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Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\). When \(5.061 \mathrm{~g}\) of this hydrate is heated to \(250^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving \(2.472 \mathrm{~g}\) of \(\mathrm{MgSO}_{4} .\) What is the value of \(x ?\)

Short Answer

Expert verified
Moles of \(\mathrm{MgSO}_{4}\) = \(\frac{2.472 \mathrm{~g}}{120.4 \mathrm{~g/mol} } = 0.0205 \mathrm{~mol}\) Moles of \(\mathrm{H}_{2} \mathrm{O}\) = \(\frac{2.589 \mathrm{~g}}{18 \mathrm{~g/mol} } = 0.1438 \mathrm{~mol}\) #tag_title#Step 3: Find the ratio of moles of water to moles of \(\mathrm{MgSO}_{4}\) #tag_content#To find the value of \(x\), we divide the moles of water by the moles of \(\mathrm{MgSO}_{4}\). \(x\) = \(\frac{\text{moles of }\mathrm{H}_{2} \mathrm{O}}{\text{moles of }\mathrm{MgSO}_{4}}\) \(x\) = \(\frac{0.1438 \mathrm{~mol}}{0.0205 \mathrm{~mol}} = 7.02\) Since \(x\) must be a whole number, we round to the nearest whole number. Therefore, the value of \(x\) in the formula for Epsom salts, \(\mathrm{MgSO}_{4}\cdot x \mathrm{H}_{2} \mathrm{O}\), is \(x = 7\).

Step by step solution

01

Calculate the mass of the water of hydration

To find the mass of the water of hydration, we subtract the mass of \(\mathrm{MgSO}_{4}\) from the mass of the Epsom salts hydrate. Mass of water of hydration = Initial mass of Epsom salts hydrate - Mass of \(\mathrm{MgSO}_{4}\) Mass of water of hydration = \(5.061 \mathrm{~g} - 2.472 \mathrm{~g} = 2.589 \mathrm{~g}\)
02

Calculate the moles of \(\mathrm{MgSO}_{4}\) and water

To find the moles of \(\mathrm{MgSO}_{4}\) and water, we divide the mass of each substance by its molar mass. The molar mass of \(\mathrm{MgSO}_{4}\) is \(\mathrm{24.3 + 32.1 + 4(16) = 120.4 ~g/mol.}\) The molar mass of \(\mathrm{H}_{2} \mathrm{O}\) is \(\mathrm{2(1)+16 = 18 ~g/mol.}\)

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Key Concepts

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

Hydrate
Hydrates are fascinating compounds where water molecules are chemically bound to another substance. This bond gives them unique properties. In the context of Epsom salts, the chemical formula can be expressed as \( \mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O} \). Here, "\(x\)" denotes the number of water molecules associated with each \( \mathrm{MgSO}_{4} \) unit.
When hydrates are heated, they typically lose their water content through a process called dehydration. This results in the anhydrous form of the compound, which in Epsom salts, is just \( \mathrm{MgSO}_{4} \).
  • Water molecules give many hydrates their crystal structure and contribute to their weight.
  • The difference in mass before and after heating allows us to calculate the amount of water, and consequently, the value of \(x\) in the formula.
This understanding is essential when working with hydrates, as it helps in determining the number of water molecules present in such compounds through experimental techniques.
Molar Mass
Molar mass is a crucial concept in stoichiometry, representing the mass of one mole of a substance. It allows us to convert between the mass of a sample and the amount of substance in moles.
For example, the molar mass of \( \mathrm{MgSO}_{4} \) is determined by adding up the atomic masses of its components: magnesium (Mg), sulfur (S), and oxygen (O). The calculations are as follows:
\[\text{Molar mass of } \mathrm{MgSO}_{4} = 24.3 + 32.1 + 4 \times 16 = 120.4 \text{ g/mol} \]
Similarly, water (\( \mathrm{H}_2 \mathrm{O} \)) has a molar mass calculated by the atomic masses of hydrogen (H) and oxygen (O):
\[\text{Molar mass of } \mathrm{H}_2 \mathrm{O} = 2 \times 1 + 16 = 18 \text{ g/mol} \]
  • Knowing the molar mass is essential for translating between grams and moles, which is a standard practice in chemical calculations.
  • It enables the calculation of the number of moles, which is necessary to find the stoichiometric ratios in reactions.
In our example, having the molar masses allowed us to use the given masses to calculate how many moles were present, a key step in determining the formula of the hydrate.
Chemical Formulas
Chemical formulas serve as the shorthand representation of the composition of molecules. In the formula for Epsom salts (\( \mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O} \)), it tells us that the compound consists of both magnesium sulfate and water molecules.
These formulas are not just a representation of elements but also give an insight into the molar ratio of components involved. For hydrates, the coefficient before the water part of the formula (often represented as "x") signifies how many moles of water are included per mole of the main compound.
  • Determining the correct chemical formula is integral to understanding the compound's behavior and its reactivity.
  • This understanding facilitates the calculation of theoretical yields, helps in predicting reaction products, and assists in interpreting experimental data.
Thus, by uncovering the numerical relationship between \( \mathrm{MgSO}_{4} \) and water in Epsom salts, we gain valuable insights into its composition and chemical properties.

