Question

Determine the empirical formula of each of the following compounds if a sample contains (a) \(3.92 \mathrm{~mol} \mathrm{C}, 5.99 \mathrm{~mol} \mathrm{H},\) and \(2.94 \mathrm{~mol} \mathrm{O} ;\) (b) \(12.0 \mathrm{~g}\) calcium and 2.8 g nitrogen; \((\mathbf{c})\) \(89.14 \%\) Au and \(10.86 \%\) O by mass.

Short answer

(a) For given moles, we find the ratios to be approximately 4:6:3. Hence, the empirical formula is \(C_4H_6O_3\). (b) For given masses, we find the ratios to be approximately 1:1. Therefore, the empirical formula is \(CaN\). (c) For given percentages, we find the ratios to be approximately 2:3. Hence, the empirical formula is \(Au_2O_3\).

Step-by-step solution

a) Calculate moles of elements and their ratios

Given the moles of elements are: - Carbon (C): 3.92 mol - Hydrogen (H): 5.99 mol - Oxygen (O): 2.94 mol Calculate the mole ratios by dividing each value by the smallest number of moles (2.94). C: \(\frac{3.92}{2.94} = 1.33\) (approx.) H: \(\frac{5.99}{2.94} = 2.04\) (approx.) O: \(\frac{2.94}{2.94} = 1\)

a) Determine the empirical formula

The ratios are close to 4:6:3, after multiplying each ratio by 3. Therefore, the empirical formula is \(C_4H_6O_3\).

b) Convert masses to moles and calculate ratios

Given the masses of elements are: - Calcium (Ca): 12.0 g - Nitrogen (N): 2.8 g First, we need to convert the masses to moles using the molar mass of each element: (Ca: 40.08 g/mol, N: 14.01 g/mol) Ca: \(\frac{12.0}{40.08}= 0.299\) mol (approx.) N: \(\frac{2.8}{14.01}= 0.200\) mol (approx.) Calculate the mole ratios by dividing each value by the smallest number of moles (0.200). Ca: \(\frac{0.299}{0.200}=1.495\) (approx.) N: \(\frac{0.200}{0.200}=1\)

b) Determine the empirical formula

The ratios are close to 1:1, so the empirical formula is \(CaN\).

c) Convert percentages to masses and moles

To solve this problem, we can assume having a 100 g sample, where the mass of each element is: - Gold (Au): 89.14 g (89.14% of 100 g) - Oxygen (O): 10.86 g (10.86% of 100 g) Now, convert the masses to moles using the molar mass of each element: (Au: 197.0 g/mol, O: 16.00 g/mol) Au: \(\frac{89.14}{197.0} = 0.452\) mol (approx.) O: \(\frac{10.86}{16.00} = 0.679\) mol (approx.) Calculate the mole ratios by dividing each value by the smallest number of moles (0.452). Au: \(\frac{0.452}{0.452} = 1\) O: \(\frac{0.679}{0.452} = 1.50\) (approx.)

c) Determine the empirical formula

The ratios are close to 2:3 after multiplying by 2. Therefore, the empirical formula is \(Au_2O_3\).

Key concepts

Mole Ratios

Understanding mole ratios helps us determine the relative amounts of each element in a compound. When we have the moles of each element, we can calculate mole ratios by dividing these by the smallest number of moles in the set. This simplifies the calculation by providing proportions that reflect how many times each element appears in the compound relative to the others.

For example, if you have a compound with 3.92 mol of Carbon, 5.99 mol of Hydrogen, and 2.94 mol of Oxygen, you start by identifying the smallest number of moles, which is 2.94 for Oxygen. You divide all quantities by this number:

• Carbon (C): \( \frac{3.92}{2.94} \approx 1.33 \)
• Hydrogen (H): \( \frac{5.99}{2.94} \approx 2.04 \)
• Oxygen (O): \( \frac{2.94}{2.94} = 1 \)

The ratios are then adjusted to the nearest whole numbers, so in this example, they become 4:6:3 after multiplying by an appropriate factor like 3 to convert them to whole numbers.

Convert Mass to Moles

Converting mass to moles is a crucial step in determining the empirical formula of a compound from its mass composition. This conversion relies on the molar mass of each element, which acts as a conversion factor. The molar mass of an element (usually given in grams per mole) allows you to convert a mass into an amount of substance measured in moles.

Let's take the example where 12.0 g of Calcium and 2.8 g of Nitrogen are given. To convert these masses to moles, we use their respective molar masses:

• Calcium (Ca): With a molar mass of 40.08 g/mol, the calculation is \( \frac{12.0}{40.08} \approx 0.299 \) moles
• Nitrogen (N): Using a molar mass of 14.01 g/mol, the calculation is \( \frac{2.8}{14.01} \approx 0.200 \) moles

Once you have the moles, you can find the mole ratios by dividing both mole quantities by the smallest number of moles. This method is universal and can be applied to any combination of elements in a compound.

Molar Mass

Molar mass, also known as molecular weight, is the mass of one mole of a substance. It provides a bridge between the macroscopic world (grams) and the microscopic world (moles). For each element, you find the molar mass on the periodic table, as it corresponds to the average atomic mass in grams per mole.

Understanding molar mass is crucial when converting between grams and moles. For example, with a molar mass of 197.0 g/mol for Gold (Au) and 16.00 g/mol for Oxygen (O), you convert the mass of each element to moles by dividing their given mass by their molar mass:

• Gold: \( \frac{89.14}{197.0} \approx 0.452 \) moles
• Oxygen: \( \frac{10.86}{16.00} \approx 0.679 \) moles

Calculating molar mass is fundamental in stoichiometry for reactions involving multiple substances, aiding in a detailed grasp of amounts required or produced in a reaction.

Mass Percentage

Mass percentage provides insight into the composition of a compound by specifying how much of each element is present in a given mass of that compound. It is expressed as a percentage of the total mass and helps deduce the empirical formula.

To calculate this for a compound like Gold and Oxygen, we treat the percentage as grams for ease of calculation by assuming a 100 g sample. Thus, if a compound contains 89.14% Gold, we calculate it as 89.14 g out of 100 g.

Once you have the masses, convert these into moles using their respective molar masses and calculate the mole ratios to establish the simplest ratio, which indicates the empirical formula. Here’s how mass percentage is instrumental:

• Assume a total mass (like 100 g) for convenience.
• Convert each element’s mass percentage into grams, then into moles.
• Use the smallest mole number to simplify mole ratios, determining the empirical formula.

Understanding mass percentage is essential to finding out how a substance's mass is distributed across its components in a chemical formula.