/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 92 Ribose, the sugar found in RNA, ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 92

Ribose, the sugar found in RNA, has been detected in experiments designed to mimic the conditions of early Earth. If ribose contains \(40.00 \% \mathrm{C}\) \(6.71 \% \mathrm{H},\) and \(53.28 \%\) O, with a molar mass of \(150.13 \mathrm{g} / \mathrm{mol}\).What are the empirical and molecular formulas for ribose?

Short Answer

Expert verified
Answer: The empirical formula of ribose is CH鈧侽, and the molecular formula is C鈧匟鈧佲個O鈧�.

Step by step solution

01

Convert percentage composition to moles

Assume we have 100g of ribose. This means that we will have 40.00g of C, 6.71g of H, and 53.28g of O. To determine the moles of each element, we will use their respective molar masses: 12.01g/mol for C, 1.01g/mol for H, and 16.00g/mol for O. Moles of C = \(\frac{40.00\,\text{g}}{12.01\,\text{g/mol}} = 3.33\) moles Moles of H = \(\frac{6.71\,\text{g}}{1.01\,\text{g/mol}} = 6.64\) moles Moles of O = \(\frac{53.28\,\text{g}}{16.00\,\text{g/mol}} = 3.33\) moles
02

Find the mole ratio

To find the mole ratio, divide the moles of each element by the smallest value (in this case, 3.33). Mole ratio of C: \(\frac{3.33\,\text{moles}}{3.33} = 1\) Mole ratio of H: \(\frac{6.64\,\text{moles}}{3.33} = 1.99 \approx 2\) (rounded to the nearest whole number) Mole ratio of O: \(\frac{3.33\,\text{moles}}{3.33} = 1\) This yields the empirical formula of C H\(_2\)O.
03

Determine the molecular formula

Calculate the molar mass of the empirical formula: Molar mass of the empirical formula, CH\(_2\)O = 12.01g/mol (C) + (2 \(\times\) 1.01g/mol) (H) + 16.00g/mol (O) = 30.03g/mol Now, divide the molar mass of ribose by the molar mass of the empirical formula to get the multiplier: Multiplier = \(\frac{150.13\,\text{g/mol}}{30.03\,\text{g/mol}} \approx 5\) Finally, multiply the empirical formula by the multiplier to obtain the molecular formula: Molecular formula = 5 \(\times\) (CH\(_2\)O) = C\(_5\)H\(_{10}\)O\(_5\) The empirical formula is CH\(_2\)O, and the molecular formula is C\(_5\)H\(_{10}\)O\(_5\).

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 91影视!

Key Concepts

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

Percentage Composition
Percentage composition is a measure of the relative amounts of each element in a compound. It is expressed as the percentage by mass of each element in the total mass of the compound. Knowing the percentage composition is crucial for determining empirical and molecular formulas as it is the first step in the calculation process.

To calculate the percentage composition, we divide the mass of each element in the compound by the total mass of the compound, and then multiply by 100 to get a percentage. For example, if a compound is made up of 40% Carbon (C), 6.71% Hydrogen (H), and 53.28% Oxygen (O), these values provide a direct roadmap for us to calculate the empirical formula by converting these percentages into moles.
Mole-to-Mole Ratio
The mole-to-mole ratio is the proportion of moles of one substance to the moles of another substance in a balanced chemical equation or in a compound. By knowing the mole-to-mole ratios, chemists can relate the amounts of reactants and products in a reaction.

When computing an empirical formula, the mole-to-mole ratio is determined by dividing the number of moles of each element by the smallest number of moles from the elements in the compound. This simplifies the ratio into the smallest whole numbers possible, which forms the basis of the empirical formula. It's analogous to simplifying a fraction to its lowest terms. Upon finding this ratio for a compound, you can easily deduce its empirical formula.
Molar Mass
Molar mass is the mass of one mole of a substance and is expressed in grams per mole (g/mol). It can be calculated by summing the atomic masses of all atoms in a molecule. The molar mass is essential for converting grams of a substance to moles, which is a step required when determining empirical and molecular formulas.

The empirical formula alone doesn't give information about the actual numbers of atoms in a molecule, just the simplest ratio. To find the molecular formula, which indicates the true numbers of atoms within a molecule, you divide the molar mass of the compound by the molar mass of the empirical formula. This gives a multiplier that is used to multiply the subscripts in the empirical formula to finally obtain the molecular formula.

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

On the seafloor, iron(II) oxide reacts with water to form \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) and hydrogen in a process called serpentization. a. Balance the following equation for serpentization: \(\mathrm{FeO}(s)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g)\). b. When \(\mathrm{CO}_{2}\) is present, the product is methane, not hydrogen. Balance the following chemical equation: \(\mathrm{FeO}(s)+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{CO}_{2}(g) \rightarrow \mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CH}_{4}(g)\).

Of the nitrogen oxides \(-\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}, \mathrm{N}_{2} \mathrm{O}_{3}, \mathrm{N}_{2} \mathrm{O}_{2}, \mathrm{NO}_{2},\) and \(\mathrm{N}_{2} \mathrm{O}_{4}-\) which are more than \(50 \%\) oxygen by mass? Which, if any, have the same empirical formula?

Aluminum reacts with elemental oxygen at high temperatures to give pure aluminum oxide. What is the coefficient of \(\mathrm{O}_{2}(\mathrm{g})\) in the balanced chemical equation?

Driving 1000 miles a month is not unusual for a short-distance commuter. If your vehicle gets 25 mpg, you would use 40 gallons \((\approx 150 \mathrm{L})\) of gasoline every month. If gasoline is approximated as \(\mathrm{C}_{8} \mathrm{H}_{18}(d=0.703 \mathrm{g} / \mathrm{mL}),\) how much carbon dioxide does your vehicle emit every month? The unbalanced chemical equation for the reaction is$$ \mathrm{C}_{8} \mathrm{H}_{18}(\ell)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$$.

Composition of Over-the-Counter Medicines Calculate the number of molecules or formula units of compound in each of the following common, over-the-counter medications: a. ibuprofen, a pain reliever and fever reducer that contains \(200.0 \mathrm{mg}\) of the active ingredient, \(\mathrm{C}_{13} \mathrm{H}_{18} \mathrm{O}_{2}\) b. an antacid containing 500.0 mg of calcium carbonate c. an allergy tablet containing 4 mg of chlorpheniramine $$\left(\mathrm{C}_{16} \mathrm{H}_{19} \mathrm{ClN}_{2}\right)$$.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

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.