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Chapter 3: Problem 135

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 \(2.00-\mathrm{g}\) sample of styrene?

Short Answer

Expert verified
In a 2.00-g sample of styrene, approximately \(1.16 \times 10^{22}\) hydrogen atoms are present.

Step by step solution

01

Determine the number of moles of styrene in the sample

To find the number of moles in the sample, we will use the equation: Number of moles = mass / molar mass Where the mass is given as 2.00 g, and the molar mass is 104.14 g/mol. Plugging in the values, we get: Number of moles = \(2.00~\text{g}\) / \(104.14\frac{\text{g}}{\text{mol}}\)
02

Calculate the number of moles of styrene

Dividing 2.00 g by 104.14 g/mol, we have: Number of moles = 0.0192 mol So, there are 0.0192 moles of styrene in the 2.00-g sample.
03

Determine the number of hydrogen atoms in one mole of styrene

Since the empirical formula is CH, there is one hydrogen atom for each styrene molecule. Therefore, there will be 1 mole of hydrogen atoms for each mole of styrene molecules.
04

Calculate the number of hydrogen atoms in the sample

We can now use Avogadro's number (6.022 x \(10^{23}\) atoms/mol) to calculate the number of hydrogen atoms in the 0.0192 moles of styrene. Number of hydrogen atoms = moles of hydrogen x Avogadro's number Number of hydrogen atoms = \(0.0192\text{ mol} \times 6.022 \times 10^{23}\text{ atoms/mol}\)
05

Determine the number of hydrogen atoms in the sample

Multiplying 0.0192 mol by 6.022 x \(10^{23}\) atoms/mol, we get: Number of hydrogen atoms = \(1.16 \times 10^{22}\) atoms So, there are approximately \(1.16 \times 10^{22}\) hydrogen atoms in the 2.00-g sample of styrene.

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Most popular questions from this chapter

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$ 2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If \(5.00 \times 10^{3} \mathrm{~kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2}\), and \(\mathrm{CH}_{4}\) are reacted, what mass of \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Balance the following equations: a. \(\mathrm{Ca}(\mathrm{OH})_{2}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) b. \(\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(t)\) c. \(\mathrm{AgNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Ag}_{2} \mathrm{SO}_{4}(s)+\mathrm{HNO}_{3}(a q)\)

When aluminum metal is heated with an element from Group \(6 \mathrm{~A}\) of the periodic table, an ionic compound forms. When the experiment is performed with an unknown Group \(6 \mathrm{~A}\) element, the product is \(18.56 \%\) Al by mass. What is the formula of the compound?

Vitamin \(\mathrm{B}_{12}\), cyanocobalamin, is essential for human nutrition. It is concentrated in animal tissue but not in higher plants. Although nutritional requirements for the vitamin are quite low. people who abstain completely from animal products may develop a deficiency anemia. Cyanocobalamin is the form used in vitamin supplements. It contains \(4.34 \%\) cobalt by mass. Calculate the molar mass of cyanocobalamin, assuming that there is one atom of cobalt in every molecule of cyanocobalamin.

Ammonia reacts with \(\mathrm{O}_{2}\) to form either \(\mathrm{NO}(\mathrm{g})\) or \(\mathrm{NO}_{2}(\mathrm{~g})\) according to these unbalanced equations: $$ \begin{array}{l} \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \\ \mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) \end{array} $$ In a certain experiment \(2.00 \mathrm{~mol} \mathrm{NH}_{3}(g)\) and \(10.00 \mathrm{~mol}\) \(\mathrm{O}_{2}(g)\) are contained in a closed flask. After the reaction is complete, \(6.75 \mathrm{~mol} \mathrm{O}_{2}(g)\) remains. Calculate the number of moles of \(\mathrm{NO}(g)\) in the product mixture: (Hint: You cannot do this problem by adding the balanced equations, because you cannot assume that the two reactions will occur with equal probability.)
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