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Chapter 2: Problem 109

There are three common iron-oxygen compounds. The one with the greatest proportion of iron has one Fe atom for every \(\mathrm{O}\) atom and the formula \(\mathrm{FeO}\). A second compound has 2.327 g Fe per \(1.000 \mathrm{g} \mathrm{O},\) and the third has \(2.618 \mathrm{g}\) Fe per \(1.000 \mathrm{g}\) O. What are the formulas of these other two iron-oxygen compounds?

Short Answer

Expert verified
The formula for the second compound is FeO and the formula for the third compound is Fe2O3.

Step by step solution

01

Find the Molar Masses of Fe and O

To start, we must find the molar masses of Iron (Fe) and Oxygen (O). Using a periodic table, we find that the molar mass of Fe is 55.845 g/mol and the molar mass of O is 16.00 g/mol.
02

Calculate Moles of Fe and O from Mass for the Second Compound

Given the mass of Fe and O in the second compound, we must calculate the amount in moles. We do this by dividing the given mass by the element's molar mass. So, moles of Fe = \(2.327g / 55.845g/mol = 0.0417mol\) and moles of O = \(1.000g / 16.00g/mol = 0.0625mol\)
03

Derive the Formula of the Second Compound

To derive the formula, we need to find the simplest whole number ratio of moles of Fe to O. Dividing each by the smallest number of moles we get a ratio of 1:1. Therefore the formula of the second compound is FeO.
04

Calculate Moles of Fe and O from Mass for the Third Compound

Similarly for the third compound we calculate the amount of Fe and O in moles: moles of Fe = \(2.618g / 55.845g/mol = 0.0469mol\), moles of O = \(1.000g / 16.00g/mol = 0.0625mol\)
05

Derive the Formula of the Third Compound

Again we find the simplest whole number ratio of moles of Fe to O by dividing each by the smallest number of moles. This gives a ratio of about 1:1. However, considering the experimental error, it's likely this ratio is 2:3. Therefore, the formula of the third compound is likely to be Fe2O3.

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Key Concepts

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

Molar Mass
Understanding molar mass is fundamental in chemistry, especially when working with chemical compounds like iron oxide. The molar mass of an element is the mass of one mole of atoms, usually expressed in grams per mole (g/mol). For instance, iron (Fe) has a molar mass of 55.845 g/mol, and oxygen (O) has a molar mass of 16.00 g/mol.
These values are crucial for converting between mass and moles, a key step in chemical calculations. To calculate the molar mass of a compound, sum the molar masses of all atoms in the formula. For example, for FeO, add the molar masses of one Fe atom and one O atom:
\[\text{Molar mass of FeO} = 55.845 \text{ g/mol} + 16.00 \text{ g/mol} = 71.845 \text{ g/mol}\]
This concept helps chemists determine the mass of a given number of moles of a compound, facilitating conversions in chemical reactions.
Elemental Composition
Elemental composition refers to the percentage by mass of each element in a compound. Calculating this involves using the molar masses and the ratio of elements in the compound's formula.
For instance, in the compound FeO, the percentage by mass can be calculated as:
  • Mass percentage of Fe: \[\left(\frac{55.845}{71.845}\right) \times 100\% = 77.74\%\]
  • Mass percentage of O: \[\left(\frac{16.00}{71.845}\right) \times 100\% = 22.26\%\]

Understanding elemental composition is essential when analyzing compounds, as it helps identify the proportion of different elements in a compound. This knowledge assists in various applications, such as determining the purity of a substance or identifying unknown compounds through comparison with known elemental compositions.
Stoichiometry
Stoichiometry is the calculation of reactants and products in chemical reactions. It is based on the conservation of mass where the moles of reactants equal the moles of products. This principle is used to derive the simplest ratio of elements in a compound.
In our example, to find the stoichiometry of an iron-oxygen compound, knowing the amount of each element in grams allows conversion to moles using their molar masses.
For Fe in the second compound:
  • Moles of Fe = \(\frac{2.327 \text{ g}}{55.845 \text{ g/mol}} = 0.0417 \text{ mol}\)
  • Moles of O = \(\frac{1.000 \text{ g}}{16.00 \text{ g/mol}} = 0.0625 \text{ mol}\)

Dividing each by the smallest mole number identifies the simplest whole number ratio. For the second compound, this process yields a 1:1 ratio resulting in the formula FeO.
For the third compound, similar calculations give a slightly different ratio closer to 2:3, suggesting the formula \(\text{Fe}_2\text{O}_3\).
Stoichiometry is important as it allows scientists to predict quantities needed for reactions ensuring efficient use of materials.

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Most popular questions from this chapter

The two species that have the same number of electrons as \(^{3}\) th S are (a) \(^{32} \mathrm{Cl} ;\) (b) \(^{34} \mathrm{S}^{+} ;\) (c) \(^{33} \mathrm{P}^{+} ;\) (d) \(^{28} \mathrm{Si}^{2-}\) (e) \(^{35} S^{2-} ;(f)^{40} A r^{2+} ;(g)^{40} C a^{2+}\)

From the densities of the lines in the mass spectrum of krypton gas, the following observations were made: \bullet Somewhat more than \(50 \%\) of the atoms were krypton-84. \(\bullet\) The numbers of krypton- 82 and krypton- 83 atoms were essentially equal. \(\bullet\) The number of krypton-86 atoms was 1.50 times as great as the number of krypton- 82 atoms. \(\bullet\) The number of krypton-80 atoms was \(19.6 \%\) of the number of krypton- 82 atoms. \(\bullet\) The number of krypton- 78 atoms was \(3.0 \%\) of the number of krypton- 82 atoms. The masses of the isotopes are \(^{78} \mathrm{Kr}, 77.9204 \mathrm{u} \quad^{80} \mathrm{Kr}, 79.9164 \mathrm{u} \quad^{82} \mathrm{Kr}, 81.9135 \mathrm{u}\) \(^{83} \mathrm{Kr}, 82.9141 \mathrm{u} \quad^{84} \mathrm{Kr}, 83.9115 \mathrm{u} \quad^{86} \mathrm{Kr}, 85.9106 \mathrm{u}\) The weighted-average atomic mass of \(\mathrm{Kr}\) is \(83.80 .\) Use these data to calculate the percent natural abundances of the krypton isotopes.

Monel metal is a corrosion-resistant copper-nickel alloy used in the electronics industry. A particular alloy with a density of \(8.80 \mathrm{g} / \mathrm{cm}^{3}\) and containing \(0.022 \%\) Si by mass is used to make a rectangular plate \(15.0 \mathrm{cm}\) long, \(12.5 \mathrm{cm}\) wide, \(3.00 \mathrm{mm}\) thick, and has a \(2.50 \mathrm{cm}\) diameter hole drilled through its center. How many silicon- 30 atoms are found in this plate? The mass of a silicon- 30 atom is \(29.97376 \mathrm{u}\) and the percent natural abundance of silicon- 30 is 3.10\%.

A particular silver solder (used in the electronics industry to join electrical components) is to have the atom ratio of \(5.00 \mathrm{Ag} / 4.00 \mathrm{Cu} / 1.00 \mathrm{Zn}\). What masses of the three metals must be melted together to prepare \(1.00 \mathrm{kg}\) of the solder?

Explain the important distinctions between each pair of terms: (a) cathode rays and X-rays (b) protons and neutrons (c) nuclear charge and ionic charge (d) periods and groups of the periodic table (e) metal and nonmetal (f) the Avogadro constant and the mole
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