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Chapter 3: Problem 147

The empirical formula of styrene is \(\mathrm{CH} ;\) the molar mass of styrene is 104.14 \(\mathrm{g} / \mathrm{mol}\) . What number of \(\mathrm{H}\) atoms are present in a 2.00 -g sample of styrene?

Short Answer

Expert verified
The number of hydrogen atoms in a 2.00-g sample of styrene is approximately 9.23 x 10²² atoms.

Step by step solution

01

Find the molar mass of the empirical formula.

First, we need to find the molar mass of the empirical formula (CH). The molar mass can be found by adding the molar mass of carbon and hydrogen. Molar mass of Carbon (C): 12.01 g/mol Molar mass of Hydrogen (H): 1.01 g/mol Molar mass of CH = 12.01 g/mol +1.01g/mol = 13.02 g/mol
02

Find the number of CH units in one molecule of styrene.

Next, we need to find the number of CH units that are present in one molecule of styrene. For that, divide the molar mass of styrene by the molar mass of the empirical formula: Number of CH units in styrene = (molar mass of styrene) / (molar mass of CH) = 104.14 g/mol / 13.02 g/mol Number of CH units in styrene ≈ 8
03

Determine the number of moles in the 2.00-g sample.

Now that we know the molar mass of styrene, we can calculate the number of moles in the 2.00-g sample: Number of moles = (mass of sample) / (molar mass of styrene) = 2.00 g / 104.14 g/mol ≈ 0.0192 moles
04

Calculate the number of hydrogen atoms in the 2.00-g sample.

Since there are 8 hydrogen atoms for every molecule of styrene, we will multiply the number of moles by 8 to get the number of moles of hydrogen: Number of hydrogen atoms = (number of moles of styrene) × (number of hydrogen atoms per molecule of styrene) = 0.0192 moles × 8 ≈ 0.1536 moles Now we can convert the number of moles to the number of atoms using Avogadro's number (6.022 x 10²³ atoms/mol): Number of hydrogen atoms = 0.1536 moles × 6.022 x 10²³ atoms/mol ≈ 9.23 x 10²² atoms The number of hydrogen atoms in a 2.00-g sample of styrene is approximately 9.23 x 10²² atoms.

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Most popular questions from this chapter

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

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