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Chapter 3: Problem 127

A hydrate of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) contains almost exactly \(50 \% \mathrm{H}_{2} \mathrm{O}\) by mass. What is the formula of this hydrate?

Short Answer

Expert verified
The formula of the sodium sulfite hydrate containing exactly 50% water by mass is \(\mathrm{Na}_{2} \mathrm{SO}_{3} \cdot 7 \mathrm{H}_{2} \mathrm{O}\).

Step by step solution

01

Calculate Molar Masses

Calculate the molar masses of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) and \( \mathrm{H}_{2} \mathrm{O}\). The molar mass of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) is \(2 * 23\)(Na) + \(32\)(S) + \(3 * 16\)(O) = 126 g/mol. The molar mass of water \( \mathrm{H}_{2} \mathrm{O}\) is \(2 * 1\)(H) + \(16\)(O) = 18 g/mol.
02

Compare the Ratios

As the mass percentage of water is exactly 50%, the mass ratio of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) to \( \mathrm{H}_{2} \mathrm{O}\) is 1:1. Therefore, they also have the same weight in grams. Let's take 126g of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) and 126g of \( \mathrm{H}_{2} \mathrm{O}\).
03

Calculate the Proportions

Convert mass into moles to find the ratio between \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) and \( \mathrm{H}_{2} \mathrm{O}\). The amount of \(\mathrm{Na}_{2} \mathrm{SO}_{3}\) and \( \mathrm{H}_{2} \mathrm{O}\) in moles is 126g / 126g/mol = 1 mol and 126g / 18g/mol = 7 mol respectively. Thus, the formula of the hydrate is \(\mathrm{Na}_{2} \mathrm{SO}_{3} \cdot 7 \mathrm{H}_{2} \mathrm{O}\).

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Key Concepts

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

Molar Mass Calculation
The concept of molar mass is fundamental in chemistry, especially when determining chemical formulas for compounds. Molar mass refers to the weight of one mole of a substance. It is measured in grams per mole (g/mol) and helps in converting between mass and the amount of substance. Here, we have two molecules to consider:
  • Na2SO3 (Sodium Sulfite)
  • H2O (Water)
Let's calculate their molar masses: - For Na2SO3, add the masses of all atoms: Two sodium (Na) atoms each weigh 23 g/mol, so together they weigh 46 g/mol. The sulfur (S) atom is 32 g/mol, and three oxygen (O) atoms equal 48 g/mol. Adding these together gives us a molar mass of 126 g/mol. - The molar mass of water, H2O, is calculated by adding two hydrogen atoms (each 1 g/mol) and one oxygen atom (16 g/mol), which totals to 18 g/mol. Using this information, we can understand how molar mass plays a crucial role in the calculations needed for hydrate experiments.
Mass Percentage
Understanding mass percentage is key to determining the composition of compounds such as hydrates. Mass percentage gives us a way to express the ratio of the mass of a part to the total mass of the compound, in percentage terms.
In the context of hydrates, this tells us how much of the total mass is due to the water present in the compound. For example, in the given problem, we are told that the mass percentage of water is 50%. This figure tells us the water constitutes half of the total mass of the hydrate.
To clarify, if the total mass of the hydrate is 100 g, exactly 50 g would be the water component, making the rest the anhydrous compound (that is, the compound without water). Understanding this percentage helps in setting up the subsequent conversions needed to determine the chemical formula of the hydrate.
Chemical Formula Determination
Determining the chemical formula of a hydrate involves knowing the ratio of the compound's parts, specifically the amount of the anhydrous compound to the water in it. After finding out the molar masses and understanding the mass percentage, we move to calculate these proportions concretely.
For our calculation, we assume a convenient sample, such as using 126 g of Na2SO3 based on its molar mass. Given a 1:1 mass ratio, the amount of water is also 126 g. To establish the formula, we convert these masses into moles: - Na2SO3: 126 g 梅 126 g/mol = 1 mol
  • This step shows we have 1 mole of the anhydrous substance.
- H2O: 126 g 梅 18 g/mol = 7 mol
  • Thus, we have 7 moles of water.
These numbers tell us the mole ratio of the anhydrous compound to water. The chemical formula of the hydrate is expressed as Na2SO3 路 7 H2O, indicating seven water molecules are associated with each formula unit of sodium sulfite. This calculation process helps in understanding how hydrates are systematically quantified and written.

