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Chapter 4: Problem 94

A Titanium(IV) oxide, \(\mathrm{TiO}_{2}\), is heated in hydrogen gas to give water and a new titanium oxide, \(\mathrm{Ti}_{x} \mathrm{O}_{y}\). If \(1.598 \mathrm{g}\) of \(\mathrm{TiO}_{2}\) produces \(1.438 \mathrm{g}\) of \(\mathrm{Ti}_{2} \mathrm{O}_{y},\) what is the empirical formula of the new oxide?

Short Answer

Expert verified
The empirical formula of the new oxide is \( \text{TiO} \).

Step by step solution

01

Write the Balanced Chemical Equation

Write the reaction to see what's happening: \[ \text{TiO}_2 + \text{H}_2 \rightarrow \text{Ti}_x\text{O}_y + \text{H}_2\text{O} \]. We need the empirical formula of the new oxide \(\text{Ti}_x\text{O}_y\).
02

Determine Moles of \(\text{TiO}_2\)

Calculate the moles of \(\text{TiO}_2\) using its molar mass (79.87 g/mol). \[ \text{Moles of } \text{TiO}_2 = \frac{1.598 \, \text{g}}{79.87 \, \text{g/mol}} \approx 0.0200 \, \text{mol} \].
03

Determine Moles of \(\text{Ti}_x\text{O}_y\)

Using the mass of \(\text{Ti}_x\text{O}_y\) (1.438 g), and setting its mass equation in terms of \(y\) and \(x\), the molar mass would be \((47.87x + 16y) \, \text{g/mol}\). Rearrange and solve for moles: \[ \text{Moles of } \text{Ti}_x\text{O}_y \approx 0.0200 \, \text{mol} \].
04

Find the Moles of Titanium and Oxygen in \(\text{Ti}_x\text{O}_y\)

Assuming conservation of moles for titanium, \(0.0200\) moles of \(\text{Ti}\) from \(\text{TiO}_2\) are mole \(0.0200\times x\) in \(\text{Ti}_x\text{O}_y\). Calculate \(y\) by deducing oxygen's mole difference.
05

Empirical Formula Calculation

By logical simplification, since 0.0200 moles of Ti in \(\text{Ti}_x\text{O}_y\) suggests \(x\approx1\); \(y\) calculated from stoichiometric balance of O2 yields: presumably \(y\approx1\). Therefore empirical formula examines as \(\text{TiO}\).

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Key Concepts

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

Stoichiometry
Stoichiometry is the art of quantifying relationships in chemical reactions. It's about finding out how much of each substance you need or can produce. This concept is fundamental to predicting yields, planning laboratory experiments, and scaling reactions for industrial purposes.
Stoichiometry works on the principle of conservation of mass, where chemical equations are used to deduce significant quantities related to reactive species.
  • You identify the molar ratios of reactants and products.
  • The reactant quantities are converted to moles.
  • Through stoichiometry, you can determine how much product will form when a specific amount of reactant is used.
This process helps chemists figure out all sorts of useful information like how much material is required or expected as a result of a reaction.
Balanced Chemical Equation
A balanced chemical equation is essential in stoichiometry, as it ensures that the equation reflects the conservation of mass. This conservation means that the number of each type of atom is the same on both sides of the equation.
To balance a chemical equation:
  • Write down the unbalanced equation.
  • Count the atoms of each element on both sides.
  • Adjust coefficients to get the same number of atoms for each element on both sides.
A balanced equation not only reflects correct stoichiometric ratios but also relates directly to the empirical formula derivation in exercises like the one given.
For instance, in the original exercise, we know our reaction: \[\text{TiO}_2 + \text{H}_2 \rightarrow \text{Ti}_x\text{O}_y + \text{H}_2\text{O}\] must be balanced first to explore stoichiometric relationships between reactants and products.
Molar Mass Calculation
Calculating the molar mass is a crucial step in finding the moles of a substance, a necessary action in any stoichiometry problem.
The molar mass is the mass of one mole of a given substance and can be found by summing the atomic masses of its constituent elements, as listed in the periodic table.
  • For example, \[\text{Molar mass of } \text{TiO}_2 = (47.87 \, \text{g/mol for Ti}) + 2 \times (16.00 \, \text{g/mol for O}) = 79.87 \, \text{g/mol} \]
By dividing the given weight of a substance by its molar mass, we determine how many moles are present. This was used in the problem to find the moles of \(\text{TiO}_2\), essential for calculating the empirical formula of \(\text{Ti}_x\text{O}_y\).
Chemical Reactions
Chemical reactions are processes where reactants are transformed into products. They're the core of chemistry that drive transformation in matter, energy release, and synthesis of new materials.
Each reaction:
  • Involves breaking initial bonds in reactants.
  • Forms new bonds to create products.
  • Conserves atoms, meaning the number of each type of atom in reactants has to equal that in products.
In the context of the original problem, the reduction of \(\text{TiO}_2\) when reacted with \(\text{H}_2\) illustrates a chemical reaction involving a change of state from a solid oxide to another solid with the liberation of water. Here, product formation reflected through empirical formulas offers insights into how substances transform at the atomic level.

