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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) CH₂, 14.03 g/molb) CH₃O, 31.04 g/molc) N₂O₅, 108.02 g/mold) Ba₃(PO₄)₂, 601.93 g/mole) TeI₄, 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

Bismuth oxide reacts with carbon to form bismuth metal: $$ \mathrm{Bi}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Bi}(s)+3 \mathrm{CO}(g) $$ When \(0.607 \mathrm{~mol}\) of \(\mathrm{Bi}_{2} \mathrm{O}_{3}\) reacts with excess carbon, how many (a) moles of Bi and (b) grams of CO form?

Nitrogen (N), phosphorus (P), and potassium (K) are the main nutrients in plant fertilizers. By industry convention, the numbers on a label refer to the mass percents of \(\mathrm{N}, \mathrm{P}_{2} \mathrm{O}_{5},\) and \(\mathrm{K}_{2} \mathrm{O},\) in that order. Calculate the N/P/K ratio of a \(30 / 10 / 10\) fertilizer in terms of moles of each element, and express it as \(x / y / 1.0\).

Menthol \((\mathscr{A} \ell=156.3 \mathrm{~g} / \mathrm{mol}),\) the strong- smelling substance in many cough drops, is a compound of carbon, hydrogen, and oxygen. When \(0.1595 \mathrm{~g}\) of menthol was burned in a combustion apparatus, \(0.449 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.184 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) formed. What is menthol's molecular formula?

A 0.652 -g sample of a pure strontium halide reacts with excess sulfuric acid. The solid strontium sulfate formed is separated, dried, and found to weigh \(0.755 \mathrm{~g}\). What is the formula of the original halide?

A 0.370 -mol sample of a metal oxide \(\left(\mathrm{M}_{2} \mathrm{O}_{3}\right)\) weighs \(55.4 \mathrm{~g}\). (a) How many moles of \(\mathrm{O}\) are in the sample? (b) How many grams of \(\mathrm{M}\) are in the sample? (c) What element is represented by the symbol M?
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