a substance. To perform a molar mass calculation, you'll need to refer to the periodic table, where each element's atomic mass is listed. The molar mass is calculated by adding up the atomic masses of all the atoms in a compound's formula.
For example, with potassium aluminum sulfate \((\mathrm{KAl(SO}_4)_2)\), you would find the molar mass by summing:
  • the atomic mass of potassium (K): 39.10
  • the atomic mass of aluminum (Al): 26.98
  • two sulfate ions \((\mathrm{SO}_4)\)\(\Rightarrow 2 \times(32.07 + (4 \times 16.00))\)
This results in a molar mass of 258.21 g/mol for \(\mathrm{KAl(SO}_4)_2 \). Understanding how to calculate molar mass is essential for converting a substance's mass in grams to moles, which is often necessary for stoichiometric calculations.
Water of Crystallization
Water of crystallization refers to water molecules that are part of a compound's crystalline structure. It is commonly seen in hydrates, where water molecules are integrated into the crystal lattice.
A classic example is found in the hydrated form of potassium aluminum sulfate, where water molecules are a key part of the compound structure. As you determine the hydrate formula, you are essentially finding out how many water molecules are associated with each formula unit of the compound.
In practice, this might involve heating the hydrate to remove the water and analyzing the change in mass to understand the role water played in the crystalline structure. By losing water upon heating, one can determine the amount of water initially present, thus indicating how many water molecules were involved in the formation of the crystalline structure, which is crucial for determining the \(x\) value in formulas such as \(\mathrm{KAl(SO}_4)_2 \cdot x \mathrm{H}_2\mathrm{O} \). This process helps understand the stoichiometry and consumption of water molecules in such compounds.
Anhydrous Compound Analysis
Analyzing an anhydrous compound involves studying a substance that has had all its water of crystallization removed. This condition allows us to focus on the "dry" part of the compound, without the influence of water content.
When dealing with hydrated crystals, such as \(\mathrm{KAl(SO}_4)_2 \cdot x \mathrm{H}_2\mathrm{O}\), the goal of heating is to drive off the water, leaving you with an anhydrous form. By measuring the weight before and after heating, you can determine the mass of the anhydrous compound.
  • Start with the initial mass of the hydrate.
  • Subtract the mass of water lost to find the mass of the anhydrate.
This subtraction yields the mass of the anhydrous compound, crucial for further calculations like establishing the molar ratio between the anhydrous compound and the water of crystallization. Such analyses not only aid in determining the value of \(x\) in hydrates but also provide insights into the stability and composition of the anhydrous form compared to its hydrated counterpart.

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Most popular questions from this chapter

Crossword Puzzle: In the \(2 \times 2\) box shown here, each answer must be correct four ways: horizontally, vertically, diagonally, and by itself. Instead of words, use symbols of elements. When the puzzle is complete, the four spaces will contain the overlapping symbols of 10 elements. There is only one correct solution. $$\begin{array}{|l|l|}\hline 1 & 2 \\\\\hline 3 & 4 \\\\\hline\end{array}$$ Horizontal \(1-2:\) two-letter symbol for a metal used in ancient times \(3-4:\) two-letter symbol for a metal that burns in air and is found in Group \(5 \mathrm{A}\) Vertical \(1-3:\) two-letter symbol for a metalloid \(2-4:\) two-letter symbol for a metal used in U.S. coins single squares: All one-letter symbols 1: a colorful nonmetal 2: colorless, gaseous nonmetal 3: an element that makes fireworks green 4: an element that has medicinal uses Diagonal \(1-4:\) two-letter symbol for an element used in electronics \(2-3:\) two-letter symbol for a metal used with Zr to make wires for superconducting magnets This puzzle first appeared in Chemical \& Engineering News, p. \(86,\) December 14,1987 (submitted by S. J. Cyvin) and in Chem Matters, October 1988.

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