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Chapter 3: Problem 76

Determine the molecular formulas to which the following empirical formulas and molar masses pertain. a. SNH \((188.35 \mathrm{~g} / \mathrm{mol})\) b. \(\mathrm{NPCl}_{2}(347.64 \mathrm{~g} / \mathrm{mol})\) c. \(\mathrm{CoC}_{4} \mathrm{O}_{4}(341.94 \mathrm{~g} / \mathrm{mol})\) d. SN \((184.32 \mathrm{~g} / \mathrm{mol})\)

Short Answer

Expert verified
The molecular formulas for the given empirical formulas and molar masses are: a. S₄N₄H₄ b. N₃P₃Cl₆ c. Co₂C₈O₈ d. S₄N₄

Step by step solution

01

Calculate the molar mass of the empirical formula

First, find the molar mass of the empirical formula, SNH. Use the periodic table to find the atomic masses of each element: S (32.07 g/mol), N (14.01 g/mol), H (1.01 g/mol). Now, add the atomic masses together to determine the molar mass of the empirical formula: \(32.07 + 14.01 + 1.01 = 47.09 \mathrm{~g/mol}\)
02

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula to calculate the ratio: \(188.35 \mathrm{~g/mol} \div 47.09 \mathrm{~g/mol} = 4\)
03

Determine the molecular formula

Multiply the empirical formula by the ratio obtained in step 2 to obtain the molecular formula: SNH × 4 = S₄N₄H₄. So the molecular formula is S₄N₄H₄. #b. NPCl₂(347.64 g/mol)#
04

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula, NPCl₂: N (14.01 g/mol), P (30.97 g/mol), Cl (35.45 g/mol). Add the atomic masses together: \(14.01 + 30.97 + 2(35.45) = 116.33 \mathrm{~g/mol}\)
05

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula: \(347.64 \mathrm{~g/mol} \div 116.33 \mathrm{~g/mol} = 3\)
06

Determine the molecular formula

Multiply the empirical formula by the ratio: NPCl₂ × 3 = N₃P₃Cl₆. The molecular formula is N₃P₃Cl₆. #c. CoC₄O₄(341.94 g/mol)#
07

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula, CoC₄O₄: Co (58.93 g/mol), C (12.01 g/mol), O (16.00 g/mol). Add the atomic masses together: \(58.93 + 4(12.01) + 4(16.00) = 170.97 \mathrm{~g/mol}\)
08

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula: \(341.94 \mathrm{~g/mol} \div 170.97 \mathrm{~g/mol} = 2\)
09

Determine the molecular formula

Multiply the empirical formula by the ratio: CoC₄O₄ × 2 = Co₂C₈O₈. The molecular formula is Co₂C₈O₈. #d. SN (184.32 g/mol)#
10

Calculate the molar mass of the empirical formula

Calculate the molar mass of the empirical formula, SN: S (32.07 g/mol), N (14.01 g/mol). Add the atomic masses together: \(32.07 + 14.01 = 46.08 \mathrm{~g/mol}\)
11

Find the ratio

Divide the given molar mass by the molar mass of the empirical formula: \(184.32 \mathrm{~g/mol} \div 46.08 \mathrm{~g/mol} = 4\)
12

Determine the molecular formula

Multiply the empirical formula by the ratio: SN × 4 = S₄N₄. The molecular formula is S₄N₄.

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Most popular questions from this chapter

Consider the following balanced chemical equation: $$ A+5 B \longrightarrow 3 C+4 D $$ a. Equal masses of \(\mathrm{A}\) and \(\mathrm{B}\) are reacted. Complete each of the following with either "A is the limiting reactant because \(" ;{ }^{\prime *} \mathrm{~B}\) is the limiting reactant because \("\) or "we cannot determine the limiting reactant because i. If the molar mass of \(\mathrm{A}\) is greater than the molar mass of B, then ii. If the molar mass of \(\mathrm{B}\) is greater than the molar mass of A, then b. The products of the reaction are carbon dioxide (C) and water (D). Compound \(\mathrm{A}\) has the same molar mass as carbon dioxide. Compound \(\mathrm{B}\) is a diatomic molecule. Identify \(\mathrm{com}-\) pound \(\mathrm{B}\) and support your answer. c. Compound \(\mathrm{A}\) is a hydrocarbon that is \(81.71 \%\) carbon by mass. Detemine its empirical and molecular formulas.

The Freons are a class of compounds containing carbon, chlorine, and fluorine. While they have many valuable uses, they have been shown to be responsible for depletion of the ozone in the upper atmosphere. In 1991 , two replacement compounds for Freons went into production: \(\mathrm{HFC}-134 \mathrm{a}\left(\mathrm{CH}_{2} \mathrm{FCF}_{3}\right)\) and \(\mathrm{HCFC}-124\) (CHCIFCF \(_{3}\) ). Calculate the molar masses of these two compounds.

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, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

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?

Boron consists of two isotopes, \({ }^{10} \mathrm{~B}\) and \({ }^{11} \mathrm{~B}\). Chlorine also has two isotopes, \({ }^{35} \mathrm{Cl}\) and \({ }^{37} \mathrm{Cl}\). Consider the mass spectrum of \(\mathrm{BCl}_{3}\). How many peaks would be present, and what approximate mass would each peak correspond to in the \(\mathrm{BCl}_{3}\) mass spectrum?
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