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Chapter 3: Problem 36

What are the empirical formula and empirical formula mass for each of the following compounds? (a) \(\mathrm{C}_{2} \mathrm{H}_{4}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\) (c) \(\mathrm{N}_{2} \mathrm{O}_{5}\) (d) \(\mathrm{Ba}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) (e) \(\mathrm{Te}_{4} \mathrm{I}_{16}\)

Short Answer

Expert verified
a) CH鈧�, 14.03 g/molb) CH鈧僌, 31.04 g/molc) N鈧侽鈧�, 108.02 g/mold) Ba鈧�(PO鈧�)鈧�, 601.93 g/mole) TeI鈧�, 635.20 g/mol

Step by step solution

01

Determine the Empirical Formula of \(\text{C}_{2} \text{H}_{4}\)

The empirical formula represents the simplest whole-number ratio of atoms in a compound. For \(\text{C}_{2} \text{H}_{4}\), the ratio of carbon to hydrogen atoms is 2:4, which simplifies to 1:2. Therefore, the empirical formula is \(\text{CH}_{2}\).
02

Calculate the Empirical Formula Mass of \(\text{CH}_{2}\)

Find the atomic masses of C and H. The atomic mass of carbon \( \text{C} \) is 12.01 g/mol, and hydrogen \( \text{H} \) is 1.01 g/mol.Empirical formula mass of \(\text{CH}_{2}\) = (1 \( \times \) 12.01) + (2 \( \times \) 1.01) = 12.01 + 2.02 = 14.03 g/mol.
03

Determine the Empirical Formula of \(\text{C}_{2} \text{H}_{6} \text{O}_{2}\)

For \(\text{C}_{2} \text{H}_{6} \text{O}_{2}\), the ratio of carbon, hydrogen, and oxygen atoms is 2:6:2, which simplifies to 1:3:1. Therefore, the empirical formula is \(\text{CH}_{3} \text{O}\).
04

Calculate the Empirical Formula Mass of \(\text{CH}_{3} \text{O}\)

Find the atomic masses of C, H, and O. The atomic mass of oxygen \( \text{O} \) is 16.00 g/mol.Empirical formula mass of \(\text{CH}_{3} \text{O}\) = (1 \( \times \) 12.01) + (3 \( \times \) 1.01) + (1 \( \times \) 16.00) = 12.01 + 3.03 + 16.00 = 31.04 g/mol.
05

Determine the Empirical Formula of \(\text{N}_{2} \text{O}_{5}\)

For \(\text{N}_{2} \text{O}_{5}\), the ratio of nitrogen to oxygen atoms is already in the simplest whole-number ratio of 2:5. Therefore, the empirical formula is \(\text{N}_{2} \text{O}_{5}\).
06

Calculate the Empirical Formula Mass of \(\text{N}_{2} \text{O}_{5}\)

Find the atomic masses of N and O.Empirical formula mass of \(\text{N}_{2} \text{O}_{5}\) = (2 \( \times \) 14.01) + (5 \( \times \) 16.00) = 28.02 + 80.00 = 108.02 g/mol.
07

Determine the Empirical Formula of \(\text{Ba}_{3}(\text{PO}_{4})_{2}\)

For \(\text{Ba}_{3}(\text{PO}_{4})_{2}\), the ratio of barium, phosphorus, and oxygen atoms in the simplest form is already 3:2:8. Therefore, the empirical formula is \(\text{Ba}_{3}(\text{PO}_{4})_{2}\).
08

Calculate the Empirical Formula Mass of \(\text{Ba}_{3}(\text{PO}_{4})_{2}\)

Find the atomic masses of Ba, P, and O. The atomic mass of barium \( \text{Ba} \) is 137.33 g/mol and phosphorus \( \text{P} \) is 30.97 g/mol.Empirical formula mass of \(\text{Ba}_{3}(\text{PO}_{4})_{2}\) = (3 \( \times \) 137.33) + (2 \( \times \) 30.97) + (8 \( \times \) 16.00) = 411.99 + 61.94 + 128.00 = 601.93 g/mol.
09

Determine the Empirical Formula of \(\text{Te}_{4} \text{I}_{16}\)

For \(\text{Te}_{4} \text{I}_{16}\), the ratio of tellurium to iodine atoms is 4:16, which simplifies to 1:4. Therefore, the empirical formula is \(\text{Te} \text{I}_{4}\).
10

Calculate the Empirical Formula Mass of \(\text{TeI}_{4}\)

Find the atomic masses of Te and I. The atomic mass of tellurium \( \text{Te} \) is 127.60 g/mol and iodine \( \text{I} \) is 126.90 g/mol. Empirical formula mass of \(\text{TeI}_{4}\) = (1 \( \times \) 127.60) + (4 \( \times \) 126.90) = 127.60 + 507.60 = 635.20 g/mol.