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Most popular questions from this chapter

An element \(\mathrm{X}\) forms an iodide \(\left(\mathrm{XI}_{3}\right)\) and a chloride \(\left(\mathrm{XCl}_{3}\right)\). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: $$ 2 \mathrm{XI}_{3}+3 \mathrm{Cl}_{2} \longrightarrow 2 \mathrm{XCl}_{3}+3 \mathrm{I}_{2} $$ If \(0.5000 \mathrm{~g}\) of \(\mathrm{XI}_{3}\) is treated with chlorine, \(0.2360 \mathrm{~g}\) of \(\mathrm{XCl}_{3}\) is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.

The complete combustion of octane, \(\mathrm{C}_{8} \mathrm{H}_{18},\) a component of gasoline, proceeds as follows: $$ 2 \mathrm{C}_{8} \mathrm{H}_{18}(l)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(g) $$ (a) How many moles of \(\mathrm{O}_{2}\) are needed to burn \(1.50 \mathrm{~mol}\) of \(\mathrm{C}_{8} \mathrm{H}_{18} ?\) (b) How many grams of \(\mathrm{O}_{2}\) are needed to burn \(10.0 \mathrm{~g}\) of \(\mathrm{C}_{8} \mathrm{H}_{18}\) ? (c) Octane has a density of \(0.692 \mathrm{~g} / \mathrm{mL}\) at \(20^{\circ} \mathrm{C}\). How many grams of \(\mathrm{O}_{2}\) are required to burn 15.0 gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\) (the capacity of an average fuel tank)? (d) How many grams of \(\mathrm{CO}_{2}\) are produced when 15.0 gal of \(\mathrm{C}_{8} \mathrm{H}_{18}\) are combusted?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) together with other substances. Reaction of the ore with CO produces iron metal: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) $$ (a) Balance this equation. (b) Calculate the number of grams of CO that can react with \(0.350 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) (c) Calculate the number of grams of Fe and the number of grams of \(\mathrm{CO}_{2}\) formed when \(0.350 \mathrm{~kg}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacts. (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

A key step in balancing chemical equations is correctly identifying the formulas of the reactants and products. For example, consider the reaction between calcium oxide, \(\mathrm{CaO}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) to form aqueous calcium hydroxide. (a) Write a balanced chemical equation for this combination reaction, having correctly identified the product as \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)\) (b) Is it possible to balance the equation if you incorrectly identify the product as \(\mathrm{CaOH}(a q)\), and if so, what is the equation?

Write balanced chemical equations to correspond to each of the following descriptions: (a) When sulfur trioxide gas reacts with water, a solution of sulfuric acid forms. (b) Boron sulfide, \(\mathrm{B}_{2} \mathrm{~S}_{3}(s),\) reacts violently with water to form dissolved boric acid, \(\mathrm{H}_{3} \mathrm{BO}_{3},\) and hydrogen sulfide gas. (c) Phosphine, \(\mathrm{PH}_{3}(g)\), combusts in oxygen gas to form water vapor and solid tetraphosphorus decaoxide. (d) When solid mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and oxygen. (e) Copper metal reacts with hot concentrated sulfuric acid solution to form aqueous copper(II) sulfate, sulfur dioxide gas, and water.

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