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Chapter 3: Problem 49

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 of the compounds are: (a) \(\mathrm{C}_6\mathrm{H}_{12}\) and (b) \(\mathrm{NH}_2\mathrm{Cl}\).

Step by step solution

01

Determine the molar mass of the empirical formula

First, find the molar mass of the empirical formula, CH₂. The molar mass of carbon (C) is 12 g/mol, and the molar mass of hydrogen (H) is 1 g/mol. So, the total molar mass of CH₂ is: \[12 + 2(1) = 14 \mathrm{~g/mol}\]
02

Determine the whole-number multiple

Next, divide the given molar mass of the compound by the molar mass of the empirical formula: \[\frac{84 \mathrm{~g/mol}}{14 \mathrm{~g/mol}} = 6\] So, the molecular formula is 6 times larger than the empirical formula.
03

Calculate the molecular formula

Multiply each element in the empirical formula, CH₂, by the whole-number multiple, 6: \[6(CH_2) \Rightarrow C_6H_{12}\] So, the molecular formula of this compound is \(\mathrm{C}_6\mathrm{H}_{12}\). #Part (b):#
04

Determine the molar mass of the empirical formula

First, find the molar mass of the empirical formula, NH₂Cl. The molar mass of nitrogen (N) is 14 g/mol, the molar mass of hydrogen (H) is 1 g/mol, and the molar mass of chlorine (Cl) is 35.5 g/mol. So, the total molar mass of NH₂Cl is: \[14 + 2(1) + 35.5 = 51.5 \mathrm{~g/mol}\]
05

Determine the whole-number multiple

Next, divide the given molar mass of the compound by the molar mass of the empirical formula: \[\frac{51.5 \mathrm{~g/mol}}{51.5 \mathrm{~g/mol}} = 1\] So, the molecular formula is 1 time larger than the empirical formula (meaning they are the same).
06

Calculate the molecular formula

Multiply each element in the empirical formula, NH₂Cl, by the whole-number multiple, 1: \[1(NH_2Cl) \Rightarrow NH_2Cl\] So, the molecular formula of this compound is \(\mathrm{NH}_2\mathrm{Cl}\).

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Most popular questions from this chapter

A bottling plant has 126,515 bottles with a capacity of \(355 \mathrm{~mL}\), 108,500 caps, and \(48,775 \mathrm{~L}\) of beverage. (a) How many bottles can be filled and capped? (b) How much of each item is left over? (c) Which component limits the production?

What is the mass, in kilograms, of an Avogadro's number of people, if the average mass of a person is \(160 \mathrm{lb}\) ? How does this compare with the mass of Earth, \(5.98 \times 10^{24} \mathrm{~kg}\) ?

Without doing any detailed calculations (but using a periodic table to give atomic weights), rank the following samples in order of increasing number of atoms: \(0.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\), \(23 \mathrm{~g} \mathrm{Na}, 6.0 \times 10^{23} \mathrm{~N}_{2}\) molecules.

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (e) \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)\)

Calculate the percentage by mass of oxygen in the following compounds: (a) morphine, \(\quad \mathrm{C}_{17} \mathrm{H}_{19} \mathrm{NO}_{3}\); (b) codeine, \(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3} \quad\) (c) cocaine, \(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}\); (d) tetracycline, \(\mathrm{C}_{22} \mathrm{H}_{24} \mathrm{~N}_{2} \mathrm{O}_{8} ;\) (e) digitoxin, \(\mathrm{C}_{41} \mathrm{H}_{64} \mathrm{O}_{13} ;\) (f) vancomycin, \(\mathrm{C}_{66} \mathrm{H}_{75} \mathrm{Cl}_{2} \mathrm{~N}_{9} \mathrm{O}_{24}\)
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