/*! 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 48 Determine the empirical formulas... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 48

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(42.1 \% \mathrm{Na}, 18.9 \% \mathrm{P}\), and \(39.0 \% \mathrm{O}\) (b) \(18.7 \% \mathrm{Li}, 16.3 \% \mathrm{C},\) and \(65.0 \% \mathrm{O}\) (c) \(60.0 \% \mathrm{C}, 4.4 \% \mathrm{H},\) and the remainder \(\mathrm{O}\)

Short Answer

Expert verified
(a) \(\mathrm{Na}_3\mathrm{PO}_4\); (b) \(\mathrm{Li}_2\mathrm{CO}_3\); (c) \(\mathrm{C}_5\mathrm{H}_4\mathrm{O}\).

Step by step solution

01

Assume 100 g sample

To simplify calculations, assume you have 100 grams of each compound. This makes the mass of each element equal to its percentage in grams, e.g., in part (a), you have 42.1 g of Na, 18.9 g of P, and 39.0 g of O.
02

Convert masses to moles

Divide each mass by the molar mass of the respective element to find the number of moles. (a) For Na: \(\frac{42.1}{22.99} \approx 1.83\), P: \(\frac{18.9}{30.97} \approx 0.61\), O: \(\frac{39.0}{16.00} \approx 2.44\)(b) For Li: \(\frac{18.7}{6.94} \approx 2.69\), C: \(\frac{16.3}{12.01} \approx 1.36\), O: \(\frac{65.0}{16.00} \approx 4.06\)(c) For C: \(\frac{60.0}{12.01} \approx 4.99\), H: \(\frac{4.4}{1.008} \approx 4.37\), O: Assume 35.6% left for O: \(\frac{35.6}{16.00} \approx 2.23\)
03

Find simplest mole ratio

Divide the number of moles of each element by the smallest number of moles in the set.(a) Na: \(\frac{1.83}{0.61} = 3\), P: \(\frac{0.61}{0.61} = 1\), O: \(\frac{2.44}{0.61} \approx 4\) leads to the ratio 3:1:4.(b) Li: \(\frac{2.69}{1.36} \approx 2\), C: \(\frac{1.36}{1.36} = 1\), O: \(\frac{4.06}{1.36} \approx 3\) gives the ratio 2:1:3.(c) C: \(\frac{4.99}{2.23} \approx 2.24\approx 5\), H: \(\frac{4.37}{2.23} \approx 1.96 \approx 4\), O: \(\frac{2.23}{2.23} = 1\) leads to the ratio 5:4:1.
04

Write the empirical formula

Use the mole ratios to write the empirical formula for each compound. (a) The empirical formula is \(\mathrm{Na}_3\mathrm{PO}_4\).(b) The empirical formula is \(\mathrm{Li}_2\mathrm{CO}_3\).(c) The empirical formula is \(\mathrm{C}_5\mathrm{H}_4\mathrm{O}_1\) or \(\mathrm{C}_5\mathrm{H}_4\mathrm{O}\).

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.

Understanding Mole Ratios
Mole ratios are an essential concept in chemistry, acting as the backbone of empirical formula determination. They simplify how we express the relative amounts of different substances reacting or produced in a chemical reaction.

To understand mole ratios, start by converting the mass of each element in a compound to moles. This is accomplished using the formula \( \, \text{moles of element} = \frac{\text{mass of element}}{\text{molar mass of element}} \).

Once you have calculated the moles for each element, it is important to determine the simplest whole number ratio of these moles. This ratio is derived by dividing the mole amount of each element by the smallest number of moles calculated among them. For instance, if you have 1 mole of Carbon and 2 moles of Oxygen, the simplest ratio would be 1:2.

Understanding the significance of mole ratios helps in writing balanced chemical equations, deducing empirical formulas, and gaining insights into the stoichiometric relationships between reactants and products in chemical reactions.
Exploring Mass Composition
Mass composition is the percentage by mass of each element in a compound. When determining empirical formulas, knowing the mass composition of a substance is your first step. It's like creating a recipe where each ingredient represents an element, and the percentages are how much of each you have.

Typically, you begin by assuming you have 100 grams of a compound. This assumption is crucial because it converts the percentage composition directly into grams. For example, if a compound has 18% of an element, in a 100 g sample, that element contributes 18 g.
  • This process makes calculations simpler and more intuitive.
  • It provides a straightforward path to convert mass into moles.
Once the mass is known, use the molar mass of each element to convert grams to moles, facilitating the development of mole ratios. Thus, mastering mass composition serves as a bedrock for further chemical analysis.
The Significance of Chemical Formulas
Chemical formulas are the symbolic representation of a compound's composition. They provide insights into the ratios and arrangements of elements in chemicals, helping chemists understand and predict the behavior of substances.

