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Chapter 7: Problem 88

Researchers from the University of Cape Town in South Afric have discovered that penguins are able to detect a che ::: released by plankton as they are being consumed by schor fish. By following this scent they are able to find these sc of fish and feast on them. The chemical they smell is comp of \(38.65 \%\) carbon, \(9.74 \%\) hydrogen, and \(51.61 \%\) sulfur. D mine the empirical formula of this compound.

Short Answer

Expert verified
The empirical formula is \(C_2H_6S\).

Step by step solution

01

Convert Percentages to Grams

Assume you have 100 grams of the compound. This allows you to directly convert the percentages to grams: - Carbon: 38.65 grams - Hydrogen: 9.74 grams - Sulfur: 51.61 grams
02

Convert Grams to Moles

Use the molar mass of each element to convert the mass to moles. The molar masses are approximately: - Carbon: 12.01 g/mol - Hydrogen: 1.008 g/mol - Sulfur: 32.07 g/mol Calculate the moles for each element: o Carbon: \[ \frac{38.65 \text{ g}}{12.01 \text{ g/mol}} \text{ = 3.22 mol} \] o Hydrogen: \[ \frac{9.74 \text{ g}}{1.008 \text{ g/mol}} \text{ = 9.66 mol} \] o Sulfur: \[ \frac{51.61 \text{ g}}{32.07 \text{ g/mol}} \text{ = 1.61 mol} \]
03

Determine the Simplest Whole Number Ratio

Find the ratio of the moles by dividing each by the smallest number of moles calculated in the previous step. o For Carbon: \[ \frac{3.22 \text{ mol}}{1.61 \text{ mol}} = 2 \] o For Hydrogen: \[ \frac{9.66 \text{ mol}}{1.61 \text{ mol}} = 6 \] o For Sulfur: \[ \frac{1.61 \text{ mol}}{1.61 \text{ mol}} = 1 \]
04

Write the Empirical Formula

Using the simplest whole number ratio derived: o Carbon (C): 2 o Hydrogen (H): 6 o Sulfur (S): 1 The empirical formula is \(C_2H_6S\).

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Key Concepts

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

Molar Mass
In chemistry, molar mass is a fundamental concept. It refers to the mass of one mole of a substance (typically in grams per mole, g/mol). Understanding molar mass helps in converting grams to moles and vice versa. Every element has a distinct molar mass. For example, the molar masses used in the problem are:
  • Carbon: 12.01 g/mol
  • Hydrogen: 1.008 g/mol
  • Sulfur: 32.07 g/mol
Keenly using these values, we can transform the given mass of an element into its equivalent moles. This conversion is a crux step in calculating empirical formulas.
Chemical Composition
Chemical composition is the breakdown of a substance into its constituent elements or compounds. In this problem, we're given the chemical composition of a mysterious compound detected by penguins: it has 38.65% carbon, 9.74% hydrogen, and 51.61% sulfur. Understanding chemical composition is critical in determining how much of each element is present in a sample.
  • These percentages can be converted into grams (assuming a 100-gram sample as done in the problem, making the calculations straightforward).
  • The elements' gram values are then used for further mole calculations and to figure out the exact formula of the compound.
Mole Ratio Analysis
Mole ratio analysis involves comparing the moles of different elements in a compound to get a simple ratio. These ratios are essential in the calculation of the empirical formula.
In our example, after converting grams to moles:
  • Carbon: 3.22 mol
  • Hydrogen: 9.66 mol
  • Sulfur: 1.61 mol
We need the simplest whole number ratio between these moles, done by dividing all values by the smallest number among them (in this case, 1.61).
  • For Carbon: \(\frac{3.22}{1.61} = 2\)
  • For Hydrogen: \(\frac{9.66}{1.61} = 6\)
  • For Sulfur: \(\frac{1.61}{1.61} = 1\)

    • These ratios, 2:6:1, tell us how many atoms of each element are in the simplest form of the compound.
Chemistry Problem Solving
Solving chemistry problems involves a systematic approach to dismantle and understand the problem step by step:
  • First, identify what information is given and what is required. Here, we start with percentages of elements.
  • Convert percentages to grams and grams to moles using the molar mass.
  • Determine the mole ratio to find the simplest formula.
  • Verify the results and write the empirical formula - in this case, it's \(C_2H_6S\).
This approach ensures a structured and effective way to solve empirical formula problems. Remember, practice makes perfect when it comes to chemistry problem solving!

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Most popular questions from this chapter

\- Copper is often added to paint for oceangoing ships and boats to discourage the growth of marine organisms on their hulls. Unfortunately the copper dissolves into the water and is toxic to the marine life in harbors where these boats are docked. When copper(II) ions reach a concentration of \(9.0\) microgram/L, the growth of some marine life slows. How many copper ions are found in a liter of water containing \(9.0\) micrograms of \(\mathrm{Cu}^{2+} / \mathrm{L}\) ?

A substance whose formula is \(\mathrm{A}_{2} \mathrm{O}\) (A is a mystery element) is \(60.0 \% \mathrm{~A}\) and \(40.0 \% \mathrm{O}\). Identify the element \(\mathrm{A}\).

Which of the following chlorides has the highest and which ha the lowest percentage of chlorine, by mass, in its formula? (a) \(\mathrm{KCl}\) (c) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{BaCl}_{2}\) (d) \(\mathrm{LiCl}\)Which of the following chlorides has the highest and which ha the lowest percentage of chlorine, by mass, in its formula? (a) \(\mathrm{KCl}\) (c) \(\mathrm{SiCl}_{4}\) (b) \(\mathrm{BaCl}_{2}\) (d) \(\mathrm{LiCl}\)

For the following compounds whose molecular formulas are given, indicate the empirical formula: (a) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \quad\) glucose (b) \(\mathrm{C}_{8} \mathrm{H}_{18} \quad\) octane (c) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3} \quad\) lactic acid (d) \(\mathrm{C}_{25} \mathrm{H}_{52} \quad\) paraffin (e) \(\mathrm{C}_{12} \mathrm{H}_{4} \mathrm{Cl}_{4} \mathrm{O}_{2} \quad\) dioxin (a powerful poison)

The compound \(\mathrm{X}_{2}\left(\mathrm{YZ}_{3}\right)_{3}\) has a molar mass of \(282.23 \mathrm{~g}\) and a percent composition (by mass) of \(19.12 \% \mathrm{X}, 29.86 \% \mathrm{Y}\), and \(51.02 \% \mathrm{Z}\). What is the formula of the compound?
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