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Chapter 3: Problem 53

Determine the empirical formulas of the compounds with the following compositions: (a) 40.1 percent \(\mathrm{C}\), 6.6 percent \(\mathrm{H}, 53.3\) percent \(\mathrm{O} ;\) (b) 18.4 percent \(\mathrm{C}\), 21.5 percent \(\mathrm{N}, 60.1\) percent \(\mathrm{K}\)

Short Answer

Expert verified
Compound (a): \( \text{CH}_2\text{O} \); Compound (b): \( \text{CNK} \).

Step by step solution

01

Convert Percentages to Grams

Assume you have 100 g of each compound. This means you have 40.1 g of C, 6.6 g of H, and 53.3 g of O for compound (a); 18.4 g of C, 21.5 g of N, and 60.1 g of K for compound (b).
02

Convert Grams to Moles

For compound (a), calculate moles of each element:- For C: \( \frac{40.1 \text{ g}}{12.01 \text{ g/mol}} = 3.34 \text{ mol} \)- For H: \( \frac{6.6 \text{ g}}{1.01 \text{ g/mol}} = 6.53 \text{ mol} \)- For O: \( \frac{53.3 \text{ g}}{16.00 \text{ g/mol}} = 3.33 \text{ mol} \)For compound (b), calculate moles of each element:- For C: \( \frac{18.4 \text{ g}}{12.01 \text{ g/mol}} = 1.53 \text{ mol} \)- For N: \( \frac{21.5 \text{ g}}{14.01 \text{ g/mol}} = 1.54 \text{ mol} \)- For K: \( \frac{60.1 \text{ g}}{39.10 \text{ g/mol}} = 1.54 \text{ mol} \)
03

Determine the Mole Ratio

For each compound, calculate the mole ratio by dividing each amount of moles by the smallest value of moles computed:For compound (a):- C: \( \frac{3.34}{3.33} = 1.00 \)- H: \( \frac{6.53}{3.33} = 1.96 \)- O: \( \frac{3.33}{3.33} = 1.00 \)For compound (b):- C: \( \frac{1.53}{1.53} = 1.00 \)- N: \( \frac{1.54}{1.53} = 1.01 \)- K: \( \frac{1.54}{1.53} = 1.01 \)
04

Write the Empirical Formula

For compound (a), round the numbers to the nearest whole number:- C: 1, H: 2, O: 1 leading to the empirical formula \( \text{CH}_2\text{O} \).For compound (b), round the numbers to the nearest whole number:- C: 1, N: 1, K: 1 leading to the empirical formula \( \text{CNK} \).

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Key Concepts

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

Stoichiometry
Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is based on the law of conservation of mass, where matter cannot be created or destroyed in an isolated system. In the context of determining empirical formulas, stoichiometry helps us understand the quantitative relationship between the different elements in a compound.

The process generally involves:
  • Knowing the percent composition of a compound
  • Converting these percentages to masses (assuming a baseline of 100 grams)
  • Converting those masses to moles, using the molar mass of each element
  • Finding the simplest whole-number ratio of moles of each element present
This systematic approach allows us to write a chemical formula that represents the simplest ratio of the atoms in a compound. By adhering to stoichiometry, we ensure that our calculations are grounded in the principles of chemical reactions and compositions, making it an essential skill in chemistry.
Mole Calculation
Mole calculation is a key component in stoichiometry and chemistry as a whole. It involves using the mole concept to relate a given mass of a substance to the number of entities (like atoms or molecules) it contains. The mole is a bridge between the atomic scale and the macroscopic scale of laboratory chemistry.

To perform mole calculations, follow these guidelines:
  • Determine the molar mass of each element, which you can find on the periodic table expressed in g/mol
  • Convert the mass of each element in a given problem to moles by dividing the mass by the element's molar mass
  • Use these mole values to find the smallest whole-number ratio when determining empirical formulas
The equation used is:\(\text{Moles of } A = \frac{\text{Mass of } A}{\text{Molar Mass of } A}\)This calculation gives you a quantitative understanding of how much of a substance you're working with, which is crucial for predicting the amount of products in chemical reactions.
Elemental Composition
Understanding elemental composition is critical when determining the empirical formula of a compound. Elemental composition tells you the percentage by mass of each element in a compound, providing valuable insight into its structure and properties.

During calculations:
  • Start by assuming you have a 100g sample of the compound. This makes it easy to convert percentage data into grams.
  • Use these masses to calculate moles of each element.
  • From here, convert these values to the simplest whole-number ratio to find the empirical formula.
Elemental composition not only helps in deducing the empirical formula, but also plays a role when you're advancing toward understanding molecular formulas and reaction stoichiometry. By mastering these concepts, students can accurately describe the composition of compounds, essential for deeper chemical analysis and synthesis.

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Most popular questions from this chapter

A common laboratory preparation of oxygen gas is the thermal decomposition of potassium chlorate \(\left(\mathrm{KClO}_{3}\right)\). Assuming complete decomposition, calculate the number of grams of \(\mathrm{O}_{2}\) gas that can be obtained from \(46.0 \mathrm{~g}\) of \(\mathrm{KClO}_{3}\). (The products are \(\mathrm{KCl}\) and \(\mathrm{O}_{2}\).)

A mixture of \(\mathrm{NaBr}\) and \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) contains 29.96 percent Na by mass. Calculate the percent by mass of each compound in the mixture.

Peroxyacylnitrate (PAN) is one of the components of smog. It is a compound of \(\mathrm{C}, \mathrm{H}, \mathrm{N},\) and \(\mathrm{O} .\) Determine the percent composition of oxygen and the empirical formula from the following percent composition by mass: 19.8 percent \(\mathrm{C}, 2.50\) percent \(\mathrm{H}, 11.6\) percent \(\mathrm{N}\). What is its molecular formula given that its molar mass is about \(120 \mathrm{~g}\) ?

Nitroglycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9}\right)\) is a powerful explosive. Its decomposition may be represented by $$ 4 \mathrm{C}_{3} \mathrm{H}_{5} \mathrm{~N}_{3} \mathrm{O}_{9} \longrightarrow 6 \mathrm{~N}_{2}+12 \mathrm{CO}_{2}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2} $$ This reaction generates a large amount of heat and gaseous products. It is the sudden formation of these gases, together with their rapid expansion, that produces the explosion. (a) What is the maximum amount of \(\mathrm{O}_{2}\) in grams that can be obtained from \(2.00 \times 10^{2} \mathrm{~g}\) of nitroglycerin? (b) Calculate the percent yield in this reaction if the amount of \(\mathrm{O}_{2}\) generated is found to be \(6.55 \mathrm{~g}\).

A sample of \(10.0 \mathrm{~g}\) of sodium reacts with oxygen to form \(13.83 \mathrm{~g}\) of sodium oxide \(\left(\mathrm{Na}_{2} \mathrm{O}\right)\) and sodium peroxide \(\left(\mathrm{Na}_{2} \mathrm{O}_{2}\right) .\) Calculate the percent composition of the product mixture.
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