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Chapter 3: 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 is approximately 1355.52 u.

Step by step solution

01

Find the atomic mass of cobalt

First, we need to determine the atomic mass of cobalt. You can find this information on the periodic table. The atomic mass of cobalt (Co) is approximately 58.93 u (unified atomic mass units).
02

Set up the equation

Now we will set up the equation that connects the percentage of cobalt by mass, the mass of cobalt atom, and the molar mass of cyanocobalamin (which we want to find). Let's denote the molar mass of cyanocobalamin as M: \( \frac{58.93}{M} \times 100 = 4.34 \)
03

Solve for M

To find the molar mass of cyanocobalamin, M, we will solve the equation we set up in Step 2: \( \frac{58.93}{M} = \frac{4.34}{100} \) Now, let's multiply both sides by M and then by 100 to isolate M: \(58.93 \times 100 = 4.34 \times M\)
04

Calculate the molar mass

Now, we will solve for M by dividing both sides of the equation by 4.34: \(M = \frac{58.93 \times 100}{4.34}\) \(M \approx 1355.52 \, u\) Now we have found the molar mass of cyanocobalamin, which is approximately 1355.52 u.

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Most popular questions from this chapter

In 1987 the first substance to act as a superconductor at a temperature above that of liquid nitrogen \((77 \mathrm{~K})\) was discovered. The approximate formula of this substance is \(\mathrm{YBa}_{2} \mathrm{Cu}_{3} \mathrm{O}_{7}\). Calculate the percent composition by mass of this material.

The compound \(\mathrm{As}_{2} \mathrm{I}_{4}\) is synthesized by reaction of arsenic metal with arsenic triiodide. If a solid cubic block of arsenic \(\left(d=5.72 \mathrm{~g} / \mathrm{cm}^{3}\right)\) that is \(3.00 \mathrm{~cm}\) on edge is allowed to react with \(1.01 \times 10^{24}\) molecules of arsenic triiodide, what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) can be prepared? If the percent yield of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was \(75.6 \%\), what mass of \(\mathrm{As}_{2} \mathrm{I}_{4}\) was actually isolated?

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O}\). What is the empirical formula of urea?

Consider a gaseous binary compound with a molar mass of \(62.09 \mathrm{~g} / \mathrm{mol}\). When \(1.39 \mathrm{~g}\) of this compound is completely burned in excess oxygen, \(1.21 \mathrm{~g}\) of water is formed. Determine the formula of the compound. Assume water is the only product that contains hydrogen.

A sample of urea contains \(1.121 \mathrm{~g} \mathrm{~N}, 0.161 \mathrm{~g} \mathrm{H}, 0.480 \mathrm{~g} \mathrm{C}\), and \(0.640 \mathrm{~g} \mathrm{O}\). What is the empirical formula of urea?
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