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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Most popular questions from this chapter

One of the most bizarre reactions in chemistry is called the Ugi reaction: $\mathrm{R}_{1} \mathrm{C}(=\mathrm{O}) \mathrm{R}_{2}+\mathrm{R}_{3}-\mathrm{NH}_{2}+\mathrm{R}_{4} \mathrm{COOH}+\mathrm{R}_{5} \mathrm{NC} \rightarrow$ $\mathrm{R}_{4} \mathrm{C}(=\mathrm{O}) \mathrm{N}\left(\mathrm{R}_{3}\right) \mathrm{C}\left(\mathrm{R}_{1} \mathrm{R}_{2}\right) \mathrm{C}=\mathrm{ONHR}_{5}+\mathrm{H}_{2} \mathrm{O}$ (a) Write out the balanced chemical equation for the Ugi reaction, for the case where $\mathrm{R}=\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2}-$ (this is called the hexyl group) for all compounds. (b) What mass of the "hexyl Ugi product" would you form if \(435.0 \mathrm{mg}\) of $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}$ was the limiting reactant?

Calculate the percentage by mass of oxygen in the following compounds: (a) vanillin, \(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}_{3} ;(\mathbf{b})\) isopropyl alcohol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\); (c) acetaminophen, \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{NO}_{2} ;\) (d) cyclopropanone, \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}\); (e) dioxin, \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2} ;\) (f) penicillin, $\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{4} \mathrm{~S}$.

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?

Detonation of nitroglycerin proceeds as follows: $$ \begin{aligned} 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}(l) & \longrightarrow \\ & 12 \mathrm{CO}_{2}(g)+6 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)+10 \mathrm{H}_{2} \mathrm{O}(g) \end{aligned} $$ (a) If a sample containing \(2.00 \mathrm{~mL}\) of nitroglycerin (density \(=\) \(1.592 \mathrm{~g} / \mathrm{mL}\) ) is detonated, how many moles of gas are produced? (b) If each mole of gas occupies \(55 \mathrm{~L}\) under the conditions of the explosion, how many liters of gas are produced? (c) How many grams of \(\mathrm{N}_{2}\) are produced in the detonation?

A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bubbler" containing sodium iodide, which removes the ozone according to the following equation: $\mathrm{O}_{3}(g)+2 \mathrm{NaI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow$ $$ \mathrm{O}_{2}(g)+\mathrm{I}_{2}(s)+2 \mathrm{NaOH}(a q) $$ (a) How many moles of sodium iodide are needed to remove $5.95 \times 10^{-6} \mathrm{~mol}\( of \)\mathrm{O}_{3} ?(\mathbf{b})$ How many grams of sodium iodide are needed to remove \(1.3 \mathrm{mg}\) of \(\mathrm{O}_{3}\) ?

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