/*! 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 63 Acrylic acid, used in the manufa... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 63

Acrylic acid, used in the manufacture of acrylic plastics, has the composition \(50.0 \% \mathrm{C}, 5.6 \% \mathrm{H}\), and \(44.4 \% \mathrm{O}\). What is its empirical formula?

Short Answer

Expert verified
The empirical formula for acrylic acid is \(\text{C}_3\text{H}_4\text{O}_2\).

Step by step solution

01

Convert Percentage to Grams

Assume 100 grams of acrylic acid, so the composition is 50.0 grams of carbon (C), 5.6 grams of hydrogen (H), and 44.4 grams of oxygen (O).
02

Convert Grams to Moles

Use the molar masses of the elements: Carbon has a molar mass of 12.01 g/mol, Hydrogen is 1.01 g/mol, and Oxygen is 16.00 g/mol. Calculate the moles of each element as follows: \[\text{Moles of C} = \frac{50.0 \text{ g C}}{12.01 \text{ g/mol}} = 4.16 \text{ moles of C}\]\[\text{Moles of H} = \frac{5.6 \text{ g H}}{1.01 \text{ g/mol}} = 5.54 \text{ moles of H}\]\[\text{Moles of O} = \frac{44.4 \text{ g O}}{16.00 \text{ g/mol}} = 2.78 \text{ moles of O}\]
03

Determine the Simplest Whole-Number Ratio

Divide the moles of each element by the smallest number of moles calculated in Step 2 (2.78 moles of O):\[\text{Ratio for C} = \frac{4.16}{2.78} = 1.50\]\[\text{Ratio for H} = \frac{5.54}{2.78} = 1.99\]\[\text{Ratio for O} = \frac{2.78}{2.78} = 1.00\]
04

Adjust Ratios to Whole Numbers

Multiply all ratios by 2 to eliminate fractions:\[\text{C: } 1.50 \times 2 = 3.00\]\[\text{H: } 1.99 \times 2 = 3.98 \approx 4.00 \]\[\text{O: } 1.00 \times 2 = 2.00\]
05

Write Empirical Formula

With the whole numbers obtained, the empirical formula is derived. The empirical formula for acrylic acid is \(\text{C}_3\text{H}_4\text{O}_2\).

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.

Molar Mass
When dealing with chemical formulas and compositions, the molar mass is a key concept to understand. Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule. For example, to find the molar mass of a compound, you need to consider each element's atomic mass.

Consider acrylic acid as an illustrative case. It comprises carbon (C), hydrogen (H), and oxygen (O). The atomic mass of carbon is approximately 12.01 g/mol, hydrogen is 1.01 g/mol, and oxygen is 16.00 g/mol.

When given a percentage composition, as often seen in exercises, converting these percentages to mass (assuming a certain sample size like 100 grams) assists in calculating the amount of each element's substance in terms of moles.
Elemental Composition
Understanding elemental composition is crucial in chemistry, as it tells you what elements constitute a compound and in what proportion. When you know the percentage of each element in a compound, you can find out how much of each element is present, usually by assuming a fixed quantity like 100 grams of the compound. For acrylic acid, the composition included:
  • 50.0% Carbon (C)
  • 5.6% Hydrogen (H)
  • 44.4% Oxygen (O)
This means in a 100-gram sample, you'd have 50 grams of carbon, 5.6 grams of hydrogen, and 44.4 grams of oxygen.

Such information is useful to determine the empirical formula, which is the simplest integer ratio of atoms in a compound. This process involves converting these masses to moles, which helps in finding out the ratio of the different elements.
Mole Calculation
Moles are a fundamental unit in chemistry used to express quantities of a chemical substance. Calculating the number of moles of each element in a compound can be done using the formula:
\[m = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\]
This aids in deriving the compound's empirical formula, which is the focus of many chemistry problems.

For acrylic acid:
  • Carbon: Convert grams to moles using its molar mass: \( \frac{50.0 \text{ g}}{12.01 \text{ g/mol}} = 4.16 \text{ moles} \)
  • Hydrogen: \( \frac{5.6 \text{ g}}{1.01 \text{ g/mol}} = 5.54 \text{ moles} \)
  • Oxygen: \( \frac{44.4 \text{ g}}{16.00 \text{ g/mol}} = 2.78 \text{ moles} \)
These mole calculations lay the groundwork for determining the simplest whole-number ratio of elements present, leading to identifying the empirical formula. The aim is to translate mass data into a chemical formula that portrays the actual ratio of elements in a compound.

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

An alloy of iron \((71.0 \%)\), cobalt \((12.0 \%)\), and molybdenum ( \(17.0 \%\) ) has a density of \(8.20 \mathrm{~g} / \mathrm{cm}^{3}\). How many cobalt atoms are there in a cylinder with a radius of \(2.50 \mathrm{~cm}\) and a length of \(10.0 \mathrm{~cm} ?\)

White phosphorus, \(\mathrm{P}_{4}\), is prepared by fusing calcium phosphate, \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\), with carbon, \(\mathrm{C}\), and sand, \(\mathrm{SiO}_{2}\), in an electric furnace. \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s)\) $$ \mathrm{P}_{4}(g)+6 \mathrm{CaSiO}_{3}(l)+10 \mathrm{CO}(g) $$ How many grams of calcium phosphate are required to give \(15.0 \mathrm{~g}\) of phosphorus?

A sample of gas mixture from a neon sign contains \(0.0856\) mol Ne and \(0.0254\) mol Kr. What are the mass percentages of \(\mathrm{Ne}\) and \(\mathrm{Kr}\) in the gas mixture? Chemical Formulas

A friend is doing his chemistry homework and is working with the following chemical reaction. $$ 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ He tells you that if he reacts 2 moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) with 4 moles of \(\mathrm{O}_{2}\) that the \(\mathrm{C}_{2} \mathrm{H}_{2}\) is the limiting reactant since there are fewer moles of \(\mathrm{C}_{2} \mathrm{H}_{2}\) than \(\mathrm{O}_{2}\). a. How would you explain to him where he went wrong with his reasoning (what concept is he missing)? b. After providing your friend with the explanation from part a, he still doesn't believe you because he had a homework problem where 2 moles of calcium were reacted with 4 moles of sulfur and he needed to determine the limiting reactant. The reaction is $$ \mathrm{Ca}(s)+\mathrm{S}(s) \longrightarrow \mathrm{CaS}(s) $$ He obtained the correct answer, \(\mathrm{Ca}\), by reasoning that since there were fewer moles of calcium reacting, calcium had to be the limiting reactant. How would you explain his reasoning flaw and why he got "lucky" in choosing the answer that he did?

Calcium carbide, \(\mathrm{CaC}_{2}\), used to produce acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}\), is prepared by heating calcium oxide, \(\mathrm{CaO}\), and carbon, \(\mathrm{C}\), to high temperature. $$ \mathrm{CaO}(s)+3 \mathrm{C}(s) \longrightarrow \mathrm{CaC}_{2}(s)+\mathrm{CO}(g) $$ If a mixture contains \(2.60 \mathrm{~kg}\) of each reactant, how many grams of calcium carbide can be prepared?
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Analysis

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.