The empirical formula, specifically, gives the simplest whole number ratio of atoms of each element present in a compound. To determine this, you use the mole ratios derived from mass composition calculations. For instance, given the mole ratio of 3:1:4 for a compound consisting of Sodium, Phosphorus, and Oxygen, the empirical formula would be \(\text{Na}_3\text{PO}_4\).

The critical component in these formulas is the subscript next to each chemical symbol, indicating the number of moles of each element in the simplest form.
  • Empirical formulas aid in understanding fundamental material properties.
  • They are foundational in stoichiometry and chemical reaction predictions.
Mastering empirical formulas cultivates a deeper comprehension of chemical composition, enabling diverse applications from basic laboratory work to complex industrial processes.

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

When a mixture of \(10.0 \mathrm{~g}\) of acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and \(10.0 \mathrm{~g}\) of oxygen \(\left(\mathrm{O}_{2}\right)\) is ignited, the resulting combustion reaction produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) (a) Write the balanced chemical equation for this reaction. (b) Which is the limiting reactant? (c) How many grams of \(\mathrm{C}_{2} \mathrm{H}_{2}, \mathrm{O}_{2}, \mathrm{CO}_{2},\) and \(\mathrm{H}_{2} \mathrm{O}\) are present after the reaction is complete?

Very small semiconductor crystals, composed of approximately 1000 to 10,000 atoms, are called quantum dots. Quantum dots made of the semiconductor CdSe are now being used in electronic reader and tablet displays because they emit light efficiently and in multiple colors, depending on dot size. The density of CdSe is \(5.82 \mathrm{~g} / \mathrm{cm}^{3}\). (a) What is the mass of one \(2.5-\mathrm{nm}\) CdSe quantum dot? (b) CdSe quantum dots that are \(2.5 \mathrm{nm}\) in diameter emit blue light upon stimulation. Assuming that the dot is a perfect sphere and that the empty space in the dot can be neglected, calculate how many Cd atoms are in one quantum dot of this size. (c) What is the mass of one \(6.5-\mathrm{nm}\) CdSe quantum dot? (d) CdSe quantum dots that are \(6.5 \mathrm{nm}\) in diameter emit red light upon stimulation. Assuming that the dot is a perfect sphere, calculate how many Cd atoms are in one quantum dot of this size. (e) If you wanted to make one \(6.5-\mathrm{nm}\) dot from multiple \(2.5-\mathrm{nm}\) dots, how many \(2.5-\mathrm{nm}\) dots would you need, and how many CdSe formula units would be left over, if any?

A mixture of \(\mathrm{N}_{2}(g)\) and \(\mathrm{H}_{2}(g)\) reacts in a closed container to form ammonia, \(\mathrm{NH}_{3}(g)\). The reaction ceases before either reactant has been totally consumed. At this stage \(3.0 \mathrm{~mol} \mathrm{~N}_{2}, 3.0 \mathrm{~mol} \mathrm{H}_{2}\), and \(3.0 \mathrm{~mol} \mathrm{NH}_{3}\) are present. Howmany moles of \(\mathrm{N}_{2}\) and \(\mathrm{H}_{2}\) were present originally?

Give the empirical formula of each of the following compounds if a sample contains (a) \(0.052 \mathrm{~mol} \mathrm{C}, 0.103 \mathrm{~mol}\) \(\mathrm{H}\), and \(0.017 \mathrm{~mol} \mathrm{O} ;\) (b) \(2.10 \mathrm{~g}\) nickel and \(0.58 \mathrm{~g}\) oxygen; (c) \(26.56 \% \mathrm{~K}, 35.41 \% \mathrm{Cr}\), and \(38.03 \% \mathrm{O}\) by mass.

If \(2.0 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}, 2.0 \mathrm{~mol} \mathrm{C}_{4} \mathrm{H}_{10},\) and \(2.0 \mathrm{~mol}\) \(\mathrm{C}_{6} \mathrm{H}_{6}\) are completely combusted in oxygen, which one produces the largest number of moles of \(\mathrm{H}_{2} \mathrm{O}\) ? Which one produces the least? Explain.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Kinetics

Read Explanation

Physical Chemistry

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

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.