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Chapter 5: Problem 135

Vitamin \(\mathrm{B}_{12}\), cyanocobalamin, is essential for human nutrition. It is concentrated in animal tissue but not in higher plants. Although nutritional requirements for the vitamin are quite low, people who abstain completely from animal products may develop a deficiency anemia. Cyanocobalamin is the form used in vitamin supplements. It contains \(4.34 \%\) cobalt by mass. Calculate the molar mass of cyanocobalamin, assuming that there is one atom of cobalt in every molecule of cyanocobalamin.

Short Answer

Expert verified
The molar mass of cyanocobalamin (Vitamin B12) is approximately 1359 g/mol.

Step by step solution

01

Understand the information given in the problem

The problem states that cyanocobalamin contains 4.34% cobalt by mass. This means that out of 100g of cyanocobalamin, there are 4.34g of cobalt. We are also provided with the molar mass of cobalt, which we will need to use later in our calculations.
02

Calculate the mass of cobalt

We will start by calculating the mass of cobalt in 100g of cyanocobalamin. Given the percentage of cobalt by mass in cyanocobalamin is 4.34%, we can calculate the mass of cobalt as follows: Mass of cobalt = Percentage of cobalt by mass * Mass of cyanocobalamin Since we are considering 100g of cyanocobalamin, the mass of cobalt in it will be: Mass of cobalt = \(4.34\% * 100g = 4.34g\)
03

Calculate the moles of cobalt

As we know the mass of cobalt, we can now calculate the moles of cobalt present in cyanocobalamin. To do this, we need the molar mass of cobalt, which is approximately 58.93 g/mol. Moles of cobalt = Mass of cobalt / Molar mass of cobalt Moles of cobalt = \(4.34g / 58.93 g/mol = 0.0736 mol\)
04

Calculate the molar mass of cyanocobalamin

Since there is one atom of cobalt present in each cyanocobalamin molecule, the moles of cobalt in cyanocobalamin will be equal to the moles of cyanocobalamin. This means we have 0.0736 moles of cyanocobalamin present in 100g sample. To find the molar mass of cyanocobalamin, we will now divide the mass of the 100g sample by the moles of cyanocobalamin: Molar mass of cyanocobalamin = Mass of cyanocobalamin / Moles of cyanocobalamin Molar mass of cyanocobalamin = \(100g / 0.0736 mol ≈ 1359 g/mol\) So, the molar mass of cyanocobalamin (Vitamin B12) is approximately 1359 g/mol.

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Most popular questions from this chapter

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen) contains \(2.59 \times 10^{23}\) atoms of hydrogen and is \(17.3 \%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and \(65 \mathrm{g} / \mathrm{mol}\), what amount (moles) of compound is present, and what is the mass of the sample?

A given sample of a xenon fluoride compound contains molecules of the type \(\mathrm{XeF}_{n},\) where \(n\) is some whole number. Given that \(9.03 \times 10^{20}\) molecules of \(\mathrm{XeF}_{n}\) weigh \(0.368 \mathrm{g},\) determine the value for \(n\) in the formula.

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When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). In a certain experiment, \(20.00 \mathrm{g}\) iron metal was reacted with \(11.20 \mathrm{g}\) oxygen gas. After the experiment, the iron was totally consumed, and 3.24 g oxygen gas remained. Calculate the amounts of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) formed in this experiment.

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(g)\) or \(\mathrm{NO}_{2}(g)\) according to these unbalanced equations: $$\begin{array}{l}\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\end{array}$$ In a certain experiment 2.00 moles of \(\mathrm{NH}_{3}(g)\) and 10.00 moles of \(\mathbf{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, 6.75 moles of \(\mathbf{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations because you cannot assume that the two reactions will occur with equal probability.)
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