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Vitamin A has a molar mass of 286.4 g/mol and a general molecular formula of \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{E},\) where \(\mathrm{E}\) is an unknown element. If vitamin \(A\) is \(83.86 \%\) C and \(10.56 \%\) H by mass, what is the molecular formula of vitamin A?

Short Answer

Expert verified
The molecular formula of vitamin A is \(\mathrm{C}_{20} \mathrm{H}_{29} \mathrm{E}\), where E is the unknown element.

Step by step solution

01

Calculate the moles of C and H in 100 grams of vitamin A

Since the mass percentages of C and H are given, we will assume that we have 100 grams of vitamin A. This will make it easier to calculate the moles of each element. If vitamin A is 83.86% C by mass, then there are 83.86 grams of C in 100 grams of vitamin A. Similarly, if vitamin A is 10.56% H by mass, then there are 10.56 grams of H in 100 grams of vitamin A. Now, we need to find the moles of C and H in 100 grams of vitamin A. Moles of C = \(\frac{mass}{molar~mass}\) = \(\frac{83.86~g}{12.01~g/mol}\) (since molar mass of C = 12.01 g/mol) Moles of H = \(\frac{mass}{molar~mass}\) = \(\frac{10.56~g}{1.01~g/mol}\) (since molar mass of H = 1.01 g/mol)
02

Find the mole ratio of C and H in the molecule

Next, we need to find the mole ratio of C and H in the molecular formula. To do this, we will divide the moles of each element by the least number of moles to get a whole number ratio. Least number of moles (C or H) = Moles of C Mole ratio of C = \(1\) Mole ratio of H = \(\frac{Moles~of~H}{Moles~of~C}\)
03

Calculate the mass of E in 100 grams of Vitamin A

We can calculate the mass of E in 100 grams of vitamin A by subtracting the masses of C and H from the total mass (100 grams). Mass of E = Total mass - Mass of C - Mass of H
04

Find the moles of E

We know that the molar mass of vitamin A is 286.4 g/mol. In 100 grams of vitamin A, you have: Moles of vitamin A = \(\frac{mass}{molar~mass}\) = \(\frac{100~g}{286.4~g/mol}\) Since each molecule of vitamin A contains 1 atom of element E, the moles of E must be equal to the moles of vitamin A. Moles of E = Moles of vitamin A Now we have moles of C, H, and E in vitamin A.
05

Find the molecular formula of vitamin A

Now that we have the mole ratios of C, H, and E, we can write the molecular formula of vitamin A as: \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{E}\) where \(x\) = mole ratio of C, \(y\) = mole ratio of H, and E is the unknown element. Make sure to calculate the final mole ratios from step 2 and plug in those values for \(x\) and \(y\) to get the molecular formula of vitamin A.

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Key Concepts

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

Molar Mass Calculation
Understanding molar mass is critical when determining the composition of a compound like vitamin A. Molar mass is the weight of one mole of a substance, measured in grams per mole (g/mol). This reflects the average mass of all isotopes of the atoms in the formula, weighted by their natural abundance. To calculate the molar mass of a compound, you would sum up the molar masses of all the individual elements in the compound based on their atomic weights and the number of atoms of each present in the formula.

For example, in our exercise involving vitamin A, we used the molar mass of carbon (C) which is approximately 12.01 g/mol and hydrogen (H) which is about 1.01 g/mol. This calculation is foundational because it allows us to determine how many moles of each element are in a sample of a given mass. It simplifies further analysis and enables a deeper understanding of the compound's molecular formula.
Percent Composition
Percent composition plays a pivotal role in the analysis of molecular formulae. It tells us what percentage of a compound's total mass is made up by each constituent element. The percent composition values are essential when working with empirical and molecular formulas because they give insight into the proportion of different atoms in a substance.

In the exercise, vitamin A's composition by mass was provided for carbon and hydrogen. If you had 100 grams of vitamin A, 83.86 grams would be carbon and 10.56 grams would be hydrogen. These percentages are the stepping stones from which we calculate the amount in moles of each element in the compound, thus bridging the gap between qualitative and quantitative analysis of the compound's structure.
Mole Ratio
The mole ratio is a term used to express the relative quantities of reactants or products in a chemical reaction or constituents in a compound. It is derived from the coefficients of substances in a balanced chemical equation or, as in our case, the empirical formula of a compound.

Once we have determined the amount in moles of each element present in a sample, we establish the simplest whole number ratio between them. Dividing the moles of each element by the smallest mole value among them achieves this. In the example of vitamin A, once the number of moles for carbon and hydrogen were found, these were then compared to find their simplest ratio which contributes to finding the molecular formula. This ratio is crucial because it aligns with the principle that compounds are made of elements in fixed ratios.
Chemical Formula Derivation
Chemical formula derivation is the process of determining the symbols and subscript numbers that represent the types and amounts of atoms in a molecule. In practice, it involves using the percent composition to find the moles of each element, finding the mole ratio to simplify these values to the smallest whole numbers, and lastly deducing the molecular formula that represents these proportions.

In the vitamin A example, utilizing the mass percentages and molar masses allowed calculation of moles of carbon and hydrogen. By finding the least number of moles and the mole ratio of each element to carbon, we used these to derive the chemical formula for vitamin A, which included an unknown element E. Once all values for carbon, hydrogen, and element E were determined, the molecular formula could be accurately stated.

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Most popular questions from this chapter

A compound with molar mass \(180.1 \mathrm{g} / \mathrm{mol}\) has the following composition by mass: $$\begin{array}{|ll|}\hline C & 40.0 \% \\\H & 6.70 \% \\\O & 53.3 \% \\\\\hline\end{array}$$ Determine the empirical and molecular formulas of the compound.

ABS plastic is a tough, hard plastic used in applications requiring shock resistance. The polymer consists of three monomer units: acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right),\) butadiene \(\left(\mathrm{C}_{4} \mathrm{H}_{6}\right),\) and styrene \(\left(\mathrm{C}_{8} \mathrm{H}_{8}\right)\) a. A sample of ABS plastic contains \(8.80 \% \mathrm{N}\) by mass. It took \(0.605 \mathrm{g}\) of \(\mathrm{Br}_{2}\) to react completely with a \(1.20-\mathrm{g}\) sample of ABS plastic. Bromine reacts 1: 1 (by moles) with the butadiene molecules in the polymer and nothing else. What is the percent by mass of acrylonitrile and butadiene in this polymer? b. What are the relative numbers of each of the monomer units in this polymer?

A 2.25-g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

The aspirin substitute. acetaminophen \(\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}\right),\) is produced by the following three-step synthesis: I. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{3} \mathrm{N}(s)+3 \mathrm{H}_{2}(g)+\mathrm{HCl}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+2 \mathrm{H}_{2} \mathrm{O}(l)\) II. \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{ONCl}(s)+\mathrm{NaOH}(a q) \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{NaCl}(a q)\) III. \(\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{ON}(s)+\mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{3}(l) \longrightarrow\) \(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{2} \mathrm{N}(s)+\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(l)\) The first two reactions have percent yields of \(87 \%\) and \(98 \%\) by mass, respectively. The overall reaction yields 3 moles of acetaminophen product for every 4 moles of \(C_{6} H_{5} O_{3} N\) reacted. a. What is the percent yield by mass for the overall process? b. What is the percent yield by mass of Step III?

A compound \(\mathrm{XF}_{5}\) is \(42.81 \%\) fluorine by mass. Identify the element \(\mathrm{X}\) and draw the Lewis structure for the compound. What is the molecular structure of \(\mathrm{XF}_{5} ?\)

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