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Most popular questions from this chapter

A hydrocarbon mixture consists of \(60.0 \%\) by mass of \(\mathrm{C}_{3} \mathrm{H}_{8}\) and \(40.0 \%\) of \(\mathrm{C}_{x} \mathrm{H}_{y} .\) When \(10.0 \mathrm{g}\) of this mixture is burned, \(29.0 \mathrm{g} \mathrm{CO}_{2}\) and \(18.8 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\) are the only products. What is the formula of the unknown hydrocarbon?

Nitrogen forms five oxides in which its oxidation states are \(+1,+2,+3,+4,\) and \(+5,\) respectively. Write appropriate formulas for these compounds.

Calculate the total number of (a) atoms in one molecule of trinitrotoluene (TNT), \(\mathrm{CH}_{3} \mathrm{C}_{6} \mathrm{H}_{2}\left(\mathrm{NO}_{2}\right)_{3}\) (b) atoms in \(0.00102 \mathrm{mol} \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{4} \mathrm{CH}_{2} \mathrm{OH}\) (c) \(\mathrm{F}\) atoms in \(1215 \mathrm{mol} \mathrm{C}_{2} \mathrm{HBr} \mathrm{ClF}_{3}\)

A hydrate of copper(II) sulfate, when heated, goes through the succession of changes suggested by the photograph. In this photograph, (a) is the original fully hydrated copper(II) sulfate; (b) is the product obtained by heating the original hydrate to \(140^{\circ} \mathrm{C}\) (c) is the product obtained by further heating to \(400^{\circ} \mathrm{C}\) and (d) is the product obtained at \(1000^{\circ} \mathrm{C}\) A \(2.574 \mathrm{g}\) sample of \(\mathrm{CuSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O}\) was heated to \(140^{\circ} \mathrm{C},\) cooled, and reweighed. The resulting solid was reheated to \(400^{\circ} \mathrm{C},\) cooled, and reweighed. Finally, this solid was heated to \(1000^{\circ} \mathrm{C},\) cooled, and reweighed for the last time. $$ \text {Original sample } \quad \quad\quad\quad\quad \text {\(2.574 \mathrm{g}\) } $$ $$ \text {After heating to \(140^{\circ} \mathrm{C}\) } \quad \quad\quad\quad\quad \text {\(1.833 \mathrm{g}\) } $$ $$ \text {After reheating to \(400^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(1.647 \mathrm{g}\) } $$ $$ \text {After reheating to \(1000^{\circ} \mathrm{C}\)} \quad \quad\quad\quad\quad \text {\(0.812 \mathrm{g}\)} $$ (a) Assuming that all the water of hydration is driven off at \(400^{\circ} \mathrm{C},\) what is the formula of the original hydrate? (b) What is the formula of the hydrate obtained when the original hydrate is heated to only \(140^{\circ} \mathrm{C} ?\) (c) The black residue obtained at \(1000^{\circ} \mathrm{C}\) is an oxide of copper. What is its percent composition and empirical formula?

Malachite is a common copper-containing mineral with the formula \(\mathrm{CuCO}_{3} \cdot \mathrm{Cu}(\mathrm{OH})_{2} .\) (a) What is the mass percent copper in malachite? (b) When malachite is strongly heated, carbon dioxide and water are driven off, yielding copper(II) oxide as the sole product. What mass of copper(II) oxide is produced per kg of malachite?
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