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Chapter 3: Problem 69

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
C\(_3\)H\(_4\)O\(_2\).

Step by step solution

01

Convert Percentages to Grams

Assume you have 100 grams of the substance, which makes the percentage values equivalent to grams. Thus, you have 50.0 grams of carbon (C), 5.6 grams of hydrogen (H), and 44.4 grams of oxygen (O).
02

Convert Grams to Moles

Calculate the number of moles of each element using their atomic masses: - For carbon (C), use its atomic mass of 12.01 g/mol: \[ \text{Moles of C} = \frac{50.0\, \text{g}}{12.01\, \text{g/mol}} \approx 4.16 \text{ moles} \]- For hydrogen (H), use its atomic mass of 1.008 g/mol: \[ \text{Moles of H} = \frac{5.6\, \text{g}}{1.008\, \text{g/mol}} \approx 5.56 \text{ moles} \] - For oxygen (O), use its atomic mass of 16.00 g/mol: \[ \text{Moles of O} = \frac{44.4\, \text{g}}{16.00\, \text{g/mol}} \approx 2.78 \text{ moles} \]
03

Determine the Simplest Mole Ratio

Divide the number of moles of each element by the smallest number of moles calculated in Step 2. In this case, divide by the moles of oxygen (2.78).- \[\frac{4.16}{2.78} \approx 1.50 \] for carbon,- \[\frac{5.56}{2.78} \approx 2.00 \] for hydrogen,- \[\frac{2.78}{2.78} = 1.00 \] for oxygen.
04

Adjust the Mole Ratio to Whole Numbers

To eliminate the fraction in the ratio (1.5 for carbon), multiply all the ratios by 2.- Carbon: \(1.50 \times 2 = 3\) - Hydrogen: \(2.00 \times 2 = 4\) - Oxygen: \(1.00 \times 2 = 2\)Thus, the whole number mole ratio is C: 3, H: 4, O: 2.
05

Write the Empirical Formula

Using the simplest whole number mole ratio, write the empirical formula of the compound: C\(_3\)H\(_4\)O\(_2\).

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Key Concepts

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

Acrylic Acid
Acrylic acid is an organic compound that serves as a building block for various applications, particularly in the manufacture of acrylic plastics.
Acrylic acid contains carbon, hydrogen, and oxygen atoms. Its distinct properties make it extremely useful in industries such as paint, adhesive, and textile manufacturing.
Students often encounter acrylic acid in chemistry classes as an example when learning about empirical formulas.
The composition of a compound like acrylic acid provides valuable insight into its structure, which helps chemists predict its reactivity and applications.
Mole Ratio
Understanding mole ratio is crucial when determining the empirical formula of a compound.
In essence, a mole ratio is an expression of the relative number of moles of each element present in a compound.
  • To find the mole ratio, follow these steps:
First, convert the mass of each element to moles using the element's atomic mass.
Next, identify the smallest number of moles among the elements.
Divide every mole quantity by this smallest value to determine a simple ratio.
If necessary, multiply all the ratios to obtain whole numbers, which will allow you to determine the empirical formula accurately. Finally, this ratio will guide you in writing the simplest formula for the compound.
Atomic Mass
Atomic mass is a vital concept in chemistry, representing the mass of an atom, typically expressed in atomic mass units (u or amu).
It is significant when converting mass to moles, which aids in calculating mole ratios.
  • Here are the atomic masses needed for acrylic acid:
  • Carbon (C): approximately 12.01 g/mol
  • Hydrogen (H): approximately 1.008 g/mol
  • Oxygen (O): approximately 16.00 g/mol
By applying these atomic masses, you can determine the amount of substance, measured in moles, present in a sample.
Understanding atomic mass allows you to quantify elements in a chemical sample and ultimately derive the empirical formula of the compound.
Chemical Composition
Chemical composition defines the types and amounts of elements that make up a compound.
For acrylic acid, its chemical composition by percentage is defined as: 50.0% Carbon, 5.6% Hydrogen, and 44.4% Oxygen.
This information is foundational when calculating the empirical formula of a compound.
  • With chemical composition, you can:
  • Determine the amount of each element in a sample, often starting from percentage composition.
  • Convert these percentages to mass, assuming a base quantity like 100 grams for simplicity.
Understanding the chemical composition will allow you to derive empirical formulas and, eventually, even predict the compound's molecular structure.
In summary, knowledge of a compound's chemical composition is essential for analyzing its characteristics and potential chemical behavior.