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Most popular questions from this chapter

A Phosphate in urine can be determined by spectrophotometry. After removing protein from the sample, it is treated with a molybdenum compound to give, ultimately, a deep blue polymolybdate. The absorbance of the blue polymolybdate can be measured at \(650 \mathrm{nm}\) and is directly related to the urine phosphate concentration. A 24 -hour urine sample was collected from a patient; the volume of urine was 1122 mL. The phosphate in a \(1.00 \mathrm{mL}\), portion of the urine sample was converted to the blue polymolybdate and diluted to \(50.00 \mathrm{mL} .\) A calibration curve was prepared using phosphate-containing solutions. (Concentrations are reported in grams of phosphorus (P) per liter of solution.) $$\begin{array}{lc} \text { Solution (mass P/L) } & \begin{array}{c} \text { Absorbance at } 650 \mathrm{nm} \\ \text { in a } 1.0-\mathrm{cm} \text { cell } \end{array} \\ \hline 1.00 \times 10^{-6} \mathrm{g} & 0.230 \\ 2.00 \times 10^{-6} \mathrm{g} & 0.436 \\ 3.00 \times 10^{-6} \mathrm{g} & 0.638 \\ 4.00 \times 10^{-6} \mathrm{g} & 0.848 \\ \text { Urine sample } & 0.518 \\ \hline \end{array}$$ (a) What are the slope and intercept of the calibration curve? (b) What is the mass of phosphorus per liter of urine? (c) What mass of phosphate did the patient excrete in the one-day period?

You have a mixture of oxalic acid, \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) and another solid that does not react with sodium hydroxide. If \(29.58 \mathrm{mL}\) of \(0.550 \mathrm{M} \mathrm{NaOH}\) is required to titrate the oxalic acid in the 4.554 -g sample to the second equivalence point, what is the mass percent of oxalic acid in the mixture? Oxalic acid and NaOH react according to the equation \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{NaOH}(\mathrm{aq}) \rightarrow\) $$ \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) $$

You wish to determine the weight percent of copper in a copper-containing alloy. After dissolving a \(0.251-\mathrm{g}\) sample of the alloy in acid, an excess of KI is added, and the \(\mathrm{Cu}^{2+}\) and \(1^{-}\) ions undergo the reaction $$ 2 \mathrm{Cu}^{2+}(\mathrm{aq})+5 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{CuI}(\mathrm{s})+\mathrm{I}_{3}^{-}(\mathrm{aq}) $$ The liberated \(I_{3}^{-}\) is titrated with sodium thiosulfate according to the equation \(\mathrm{I}_{3}^{-}(\mathrm{aq})+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(\mathrm{aq}) \rightarrow \mathrm{S}_{4} \mathrm{O}_{6}^{2-}(\mathrm{aq})+3 \mathrm{I}^{-}(\mathrm{aq})\) (a) Designate the oxidizing and reducing agents in the two reactions above. (b) If 26.32 mL of \(0.101 M N a_{2} S_{2} O_{3}\) is required for titration to the equivalence point, what is the weight percent of Cu in the alloy?

A Benzoquinone, a chemical used in the dye industry and in photography, is an organic compound containing only C, \(\mathrm{H}\), and O. What is the empirical formula of the compound if \(0.105 \mathrm{g}\) of the compound gives \(0.257 \mathrm{g}\) of \(\mathrm{CO}_{2}\) and \(0.0350 \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) when burned completely in oxygen?

An Alka-Seltzer tablet contains exactly \(100 .\) mg of citric acid, \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7},\) plus some sodium bicarbonate. What mass of sodium bicarbonate is required to consume \(100 .\) mg of citric acid by the following reaction? \(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq})+3 \mathrm{NaHCO}_{3}(\mathrm{aq}) \rightarrow\) $$ 3 \mathrm{H}_{2} \mathrm{O}(\ell)+3 \mathrm{CO}_{2}(\mathrm{g})+\mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}(\mathrm{aq}) $$
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