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Indium oxide contains \(4.784 \mathrm{g}\) of indium for every \(1.000 \mathrm{g}\) of oxygen. In \(1869,\) when Mendeleev first presented his version of the periodic table, he proposed the formula \(\operatorname{In}_{2} \mathrm{O}_{3}\) for indium oxide. Before that time it was thought that the formula was InO. What values for the atomic mass of indium are obtained using these two formulas? Assume that oxygen has an atomic mass of \(16.00 .\)

Short Answer

Expert verified
The values for the atomic mass of indium for the two proposed formulas are: For InO: \(4.784 g/mol\) For In2O3: \(7.176 g/mol\)

Step by step solution

01

Find Mass of Oxygen in 1 Mole of Each Formulation

For each potential formula of indium oxide, we need to find the corresponding mass of oxygen. For InO, there is only 1 mole of oxygen (O) present. For In2O3, there are 3 moles of oxygen (O) present.
02

Calculate Mass of Indium for 1 Mole of Oxygen

We are given the ratio of indium to oxygen in indium oxide, i.e., \(4.784 g\) of indium for every \(1.000 g\) of oxygen. Mass of Indium for InO (1 mole of oxygen) = \(4.784 * 1 = 4.784 g\) Mass of Indium for In2O3 (3 moles of oxygen) = \(4.784 * 3 = 14.352 g\)
03

Find Number of Moles of Indium for Each Formulation

For each potential formula of indium oxide, we need to find the number of moles of indium (In). For InO, there is only 1 mole of indium (In) present. For In2O3, there are 2 moles of indium (In) present.
04

Calculate the Atomic Mass of Indium for Each Formulation

Now that we have the mass and the number of moles for indium in each formulation, we can calculate the atomic mass of indium for each formulation using the formula: Atomic mass of Indium = Mass of Indium in the Compound / Number of moles of Indium in the Compound Atomic mass of Indium for InO = \(4.784 g / 1 mole = 4.784 g/mol\) Atomic mass of Indium for In2O3 = \(14.352 g / 2 moles = 7.176 g/mol\) Thus, the values for the atomic mass of indium for the two proposed formulas are: For InO: \(4.784 g/mol\) For In2O3: \(7.176 g/mol\)

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Key Concepts

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

Indium Oxide Formula
The formula of indium oxide has evolved over time based on advancements in chemical knowledge. Originally, scientists thought the formula was simply InO, meaning each molecule consisted of one atom of indium and one atom of oxygen. This changed when Mendeleev, a pioneering chemist, suggested a new formula:
  • InO: One atom of indium and one atom of oxygen.
  • In2O3: Two atoms of indium and three atoms of oxygen.
Based on chemical ratios given in the problem, for every 1 gram of oxygen, there are 4.784 grams of indium. This allows us to calculate the hypothetical atomic masses used in these formulas.
As our understanding of chemistry improved, In2O3 became widely accepted because it accounts for the stoichiometry that correctly matches with the periodic table's later developments.
Mendeleev's Periodic Table
Mendeleev's periodic table was a groundbreaking advancement in chemistry, offering a systematic way to organize elements based on their atomic weights and properties. Back in 1869, Dmitri Mendeleev's version did not just categorize elements but also predicted the existence and properties of undiscovered elements.
This periodic arrangement helped him correctly propose formulas like In2O3 for indium oxide. However, the significant contribution of Mendeleev's table lies in its ability to order elements logically by properties and predict chemical behaviors.
For indium oxide, Mendeleev's foresight into the compound's structure illustrated the power of his periodic table in influencing the atomic mass calculations that follow from accurate molecular formulas.
Chemical Stoichiometry
Chemical stoichiometry is the pillar of understanding how compounds form and how their constituent parts relate to each other. It involves calculating the ratios of elements in compounds to predict the outcomes of chemical reactions.
  • When calculating the atomic mass of elements in a compound, stoichiometry is crucial.
  • It explains how 4.784g of indium combines with 1.000g of oxygen to form indium oxide correctly.
  • This ratio is essential to determine which formula - either InO or In2O3 - accurately represents the compound.
Thus, stoichiometry not only guides these calculations but ensures that the atomic mass of indium is interpreted correctly based on the number of atoms present as suggested by the stoichiometric coefficients in a balanced equation.
Molecular Formulas
Molecular formulas provide insight into the exact number of each type of atom in a molecule, which directly affects calculating atomic masses. They are key to understanding the composition and properties of chemical compounds.
Considering the formulas in this example, InO and In2O3 suggest different compositions:
  • InO indicates a simpler, potentially less accurate representation based on earlier assumptions.
  • In2O3 gives a more complex and accurate depiction of the bond structure and masses involved.
With the correct molecular formulas, we achieve precise calculations of atomic masses, which then allow scientists to accurately understand and predict chemical reactions and compound formation involving elements like indium.

