Question

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

The empirical formula is \(C_2H_6S\).

Step-by-step solution

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

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: \[ \frac{51.61 \text{ g}}{32.07 \text{ g/mol}} \text{ = 1.61 mol} \]

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: \[ \frac{1.61 \text{ mol}}{1.61 \text{ mol}} = 1 \]

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 \(C_2H_6S\).

Key concepts

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!