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Chapter 7: Problem 109

A common mineral in Earth's crust has the chemical composition \(34.55 \% \mathrm{Mg}, 19.96 \% \mathrm{Si},\) and \(45.49 \%\) O. What is its empirical formula?

Short Answer

Expert verified
Answer: The empirical formula of the mineral is Mg2SiO4.

Step by step solution

01

Convert percentages to moles

To convert the percentages into moles, we need to assume a total mass for the mineral. Let's assume that we have 100 grams of the mineral. This makes it easy to convert percentages into grams. For 100 grams of the mineral, we have: - 34.55 g of Mg - 19.96 g of Si - 45.49 g of O Now we convert the mass of each element into moles using their respective molar masses: - Mg: \(Molar~Mass = 24.31~g/mol\) - Si: \(Molar~Mass = 28.09~g/mol\) - O: \(Molar~Mass = 16.00~g/mol\)
02

Calculate the moles of each element

Using the molar masses: Moles of Mg = \(\frac{34.55~g}{24.31~g/mol} = 1.421~mol\) Moles of Si = \(\frac{19.96~g}{28.09~g/mol} = 0.711~mol\) Moles of O = \(\frac{45.49~g}{16.00~g/mol} = 2.843~mol\)
03

Determine the mole ratio

To find the mole ratio, divide the moles of each element by the smallest number of moles calculated. Smallest moles = 0.711 mol Mole ratio: Mg = \(\frac{1.421~mol}{0.711~mol} 鈮� 2\) Si = \(\frac{0.711~mol}{0.711~mol} = 1\) O = \(\frac{2.843~mol}{0.711~mol} 鈮� 4\)
04

Deduce the empirical formula

The mole ratio represents the relative number of atoms of each element in the mineral. Therefore, the empirical formula of the mineral is Mg\(_{2}\)SiO\(_{4}\).

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Key Concepts

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

Mole Ratio
The concept of the mole ratio is pivotal when determining empirical formulas. It helps us understand the relative number of moles of each element in a compound. By reducing the moles to the smallest whole numbers, you can express the simplest ratio of the elements. For instance, in our example involving magnesium, silicon, and oxygen, the mole ratios were calculated after finding the moles of each element.
  • Find the smallest number of moles among the elements.
  • Divide the moles of each element by this smallest number.
  • The resulting numbers are used to determine the simplest whole-number ratio.
Breaking the procedure down: magnesium's moles (1.421) divided by the smallest moles (0.711) yields approximately 2, while silicon results in 1, and oxygen gives about 4. This ratio (Mg:Si:O = 2:1:4) tells us how many atoms of each element are present in the simplest form.
Molar Mass
Molar mass is a key factor in converting masses to moles, which is crucial in determining an empirical formula. By understanding an element's molar mass, you can calculate how many moles of the element are in a given sample. The molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). Here's how to use molar mass to convert:
  • Look up the molar mass of each element from the periodic table. For instance, magnesium is 24.31 g/mol, silicon is 28.09 g/mol, and oxygen is 16.00 g/mol.
  • Take the given mass of each element in the compound (from your 100 g assumption) and divide it by the element's molar mass.
  • This will convert the mass to moles, which are necessary for finding the mole ratio.
In our mineral example, using the molar masses allows us to translate the percentage composition into mole quantities, leading to a clearer understanding of the compound's structure.
Chemical Composition
Chemical composition refers to the relative amounts of each element within a compound. It鈥檚 typically expressed as a percentage, showing the proportions of elements by mass. Knowing the chemical composition allows us to find the empirical formula, the simplest version of the compound's formula. To transform chemical composition into useful data:
  • Assume a 100 g sample to directly convert percentage into grams.
  • Use the grams to convert into moles via molar masses, translating mass into a chemical context.
  • The moles are then used to identify the simplest whole-number ratio of elements in the compound, forming the empirical formula.
In practice, chemical composition provides a stepping stone in transitioning from the macro world of percentage and mass to the micro world of moles and formulas.
Atoms to Moles Conversion
Atoms to moles conversion is a fundamental principle in chemistry that bridges the microscopic atomic world with our macroscopic understanding. This conversion is essential for determining how many moles of an element exist in a chemical sample. To convert atoms to moles:
  • Employ Avogadro's number ( (6.022 脳 1023)), which tells us how many atoms are in a mole.
  • When starting with a mass or atomic count, divide by Avogadro's number to find the number of moles.
Although our exercise deals with percentages converting directly to moles, understanding the atom-to-mole concept solidifies why such calculations are possible. It鈥檚 particularly useful when determining how many individual atoms contribute to a given mass, reinforcing the molecular story behind the data.

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Most popular questions from this chapter

Organic Compounds in Space The following compounds have been detected in space. Which contains the greatest percentage of carbon by mass? a. naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\) b. chrysene, \(\mathrm{C}_{18} \mathrm{H}_{12}\) c. pentacene, \(\mathrm{C}_{22} \mathrm{H}_{14}\) d. pyrene, \(C_{16} H_{10}\)

Fiber in the Diet Dietary fiber is a mixture of many compounds, including xylose \(\left(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{5}\right)\) and methyl galacturonate \(\left(\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{7}\right)\) a. Do these compounds have the same empirical formula? b. Write balanced chemical equations for the complete combustion of xylose and methyl galacturonate.

Pigments for Stoplights Cadmium yellow (cadmium sulfide) is a lemon-yellow pigment used in the lenses of stoplights. Its formula is CdS, and it is very insoluble in water. The recommended recipe for cadmium yellow is to mix cadmium nitrate with sodium sulfide in water. The cadmium yellow forms as a solid, while the other product, sodium nitrate, remains dissolved in the water. a. Write a balanced chemical equation for the reaction. b. Calculate the mass of cadmium nitrate you must start with to make 125 g of CdS.

Balance the following reactions for the formation of nitrogen compounds: a. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NO}(g)\) b. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}(g)\) c. \(\mathrm{NO}(g)+\mathrm{NO}_{3}(g) \rightarrow \mathrm{NO}_{2}(g)\) d. \(\mathrm{NO}(g)+\mathrm{O}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{HNO}_{2}(\ell)\)

If a reaction vessel contains equal masses of \(\mathrm{Fe}\) and \(\mathrm{S}, \mathrm{a}\) mass of FeS corresponding to which of the following could theoretically be produced? a. the sum of the masses of \(\mathrm{Fe}\) and \(\mathrm{S}\) b. more than the sum of the masses of Fe and \(S\) c. less than the sum of the masses of Fe and \(S\)
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