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

Lead is a heavy metal that remains in the bloodstream, causing mental retardation in children. It is believed that \(3 \times 10^{-7} \mathrm{~g}\) of \(\mathrm{Pb}\) in \(1.00 \mathrm{~mL}\) of blood is a health hazard. For this amount of lead (a) how many atoms of lead are there in one \(\mathrm{mL}\) of a child's blood? (b) how many moles of lead are in \(1.00 \mathrm{~L}\) of blood?

Turquoise has the following chemical formula: \(\mathrm{CuAl}_{6}\left(\mathrm{PO}_{4}\right)_{4}(\mathrm{OH})_{8}\). \(4 \mathrm{H}_{2} \mathrm{O} .\) Calculate the mass percent of each element in turquoise.

The active ingredient of some antiperspirants is \(\mathrm{Al}_{2}(\mathrm{OH})_{5} \mathrm{Cl} .\) Calculate the mass percent of each element in this ingredient.

Consider the hypothetical reaction $$ 8 \mathrm{~A}_{2} \mathrm{~B}_{3}(s)+3 \mathrm{X}_{4}(g) \longrightarrow 4 \mathrm{~A}_{4} \mathrm{X}_{3}(s)+12 \mathrm{~B}_{2}(g) $$ When \(10.0 \mathrm{~g}\) of \(\mathrm{A}_{2} \mathrm{~B}_{3}(\mathrm{MM}=255 \mathrm{~g} / \mathrm{mol})\) react with an excess of \(\mathrm{X}_{4}, 4.00 \mathrm{~g}\) of \(\mathrm{A}_{4} \mathrm{X}_{3}\) are produced. (a) How many moles of \(A_{4} X_{3}\) are produced? (b) What is the molar mass of \(\mathrm{A}_{4} \mathrm{X}_{3}\) ?

Riboflavin is one of the \(\mathrm{B}\) vitamins. It is also known as vitamin \(\mathrm{B}_{6}\) and is made up of carbon, hydrogen, nitrogen, and oxygen atoms. When \(10.00 \mathrm{~g}\) of vitamin \(\mathrm{B}_{6}\) is burned in oxygen, \(19.88 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(4.79 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are obtained. Another experiment shows that vitamin \(\mathrm{B}_{6}\) is made up of \(14.89 \% \mathrm{~N}\). What is the simplest formula for vitamin \(\mathrm{B}_{6}\) ?

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