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Chapter 3: Problem 47

What is the molecular formula of each of the following compounds? (a) empirical formula \(\mathrm{CH}_{2}\), molar mass \(=84 \mathrm{~g} / \mathrm{mol}\) (b) empirical formula \(\mathrm{NH}_{2} \mathrm{Cl}\), molar mass \(=51.5 \mathrm{~g} / \mathrm{mol}\)

Short Answer

Expert verified
The molecular formula for the compounds are: (a) \(\mathrm{C}_{6} \mathrm{H}_{12}\) (b) \(\mathrm{NH}_{2}\mathrm{Cl}\)

Step by step solution

01

(a) Step 1: Calculate the empirical formula weight for CH2

Calculate the empirical formula weight by adding the atomic weights of the elements in the empirical formula. For \(\mathrm{CH}_{2}\): Empirical Formula Weight = Atomic Weight of Carbon + (2 脳 Atomic Weight of Hydrogen) = 12.01 + (2 脳 1.01) = 12.01 + 2.02 = 14.03 g/mol
02

(a) Step 2: Find the ratio between molar mass and empirical formula weight

Divide the molar mass given by the empirical formula weight. Ratio = (Molar Mass) / (Empirical Formula Weight) = 84 / 14.03 = 6
03

(a) Step 3: Determine the molecular formula

Multiply each element in the empirical formula by the ratio found in Step 2. Molecular Formula = \(\mathrm{C}(6) \mathrm{H}_{2}(6) = \mathrm{C}_{6} \mathrm{H}_{12}\) So, the molecular formula for a compound with an empirical formula of \(\mathrm{CH}_{2}\) and molar mass of 84 g/mol is \(\mathrm{C}_{6} \mathrm{H}_{12}\).
04

(b) Step 1: Calculate the empirical formula weight for NH2Cl

Calculate the empirical formula weight by adding the atomic weights of the elements in the empirical formula. For \(\mathrm{NH}_{2} \mathrm{Cl}\): Empirical Formula Weight = Atomic Weight of Nitrogen + (2 脳 Atomic Weight of Hydrogen) + Atomic Weight of Chlorine = 14.01 + (2 脳 1.01) + 35.45 = 14.01 + 2.02 + 35.45 = 51.48 g/mol
05

(b) Step 2: Find the ratio between molar mass and empirical formula weight

Divide the molar mass given by the empirical formula weight. Ratio = (Molar Mass) / (Empirical Formula Weight) = 51.5 / 51.48 鈮� 1
06

(b) Step 3: Determine the molecular formula

Multiply each element in the empirical formula by the ratio found in Step 2. Since the ratio is approximately 1, the molecular formula is the same as the empirical formula. Molecular Formula = \(\mathrm{NH}_{2}\mathrm{Cl}\) So, the molecular formula for a compound with an empirical formula of \(\mathrm{NH}_{2} \mathrm{Cl}\) and molar mass of 51.5 g/mol is \(\mathrm{NH}_{2}\mathrm{Cl}\).

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Most popular questions from this chapter

A sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), contains \(1.250 \times 10^{21}\) carbon atoms. (a) How many atoms of hydrogen does it contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it contain? (d) What is the mass of this sample in grams?

Why is it essential to use balanced chemical equations when determining the quantity of a product formed from a given quantity of a reactant?

A mixture containing \(\mathrm{KClO}_{3}, \mathrm{~K}_{2} \mathrm{CO}_{3}, \mathrm{KHCO}_{3}\), and \(\mathrm{KCl}\) was heated, producing \(\mathrm{CO}_{2}, \mathrm{O}_{2}\), and \(\mathrm{H}_{2} \mathrm{O}\) gases according to the following equations: $$ \begin{aligned} 2 \mathrm{KClO}_{3}(s) & \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) \\\ 2 \mathrm{KHCO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{CO}_{2}(g) \\ \mathrm{K}_{2} \mathrm{CO}_{3}(s) & \longrightarrow \mathrm{K}_{2} \mathrm{O}(s)+\mathrm{CO}_{2}(g) \end{aligned} $$ The KCl does not react under the conditions of the reaction. If \(100.0 \mathrm{~g}\) of the mixture produces \(1.80 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), \(13.20 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(4.00 \mathrm{~g}\) of \(\mathrm{O}_{2}\), what was the composition of the original mixture? (Assume complete decomposition of the mixture.)

(a) Combustion analysis of toluene, a common organic solvent, gives \(5.86 \mathrm{mg}\) of \(\mathrm{CO}_{2}\) and \(1.37 \mathrm{mg}\) of \(\mathrm{H}_{2} \mathrm{O}\). If the compound contains only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of \(\mathrm{C}, \mathrm{H}\), and \(\mathrm{O}\). A \(0.1005-\mathrm{g}\) sample of menthol is combusted, producing \(0.2829 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) and \(0.1159 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\). What is the empirical formula for menthol? If menthol has a molar mass of \(156 \mathrm{~g} / \mathrm{mol}\), what is its molecular formula?

What parts of balanced chemical equations give information about the relative numbers of moles of reactants and products involved in a reaction?
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