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Most popular questions from this chapter

Come up with some examples of limiting reactants that use the concept but don't involve chemical reactions.

A fertilizer is advertised as containing \(14.0 \%\) nitrogen (by mass). How much nitrogen is there in \(4.15 \mathrm{~kg}\) of fertilizer? 3.54 Seawater contains \(0.0065 \%\) (by mass) of bromine. How many grams of bromine are there in \(2.50 \mathrm{~L}\) of seawater? The density of seawater is \(1.025 \mathrm{~g} / \mathrm{cm}^{3}\).

Moles within Moles and Molar Mass Part 1: How many hydrogen and oxygen atoms are present in 1 molecule of \(\mathrm{H}_{2} \mathrm{O}\) ? How many moles of hydrogen and oxygen atoms are present in \(1 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? What are the masses of hydrogen and oxygen in \(1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O} ?\) What is the mass of \(1.0 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) ? Part 2: Two hypothetical ionic compounds are discovered with the chemical formulas \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), where \(\mathrm{X}\) and Y represent symbols of the imaginary elements. Chemical analysis of the two compounds reveals that \(0.25 \mathrm{~mol} \mathrm{XCl}_{2}\) has a mass of \(100.0 \mathrm{~g}\) and \(0.50 \mathrm{~mol} \mathrm{YCl}_{2}\) has a mass of \(125.0 \mathrm{~g}\) What are the molar masses of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\) ? If you had 1.0 -mol samples of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), how would the number of chloride ions compare? If you had 1.0 -mol samples of \(\mathrm{XCl}_{2}\) and \(\mathrm{YCl}_{2}\), how would the masses of elements \(\mathrm{X}\) and \(\mathrm{Y}\) compare? What is the mass of chloride ions present in 1.0 \(\mathrm{mol} \mathrm{XCl}_{2}\) and \(1.0 \mathrm{~mol} \mathrm{YCl}_{2} ?\) What are the molar masses of elements \(\mathrm{X}\) and \(\mathrm{Y} ?\) How many moles of \(\mathrm{X}\) ions and chloride ions would be present in a 200.0 -g sample of \(\mathrm{XCl}_{2}\) ? How many grams of \(Y\) ions would be present in a \(250.0-\mathrm{g}\) sample of \(\mathrm{YCl}_{2} ?\) What would be the molar mass of the compound \(\mathrm{YBr}_{3} ?\) \(\operatorname{Part} 3:\) A minute sample of \(\mathrm{AlCl}_{3}\) is analyzed for chlorine. The analysis reveals that there are 12 chloride ions present in the sample. How many aluminum ions must be present in the sample? What is the total mass of \(\mathrm{AlCl}_{3}\) in this sample? How many moles of \(\mathrm{AlCl}_{3}\) are in this sample?

Which contains more sulfur, \(40.8 \mathrm{~g}\) of calcium sulfate, \(\mathrm{CaSO}_{4},\) or \(38.8 \mathrm{~g}\) of sodium sulfite, \(\mathrm{Na}_{2} \mathrm{SO}_{3} ?\)

Potassium superoxide, \(\mathrm{KO}_{2}\), is used in rebreathing gas masks to generate oxygen. $$ 4 \mathrm{KO}_{2}(s)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{KOH}(s)+3 \mathrm{O}_{2}(g) $$ If a reaction vessel contains \(0.25 \mathrm{~mol} \mathrm{KO}_{2}\) and \(0.15 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O},\) what is the limiting reactant? How many moles of oxygen can be produced?
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