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Most popular questions from this chapter

You have two distinct gaseous compounds made from element \(\mathrm{X}\) and element Y. The mass percents are as follows: Compound I: \(30.43 \%\) X, \(69.57 \%\) Y Compound II: \(63.64 \% \mathrm{X}, 36.36 \% \mathrm{Y}\) In their natural standard states, element X and element Y exist as gases. (Monatomic? Diatomic? Triatomic? That is for you to determine.) When you react "gas X" with "gas Y" to make the products, you get the following data (all at the same pressure and temperature): 1 volume "gas \(\mathrm{X}^{\prime \prime}+2\) volumes "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound I 2 volumes "gas \(\mathrm{X}^{\prime \prime}+1\) volume "gas \(\mathrm{Y}^{\prime \prime} \longrightarrow\) 2 volumes compound II Assume the simplest possible formulas for reactants and products in the chemical equations above. Then, determine the relative atomic masses of element \(X\) and element Y.

What number of protons and neutrons is contained in the nucleus of each of the following atoms? Assuming each atom is uncharged, what number of electrons is present? a. \(\frac{235}{92} \mathrm{U}\) b. \(_{13}^{2} \mathrm{Al}\) c. \(\frac{57}{26} \mathrm{Fe}\) d. \(\frac{208}{82} \mathrm{Pb}\) e. \(\frac{86}{32} \mathrm{Rb}\) f. \(\frac{41}{20} \mathrm{Ca}\)

An element's most stable ion has a \(2+\) charge. If the ion of element \(\mathrm{X}\) has a mass number of 230 and has 86 electrons, what is the identity of the element, and how many neutrons does it have?

This problem is designed to incorporate several concepts and techniques into one situation. You have gone back in time and are working with Dalton on a table of relative masses. Following are his data. \(0.602 \mathrm{g}\) gas A reacts with \(0.295 \mathrm{g}\) gas \(\mathrm{B}\) \(0.172 \mathrm{g}\) gas \(\mathrm{B}\) reacts with \(0.401 \mathrm{g}\) gas \(\mathrm{C}\) \(0.320 \mathrm{g}\) gas \(\mathrm{A}\) reacts with \(0.374 \mathrm{g}\) gas \(\mathrm{C}\) a. Assuming simplest formulas \((\mathrm{AB}, \mathrm{BC}, \text { and } \mathrm{AC}),\) construct a table of relative masses for Dalton. b. Knowing some history of chemistry, you tell Dalton that if he determines the volumes of the gases reacted at constant temperature and pressure, he need not assume simplest formulas. You collect the following data: 6 volumes gas \(A+1\) volume gas \(B \rightarrow 4\) volumes product 1 volume gas \(\mathrm{B}+4\) volumes gas \(\mathrm{C} \rightarrow 4\) volumes product 3 volumes gas \(A+2\) volumes gas \(C \rightarrow 6\) volumes product Write the simplest balanced equations, and find the actual relative masses of the elements. Explain your reasoning.

Dalton assumed that all atoms of the same element were identical in all their properties. Explain why this assumption is not valid.

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