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Chapter 3: Problem 47

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)

Short Answer

Expert verified
The empirical formulas of the given compounds are: (a) C5H7N (b) NaOH (c) N2H8S

Step by step solution

01

(a) Conversion to Moles

For the first compound with the given mass percentages, determine the moles of each element by assuming 100 grams of compound. 74.0 g of C: \(\frac{74.0~g}{12.01~g/mol} \approx 6.16~mol\) of C 8.7 g of H: \(\frac{8.7~g}{1.01~g/mol} \approx 8.61~mol\) of H 17.3 g of N: \(\frac{17.3~g}{14.01~g/mol} \approx 1.235~mol\) of N
02

(a) Calculate the Mole Ratios

Divide all the mole values obtained above by the smallest value to get the mole ratios for the elements. C: \(\frac{6.16}{1.235} \approx 5\) H: \(\frac{8.61}{1.235} \approx 7\) N: \(\frac{1.235}{1.235} \approx 1\)
03

(a) Final Empirical Formula

Write the empirical formula for the first compound by noting the mole ratios. Empirical formula of compound (a): C5H7N
04

(b) Conversion to Moles

For the second compound, we do the same process as in (a) to get the moles of each element: 57.5 g of Na: \(\frac{57.5~g}{22.99~g/mol} \approx 2.50~mol\) of Na 40.0 g of O: \(\frac{40.0~g}{16.00~g/mol} \approx 2.50~mol\) of O 2.5 g of H: \(\frac{2.5~g}{1.01~g/mol} \approx 2.475~mol\) of H
05

(b) Calculate the Mole Ratios

Divide the mole values obtained above by the smallest value. Na: \(\frac{2.50}{2.475} \approx 1\) O: \(\frac{2.50}{2.475} \approx 1\) H: \(\frac{2.475}{2.475} \approx 1\)
06

(b) Final Empirical Formula

Write the empirical formula for the second compound by considering the mole ratios: Empirical formula of compound (b): NaOH
07

(c) Determining the Percentage of S

Since the remainder of the compound is sulfur, we need to calculate its mass percentage. 100% - 41.1 (N) - 11.8 (H) = 47.1% S
08

(c) Conversion to Moles

For the third compound, similarly, obtain the moles of each element and S: 41.1 g of N: \(\frac{41.1~g}{14.01~g/mol} \approx 2.934~mol\) of N 11.8 g of H: \(\frac{11.8~g}{1.01~g/mol} \approx 11.68~mol\) of H 47.1 g of S: \(\frac{47.1~g}{32.06~g/mol} \approx 1.468~mol\) of S
09

(c) Calculate the Mole Ratios

Divide the mole values obtained above by the smallest value. N: \(\frac{2.934}{1.468} \approx 2\) H: \(\frac{11.68}{1.468} \approx 8\) S: \(\frac{1.468}{1.468} \approx 1\)
10

(c) Final Empirical Formula

Write the empirical formula for the third compound by considering the mole ratios: Empirical formula of compound (c): N2H8S So, the empirical formulas of the given compounds are: (a) C5H7N (b) NaOH (c) N2H8S

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Most popular questions from this chapter

Balance the following equations: (a) $\mathrm{CaS}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Ca}(\mathrm{HS})_{2}(a q)+\mathrm{Ca}(\mathrm{OH})_{2}(a q)$ (b) $\mathrm{NH}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (c) $\mathrm{FeCl}_{3}(s)+\mathrm{Na}_{2} \mathrm{CO}_{3}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}(s)+\mathrm{NaCl}(a q)$ (d) $\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{2}(g)$

The thermite reaction, $$ \mathrm{Fe}_{2} \mathrm{O}_{3}+\mathrm{Al} \rightarrow \mathrm{Al}_{2} \mathrm{O}_{3}+\mathrm{Fe} $$ produces so much heat that the Fe product melts. This reaction is used industrially to weld metal parts under water, where a torch cannot be employed. It is also a favorite chemical demonstration in the lecture hall (on a small scale). (a) Balance the chemical equation for the thermite reaction, and include the proper states of matter. (b) Calculate how many grams of aluminum are needed to completely react with \(500.0 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in this reaction. (c) This reaction produces \(852 \mathrm{~kJ}\) of heat per mole of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) reacted. How many grams of $\mathrm{Fe}_{2} \mathrm{O}_{3}\( are needed to produce \)1.00 \times 10^{4} \mathrm{~kJ}$ of heat? (d) If you performed the reverse reaction- aluminum oxide plus iron makes iron oxide plus aluminum-would that reaction have heat as a reactant or a product?

Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A solution containing $3.50 \mathrm{~g}\( of sodium carbonate is mixed with one containing \)5.00 \mathrm{~g}$ of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete?

Boron nitride, \(\mathrm{BN}\), is an electrical insulator with remarkable thermal and chemical stability. Its density is $2.1 \mathrm{~g} / \mathrm{cm}^{3}\(. It can be made by reacting boric acid, \)\mathrm{H}_{3} \mathrm{BO}_{3}$, with ammonia. The other product of the reaction is water. (a) Write a balanced chemical equation for the synthesis of BN. (b) If you made \(225 \mathrm{~g}\) of boric acid react with \(150 \mathrm{~g}\) ammonia, what mass of BN could you make? (c) Which reactant, if any, would be left over, and how many moles of leftover reactant would remain? (d) One application of \(\mathrm{BN}\) is as thin film for electrical insulation. If you take the mass of BN from part (a) and make a \(0.4 \mathrm{~mm}\) thin film from it, what area, in \(\mathrm{cm}^{2}\), would it cover?

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\). The density of air at \(26^{\circ} \mathrm{C}\) is $0.00118 \mathrm{~g} / \mathrm{cm}^{3} .(\mathbf{b})\( If the HCN is formed by reaction of \)\mathrm{NaCN}$ with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon \(^{\text {- }}\) or Acrilan \(^{\circledast}\) burn. Acrilan \(^{\circledast}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is \(50.9 \%\) of the formula by mass. A rug measures \(3.5 \times 4.5 \mathrm{~m}\) and contains $850 \mathrm{~g}\( of Acrilan \)^{\circledast}$ fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.
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