Allicin is responsible for the distinctive taste and odor of garlic. Its simple formula is \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~S}\). How many grams of sulfur can be obtained from \(25.0 \mathrm{~g}\) of allicin?

Short Answer

Expert verified
Answer: The mass of sulfur that can be obtained from 25.0 g of allicin is approximately 4.93 g.

Step by step solution

01

Calculate the molar mass of allicin

First, we need to find the molar mass of allicin by adding the molar masses of each element in its chemical formula. The molar masses of carbon (C), hydrogen (H), oxygen (O), and sulfur (S) are \(12.01 \mathrm{~g/mol}\), \(1.01 \mathrm{~g/mol}\), \(16.00 \mathrm{~g/mol}\), and \(32.07 \mathrm{~g/mol}\), respectively. Molar mass of allicin \(= 6 \times 12.01 + 10 \times 1.01 + 2 \times 16.00 + 32.07 = 162.33 \mathrm{~g/mol}\).
02

Convert mass of allicin to moles

Now, we can convert the mass of allicin (25.0 g) to moles by dividing the mass by its molar mass: Moles of allicin \(= \frac{25.0 \mathrm{~g}}{162.33 \mathrm{~g/mol}} = 0.154 \mathrm{~mol}\).
03

Determine the moles of sulfur in allicin

In the chemical formula of allicin \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~S}\), there is 1 mole of sulfur for every mole of allicin. Therefore, the moles of sulfur in the given mass of allicin are: Moles of sulfur \(= 0.154 \mathrm{~mol}\) (since it is in a 1:1 ratio with allicin).
04

Convert moles of sulfur to mass

Finally, to find the mass of sulfur that can be obtained from the given mass of allicin, we can convert the moles of sulfur to mass using the molar mass of sulfur (\(32.07 \mathrm{~g/mol}\)): Mass of sulfur \(= 0.154 \mathrm{~mol} \times 32.07 \mathrm{~g/mol} = 4.93 \mathrm{~g}\). Thus, the mass of sulfur that can be obtained from \(25.0 \mathrm{~g}\) of allicin is approximately \(4.93 \mathrm{~g}\).

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate the molar mass is at the core of stoichiometry in chemistry. It allows us to account for the mass of all atoms in a compound. To find the molar mass, we add together the mass of each element present in one mole of the compound.

So, when we examine allicin with the formula \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~S}\), we calculate the molar mass by multiplying the atomic mass of each element by the number of times it occurs in the formula. The atomic masses for carbon is \(12.01 \mathrm{~g/mol}\), hydrogen is \(1.01 \mathrm{~g/mol}\), oxygen is \(16.00 \mathrm{~g/mol}\), and sulfur is \(32.07 \mathrm{~g/mol}\).

By this approach, we calculate the exact amount of each element, ensuring a coherent result in the subsequent conversion processes.
Moles to Grams Conversion
The conversion of moles to grams is a straightforward process but a fundamental one for carrying out stoichiometric calculations. To accomplish the conversion, we use the molar mass as a conversion factor.

For our example with allicin, after finding the molar mass, we need to know how many grams are equivalent to a given number of moles. This is done by multiplying the number of moles of the substance by its molar mass. This step is crucial, as it is the bridge between the abstract amount of substance (moles) and its tangible mass we can measure (grams).

In essence, understanding moles to grams conversion is essential for tasks ranging from predicting reaction yields to scaling up reactions for industrial processes.
Chemical Formula Analysis
Chemical formula analysis is the examination of the elemental composition and proportion in a chemical formula. It is the blueprint for understanding a compound's make-up and is a pivotal skill for chemists.

Taking allicin as an example, its formula \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{2} \mathrm{~S}\) tells us that there are six carbon atoms, ten hydrogen atoms, two oxygen atoms, and one sulfur atom in a single molecule of the substance. Assessing the formula this way, we can relate the moles of allicin to the moles of sulfur in a 1:1 ratio, as they coexist in the same quantity within the chemical structure. This relationship guides us in further calculations and interpretations of chemical reactions or substance compositions.

Knowing how to analyze chemical formulas is fundamental for any scientist or student, who ventures into solving stoichiometric problems or theorizing about the properties of different chemicals.

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Most popular questions from this chapter

Epsom salts are hydrates of magnesium sulfate. The formula for Epsom salts is \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\). A \(7.834\) -g sample is heated until a constant mass is obtained indicating that all the water has been evaporated off. What is the mass of the anhydrous magnesium sulfate? What percentage of the hydrate is water?

The active ingredient in Pepto-Bismol \(^{\text {m }}\) (an over-the-counter remedy for upset stomach) is bismuth subsalicylate, \(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{BiO}_{4}\). Analysis of a 1.500-g sample of Pepto-Bismol yields \(346 \mathrm{mg}\) of bismuth. What percent by mass is bismuth subsalicylate? (Assume that there is no other bismuthcontaining compound in Pepto-Bismol.)

The combustion of liquid chloroethylene, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}\), yields carbon dioxide, steam, and hydrogen chloride gas. (a) Write a balanced equation for the reaction. (b) How many moles of oxygen are required to react with \(35.00 \mathrm{~g}\) of chloroethylene? (c) If \(25.00 \mathrm{~g}\) of chloroethylene reacts with an excess of oxygen, how many grams of each product are formed?

Balance the following equations: (a) \(\mathrm{CaC}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{C}_{2} \mathrm{H}_{2}(g)\) (b) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}(s) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CH}_{3} \mathrm{NH}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\)

The Space Shuttle uses aluminum metal and ammonium perchlorate in its reusable booster rockets. The products of the reaction are aluminum oxide, aluminum chloride, nitrogen oxide gas, and steam. The reaction mixture contains \(7.00 \mathrm{~g}\) of aluminum and \(9.32 \mathrm{~g}\) of ammonium perchlorate. (a) Write a balanced equation for the reaction. (b) What is the theoretical yield of aluminum oxide? (c) If \(1.56 \mathrm{~g}\) of aluminum oxide is formed, what is the percent yield? (d) How many grams of excess reactant are unused? (Assume \(100 \%\) yield.)

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