Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The mass percentage of carbon in cortisone is \(69.98 \%\). What is the molecular mass of cortisone? (a) \(180.05\) (b) \(360.1\) (c) \(312.8\) (d) \(205.8\)

Short Answer

Expert verified
Answer: (b) 360.1 g/mol

Step by step solution

01

Calculate the mass of carbon per molecule of cortisone

We are given that there are 21 carbon atoms in one molecule of cortisone. The atomic mass of carbon is approximately 12.01 g/mol. Therefore, the total mass of carbon in one molecule of cortisone can be calculated as follows: Total mass of carbon in one molecule of cortisone = (Number of carbon atoms per molecule) x (Atomic mass of carbon) Total mass of carbon in one molecule of cortisone = 21 x 12.01 g/mol
02

Set up a proportion using the mass percentage of carbon in cortisone

The mass percentage of carbon in cortisone is given as \(69.98 \%\). Let the molecular mass of cortisone be M. Since mass percentage is mass of carbon in cortisone divided by the molecular mass of cortisone, we can set up the following proportion: \(Mass\: \% \:of\: carbon = \frac{total \:mass\: of \:carbon \:in\: one\: molecule\: of\: cortisone}{molecular\: mass\: of\: cortisone}\) \(69.98 \% = \frac{21 \times 12.01}{M}\)
03

Solve for the molecular mass of cortisone (M)

Now, we just need to solve the proportion for the molecular mass of cortisone (M). We multiply M by both sides to isolate it: \(M = \frac{21 \times 12.01}{69.98 \%}\) M = 360.21 g/mol
04

Select the correct answer from the given options

Comparing the calculated molecular mass of cortisone (360.21 g/mol) with the given options, the closest answer is: (b) \(360.1\) Hence, the molecular mass of cortisone is approximately 360.1 g/mol.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One molecule of haemoglobin will combine with four molecules of oxygen. If \(1.0 \mathrm{~g}\) of haemoglobin combines with \(1.642 \mathrm{ml}\) of oxygen at body temperature \(\left(27^{\circ} \mathrm{C}\right)\) and a pressure of 760 torr, what is the molar mass of haemoglobin? (a) \(6,00,000\) (b) \(1,50,000\) (c) 15,000 (d) 60,000

A compound has carbon, hydrogen, and oxygen in \(3: 3: 1\) atomic ratio. If the number of moles in \(1 \mathrm{~g}\) of the compound is \(6.06 \times 10^{-3}\), the molecular formula of the compound will be (a) \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}\) (b) \(\mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{9} \mathrm{H}_{9} \mathrm{O}_{3}\) (d) \(\mathrm{C}_{12} \mathrm{H}_{12} \mathrm{O}_{4}\)

Iodobenzene is prepared from aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{NH}_{2}\right)\) in a two-step process as shown here: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{HNO}_{2}+\mathrm{HCl} \longrightarrow\) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+2 \mathrm{H}_{2} \mathrm{O}\) \(\mathrm{C}_{6} \mathrm{H}_{3} \mathrm{~N}_{2}^{+} \mathrm{Cl}^{-}+\mathrm{KI} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{I}+\mathrm{N}_{2}+\mathrm{KCl}\) In an actual preparation, \(9.30 \mathrm{~g}\) of aniline was converted to \(16.32 \mathrm{~g}\) of iodobenzene. The percentage yield of iodobenzene is \((I=127)\)

\(\mathrm{SO}_{2}\) gas is slowly passed through an aqueous suspension containing \(12 \mathrm{~g}\) \(\mathrm{CaSO}_{3}\) till the milkiness just disappears. What amount of \(\mathrm{SO}_{2}\) would be required? (a) \(6.4\) mole (b) \(0.3\) mole (c) \(0.1 \mathrm{~mole}\) (d) \(0.2\) mole

It was found from the chemical analysis of a gas that it has two hydrogen atoms for each carbon atom. At \(0^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), its density is \(1.25 \mathrm{~g}\) per litre. The formula of the gas would be (a) \(\mathrm{CH}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{8}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free