Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 9: Problem 163

The amino acid arginine is made up of \(41.37 \%\) by mass \(\mathrm{C}, 8.10 \%\) by mass \(\mathrm{H}, 32.16 \%\) by mass \(\mathrm{N}\), and \(18.37 \%\) by mass \(\mathrm{O}\). If the molar mass of arginine is approximately \(174 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

Short Answer

Expert verified
The molecular formula of arginine is C₃H₇N₂O.

Step by step solution

01

Determine the number of grams of each element from the mass percentages

To find out how much of each element arginine contains, we will use the mass percentage values given. Assuming a 100g sample, we can calculate the amount of each element in grams: - C: 41.37% × 100g = 41.37g - H: 8.10% × 100g = 8.10g - N: 32.16% × 100g = 32.16g - O: 18.37% × 100g = 18.37g
02

Calculate the number of moles of each element

Now that we have the amount of each element in grams, we can convert them to moles using the molar masses of each element. - Molar mass of C: 12.01 g/mol - Molar mass of H: 1.01 g/mol - Molar mass of N: 14.01 g/mol - Molar mass of O: 16.00 g/mol Number of moles for each element: - Moles of C: \( \frac{41.37g}{12.01g/mol} = 3.44mol \) - Moles of H: \( \frac{8.10g}{1.01g/mol} = 8.02mol \) - Moles of N: \( \frac{32.16g}{14.01g/mol} = 2.30mol \) - Moles of O: \( \frac{18.37g}{16.00g/mol} = 1.15mol \)
03

Determine the simplest whole number ratio between the numbers of moles of each element

To obtain the simplest whole number ratio, divide the number of moles of each element by the element with the lowest number of moles: - Ratio of C: \( \frac{3.44}{1.15} = 2.99 \approx 3 \) - Ratio of H: \( \frac{8.02}{1.15} = 6.97 \approx 7 \) - Ratio of N: \( \frac{2.30}{1.15} = 2.00 \approx 2 \) - Ratio of O: \( \frac{1.15}{1.15} = 1.00 \approx 1 \) The simplest whole number ratio is approximately 3:7:2:1.
04

Write the molecular formula

Using the simplest whole number ratio obtained in step 3, we can write the molecular formula of arginine as: C₃H₇N₂O So, the molecular formula of arginine is C₃H₇N₂O.

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.

Start your free trial

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

Consider the following unbalanced chemical equation: \(\mathrm{H}_{2}+\mathrm{N}_{2} \rightarrow \mathrm{NH}_{3}\) (a) To run this reaction in a balanced fashion, how much nitrogen is required if you start with \(10.0 \mathrm{~g}\) of \(\mathrm{H}_{2} ?\) (b) How many grams of ammonia \(\left(\mathrm{NH}_{3}\right)\) will you produce if you run the reaction with the masses calculated in part (a)? (c) How many molecules of ammonia will you produce?

A compound used as an insecticide that contains only \(C, H\), and \(C l\) is subjected to combustion analysis, yielding \(24.78 \% \mathrm{C}\) and \(2.08 \% \mathrm{H}\). (a) What is the empirical formula for this compound? (b) What is the molecular formula if its actual formula is four times the mass of its empirical formula?

Chrome yellow, a pigment used in paints, is \(64.11 \%\) by mass \(\mathrm{Pb}, 16.09 \%\) by mass \(\mathrm{Cr}\), and \(19.80 \%\) by mass O. What is the empirical formula of this compound?

For the reaction \(\mathrm{SiCl}_{4}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{SiO}_{2}+\mathrm{HCl}\), (a) Balance the equation. (b) If you want to make \(120.0 \mathrm{~g}\) of \(\mathrm{HCl}\), how many grams of \(\mathrm{SiCl}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) do you need? (c) How many grams of \(\mathrm{SiO}_{2}\) are produced from the quantities calculated in (b)?

\(5.00 \mathrm{~g}\) of solid sodium (Na) and \(30.0 \mathrm{~g}\) of liquid bromine \(\left(\mathrm{Br}_{2}\right)\) react to form solid \(\mathrm{NaBr}\). (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(14.7 \mathrm{~g}\) of \(\mathrm{NaBr}\) is recovered. What is the percent yield of the reaction?
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation
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