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Chapter 22: Problem 89

When aluminum replaces up to half of the silicon atoms in \(\mathrm{SiO}_{2},\) a mineral class called feldspars result. The feldspars are the most abundant rock-forming minerals, comprising about 50\(\%\) of the minerals in Earth's crust. Orthoclase is a feldspar in which Al replaces one-fourth of the Si atoms of \(\mathrm{SiO}_{2},\) and charge balance is completed by \(\mathrm{K}^{+}\) ions. Determine the chemical formula for orthoclase.

Short Answer

Expert verified
The chemical formula for orthoclase is \( \mathrm{K_3AlSi_4O_{10}} \).

Step by step solution

01

Identify the moles of silicon in SiO2

In 1 mole of SiO2, there are 1 mole of silicon(Si) atoms and 2 moles of Oxygen(O) atoms. The chemical formula is given as: \[ \mathrm{SiO}_{2} \]
02

Calculate the amounts of Si and Al atoms found in orthoclase

It is given that Al replaces one-fourth of the Si atoms in orthoclase. In order to maintain the proportions, we have: \[ \frac{\mathrm{Moles\,of\,Al}}{\mathrm{Moles\,of\,Si}} = \frac{1}{4} \]
03

Determine the number of moles of K+ ions needed to balance charges

In orthoclase, the charge balance is completed by K+ ions. Since Si has a 4+ charge, Al has a 3+ charge, and O has a 2- charge, we need to find the number of moles of K+ ions with a 1+ charge to balance the charges. Let x be the number of moles of Al and y be the number of moles of Si. We can write the equation representing charge balance: \[ 3x + 4y - 2(2+y+x) = 0 \]
04

Solve the equation to derive the chemical formula for orthoclase

Using the ratio we derived in step 2 and the charge balance equation from step 3, we can solve for x and y: From the ratio in step 2: \[ x = \frac{y}{4} \] Substitute this value into the charge balance equation: \[ 3\left(\frac{y}{4}\right) + 4y - 2(2+y+\frac{y}{4}) = 0 \] Solve for y: \[ y = 4 \] Now, find the value of x: \[ x = \frac{4}{4} = 1 \] Thus, we have 4 moles of Si, 1 mole of Al and 5 moles of O atoms in orthoclase. Since K+ ions are needed to balance the charges, and 1 mole of K+ ions balance 1 mole of Al with a 3+ charge, we need 3 moles of K+ ions in orthoclase. The chemical formula for orthoclase is therefore: \[ \mathrm{K_3AlSi_4O_{10}} \]

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Key Concepts

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

Chemical Formula Derivation
Understanding how to derive the chemical formula for minerals such as orthoclase is crucial for comprehending the intricate nature of mineral structures. The derivation begins by exploring the fundamental building blocks of the mineral, which are the atoms making up its composition. In the case of orthoclase, a type of feldspar, it's essential to know the replacement ratio of aluminum (Al) to silicon (Si) and how potassium (K) ions help maintain electrical neutrality.

In orthoclase, aluminum replaces one-fourth of the silicon atoms; hence, for every four silicon atoms, there is one aluminum atom. This replacement affects the overall charge of the molecule as Al has a +3 charge, compared to Si's +4. To compensate and achieve charge neutrality, potassium ions with a +1 charge are added. The amount of potassium is directly proportional to the amount of substituted aluminum to maintain overall charge balance. By balancing the total positive and negative charges, we can derive a plausible chemical formula for orthoclase, an integral part in understanding the mineral's chemistry.
Aluminum Substitution in Silicates
Silicate minerals, such as feldspars, have a rich chemistry largely due to the ability of aluminum to substitute for silicon in the crystal lattice. This process, known as 'aluminum substitution,' is a common phenomenon in rock-forming minerals.

The substitution occurs because aluminum and silicon have a similar size and both can form tetrahedral units with oxygen. However, since aluminum has a charge of +3 and silicon has a charge of +4, the substitution alters the charge balance within the mineral.

Importance of Tschermak's Substitution

A notable example of this is Tschermak's substitution, where an aluminum atom replaces a silicon atom within the tetrahedral framework of the mineral; this is partially what defines the feldspar group. One implication of aluminum substitution is that it introduces the need for extra cations like potassium, sodium, or calcium to counterbalance the reduction in positive charge.
Charge Balance in Minerals
Charge balance is a fundamental principle in mineral chemistry, ensuring that the total positive charge of cations equals the total negative charge of anions within a mineral structure. This balance is critical in maintaining the stability and electrical neutrality of minerals.

When aluminum substitutes for silicon in minerals like feldspar, it introduces a charge imbalance due to its +3 charge, compared to silicon's +4. To correct this, complimentary cations, such as potassium in orthoclase, are added to the structure. The number of these balancing cations is carefully calculated to offset the deficiency in positive charge incurred by aluminum substitution, ensuring the compound remains electrostatically neutral.

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Most popular questions from this chapter

In aqueous solution, hydrogen sulfide reduces (a) \(\mathrm{Fe}^{3+}\) to \(\mathrm{Fe}^{2+},(\mathbf{b}) \mathrm{Br}_{2}\) to \(\mathrm{Br}^{-},(\mathbf{c}) \mathrm{MnO}_{4}^{-}\) to \(\mathrm{Mn}^{2+},(\mathbf{d}) \mathrm{HNO}_{3}\) to \(\mathrm{NO}_{2}\) . In all cases, under appropriate conditions, the product is elemental sulfur. Write a balanced net ionic equation for each reaction.

(a) What is the oxidation state of \(P\) in \(P O_{4}^{3-}\) and of \(N\) in \(N O_{3}^{-} ?(\mathbf{b})\) Why doesn't \(N\) form a stable \(N O_{4}^{3-}\) analogous to P?

The dissolved oxygen present in any highly pressurized, high-temperature steam boiler can be extremely corrosive to its metal parts. Hydrazine, which is completely miscible with water, can be added to remove oxygen by reacting with it to form nitrogen and water. (a) Write the balanced equation for the reaction between gaseous hydrazine and oxygen.(b) Calculate the enthalpy change accompanying this reaction. (c) Oxygen in air dissolves in water to the extent of 9.1 ppm at \(20^{\circ} \mathrm{C}\) at sea level. How many grams of hydrazine are required to react with all the oxygen in 3.0 \(\times 10^{4} \mathrm{L}\) (the volume of a small swimming pool) under these conditions?

Indicate whether each of the following statements is true or false (a) \(\mathrm{H}_{2}(g)\) and \(\mathrm{D}_{2}(g)\) are allotropic forms of hydrogen. (b) \(\mathrm{ClF}_{3}\) is an interhalogen compound. (c) MgO(s) is an acidic anhydride. (d) \(\mathrm{SO}_{2}(g)\) is an acidic anhydride. (e) \(2 \mathrm{H}_{3} \mathrm{PO}_{4}(l) \rightarrow \mathrm{H}_{4} \mathrm{P}_{2} \mathrm{O}_{7}(l)+\mathrm{H}_{2} \mathrm{O}(g)\) is an example of a condensation reaction. (f) Tritium is an isotope of the element hydrogen. (g) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)\) is an example of a disproportionation reaction.

Consider the elements Li, \(\mathrm{K}, \mathrm{Cl}, \mathrm{C}, \mathrm{Ne},\) and Ar. From this list, select the element that (a) is most electronegative, (b) has the greatest metallic character, (c) most readily forms a positive ion, (d) has the smallest atomic radius, (e) forms \(\pi\) bonds most readily, (f) has multiple allotropes.
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