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Chapter 2: Problem 11

How does Dalton's atomic theory account for the fact that when \(1.000 \mathrm{~g}\) of water is decomposed into its elements, \(0.111 \mathrm{~g}\) of hydrogen and \(0.889 \mathrm{~g}\) of oxygen are obtained regardless of the source of the water?

Short Answer

Expert verified
Dalton's atomic theory accounts for the constant mass ratio of hydrogen and oxygen in the decomposition of water by stating that compounds are formed by the combination of atoms of different elements in simple, whole-number ratios, and chemical reactions involve the rearrangement of atoms without creation or destruction. Therefore, the mass ratio of hydrogen to oxygen in water (approximately 1:8) will always remain the same, regardless of the source of the water.

Step by step solution

01

Review the key points of Dalton's atomic theory

Dalton postulated that: 1. All matter is composed of indivisible atoms. 2. Atoms of the same element are identical in mass and properties; atoms of different elements have different masses and properties. 3. Compounds are formed by the combination of atoms of different elements in simple, whole-number ratios. 4. Chemical reactions occur when atoms are rearranged, but no atoms are created or destroyed during the process.
02

Determine the mass ratio of hydrogen and oxygen in water

According to the exercise, 1.000 g of water decomposes into 0.111 g of hydrogen and 0.889 g of oxygen. To find the mass ratio of hydrogen to oxygen in water, divide the mass of hydrogen by the mass of oxygen: \[ \frac{0.111\,\text{g}}{0.889\,\text{g}} \approx 0.125 \] This means that the mass ratio of hydrogen to oxygen in water is approximately 1:8.
03

Relate the mass ratio to Dalton's atomic theory

According to Dalton's atomic theory, compounds are formed by the combination of atoms of different elements in simple, whole-number ratios. In the case of water, this ratio is 1:8 for hydrogen and oxygen, which shows that there are two hydrogen atoms for every oxygen atom in water (since the atomic mass of hydrogen is about 1 and that of oxygen is 16, this ratio is 2:16). As Dalton's atomic theory also states that atoms are neither created nor destroyed during chemical reactions, it means that when water decomposes, the mass ratio of hydrogen to oxygen should stay the same, regardless of the source of the water.
04

Conclusion

Dalton's atomic theory accounts for the fact that when 1.000 g of water is decomposed into its elements, 0.111 g of hydrogen and 0.889 g of oxygen are obtained regardless of the source of the water. This is because compounds are formed by the combination of atoms of different elements in simple, whole-number ratios, and chemical reactions only involve the rearrangement of atoms with no creation or destruction of atoms. Hence, the mass ratio of hydrogen to oxygen in water will always remain the same, as observed in this exercise.

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Most popular questions from this chapter

Predict the chemical formulas of the compounds formed by the following pairs of ions: (a) \(\mathrm{Cr}^{3+}\) and \(\mathrm{Br}^{-}\), (b) \(\mathrm{Fe}^{3+}\) and \(\mathrm{O}^{2-}\), (c) \(\mathrm{Hg}_{2}{ }^{2+}\) and \(\mathrm{CO}_{3}{ }^{2-}\), (d) \(\mathrm{Ca}^{2+}\) and \(\mathrm{ClO}_{3}^{-}\), (e) \(\mathrm{NH}_{4}^{+}\)and \(\mathrm{PO}_{4}{ }^{3-}\).

From the following list of elements-Ar, H, Ga, Al, Ca, Br, \(\mathrm{Ge}, \mathrm{K}, \mathrm{O}\) - pick the one that best fits each description. Use each element only once: (a) an alkali metal, (b) an alkaline earth metal, (c) a noble gas, (d) a halogen, (e) a metalloid, (f) a nonmetal listed in group \(1 \mathrm{~A},(\mathrm{~g})\) a metal that forms a \(3+\) ion, (h) a nonmetal that forms a \(2-\) ion, (i) an element that resembles aluminum.

Predict the chemical formula for the ionic compound formed by (a) \(\mathrm{Ca}^{2+}\) and \(\mathrm{Br}^{-}\), (b) \(\mathrm{K}^{+}\)and \(\mathrm{CO}_{3}^{2-}\), (c) \(\mathrm{Al}^{3+}\) and \(\mathrm{CH}_{3} \mathrm{COO}^{-}\), (d) \(\mathrm{NH}_{4}^{+}\)and \(\mathrm{SO}_{4}^{2-}\), (c) \(\mathrm{Mg}^{2+}\) and \(\mathrm{PO}_{4}{ }^{3-}\).

Fill in the gaps in the following table, assuming each column represents a neutral atom. $$ \begin{array}{l|c|c|c|c|c} \hline \text { Symbol } & { }^{159} \mathrm{~Tb} & & & & \\ \text { Protons } & & 29 & & & 37 \\ \text { Neutrons } & & 34 & 53 & & \\ \text { Electrons } & & & 42 & 34 & \\ \text { Mass no. } & & & & 79 & 85 \\ \hline \end{array} $$

Write the chemical formulas for the following compounds: (a) aluminum hydroxide, (b) potassium sulfate, (c) copper(I) oxide, (d) zinc nitrate, (e) mercury(II) bromide, (f) iron(III) carbonate, (g) sodium hypobromite.
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