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Key Concepts

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

empirical formula calculation
The empirical formula of a compound shows the simplest whole-number ratio of the elements in the compound. This means you reduce the subscripts to the smallest possible numbers while keeping the same ratio between the elements. For instance, if a compound has a molecular formula of \(\text{C}_{2} \text{H}_{4}\), you would find the simplest ratio of carbon to hydrogen atoms. Here, the ratio is 2:4, which simplifies to 1:2. Therefore, the empirical formula is \(\text{CH}_{2}\). Always remember, the empirical formula may not always be the same as the molecular formula, but it represents the simplest whole-number ratio of the atoms.
elemental ratio simplification
Step two involves simplifying the ratios of each element in the compound to their lowest terms. For instance, let's consider another compound, \(\text{C}_{2} \text{H}_{6} \text{O}_{2}\). Here, the carbon to hydrogen to oxygen ratio is initially 2:6:2. Simplifying each part by dividing by the greatest common factor, which in this case is 2, simplifies the ratio to 1:3:1. So, the empirical formula simplifies to \(\text{CH}_{3} \text{O}\). Understanding how to simplify these ratios is crucial since it helps in determining the empirical formula from a given molecular formula.
atomic mass determination
To calculate the empirical formula mass, you need to know the atomic masses of all the involved elements. Let's take the empirical formula \(\text{CH}_{2}\) as an example. Carbon (C) has an atomic mass of 12.01 g/mol, and hydrogen (H) has an atomic mass of 1.01 g/mol. To find the empirical formula mass of \(\text{CH}_{2}\), calculate the sum of the atomic masses of each element multiplied by its subscript: \((1 \times 12.01) + (2 \times 1.01) = 12.01 + 2.02 = 14.03 \text{ g/mol}\). Doing this for any empirical formula will give you its empirical formula mass. This process is repeated for every compound to ensure you have the correct mass.
chemical compound analysis
Chemical compound analysis involves using both the empirical formula and atomic masses to understand the compound's composition. For example, let's take \(\text{Ba}_{3} \text{(PO}_{4} \text{)}_{2}\). Here, barium (Ba) has an atomic mass of 137.33 g/mol, phosphorus (P) is 30.97 g/mol, and oxygen (O) is 16.00 g/mol. The empirical formula mass would thus be \((3 \times 137.33) + (2 \times 30.97) + (8 \times 16.00) = 411.99 + 61.94 + 128.00 = 601.93 \text{ g/mol}\). This detailed breakdown helps in understanding the compound makeup, making it easier to manipulate during chemical reactions or in practical applications.

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Most popular questions from this chapter

Calculate each of the following quantities: (a) Amount (mol) of Mn atoms in \(62.0 \mathrm{mg}\) of Mn (b) Amount (mol) for \(1.36 \times 10^{22}\) atoms of \(\mathrm{Cu}\) (c) Mass (g) of \(8.05 \times 10^{24} \mathrm{Li}\) atoms

Percent yields are generally calculated from masses. Would the result be the same if amounts (mol) were used instead? Why?

What advantage is there to using a counting unit (the mole), rather than a mass unit, for amount of substance?

Find the empirical formula of each of the following compounds: (a) 0.063 mol of chlorine atoms combined with \(0.22 \mathrm{~mol}\) of oxygen atoms; (b) \(2.45 \mathrm{~g}\) of silicon combined with \(12.4 \mathrm{~g}\) of chlorine; (c) 27.3 mass \% carbon and 72.7 mass \% oxygen

Aspirin (acetylsalicylic acid, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}\) ) is made by reacting salicylic acid \(\left(\mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}\right)\) with acetic anhydride \(\left[\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\right]:\) $$ \mathrm{C}_{7} \mathrm{H}_{6} \mathrm{O}_{3}(s)+\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4}(s)+\mathrm{CH}_{3} \mathrm{COOH}(l) $$ In one preparation, \(3.077 \mathrm{~g}\) of salicylic acid and \(5.50 \mathrm{~mL}\) of acetic anhydride react to form \(3.281 \mathrm{~g}\) of aspirin. (a) Which is the limiting reactant (the density of acetic anhydride is \(1.080 \mathrm{~g} / \mathrm{mL}) ?\) (b) What is the percent yield of this reaction? (c) What is the percent atom economy of this reaction?
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