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Chapter 3: Problem 82

Determine the molecular formula of a compound that contains \(26.7 \% \mathrm{P}, 12.1 \% \mathrm{~N}\), and \(61.2 \% \mathrm{Cl}\), and has a molar mass of \(580 \mathrm{~g} / \mathrm{mol}\).

Short Answer

Expert verified
The molecular formula of the given compound is \(P_5N_5Cl_{10}\).

Step by step solution

01

Convert the percentage composition to grams

We can assume that we have 100 grams of the compound, making the conversion simpler. Therefore, we have: - 26.7 g of Phosphorus (P) - 12.1 g of Nitrogen (N) - 61.2 g of Chlorine (Cl)
02

Convert grams to moles

Now, convert the mass of each element in grams to moles using their respective molar masses: Molar mass of Phosphorus (P): 30.97 g/mol Molar mass of Nitrogen (N): 14.01 g/mol Molar mass of Chlorine (Cl): 35.45 g/mol \(moles_{P} = \frac{26.7 g}{30.97 g/mol} = 0.862 mol\) \(moles_{N} = \frac{12.1 g}{14.01 g/mol} = 0.863 mol\) \(moles_{Cl} = \frac{61.2 g}{35.45 g/mol} = 1.726 mol\)
03

Find the mole ratio

Divide each of the obtained moles by the smallest value among them to find the mole ratio for each element: \(Ratio_{P} = \frac{0.862}{0.862} = 1\) \(Ratio_{N} = \frac{0.863}{0.862} = 1\) \(Ratio_{Cl} = \frac{1.726}{0.862} = 2\) Hence, the empirical formula is P1N1Cl2 (PNCl2).
04

Calculate the molar mass of the empirical formula

Next, we need to calculate the molar mass of the empirical formula (PNCl2) using the molar masses of the individual elements: \(Molar\,mass \,of\, PNCl2 = 30.97 + 14.01 + 2(35.45) = 115.88 g/mol\)
05

Find the molecular formula using the molar mass provided

Now, we'll use the provided molar mass (580 g/mol) to find the molecular formula. Divide the given molar mass by the molar mass of the empirical formula and round it to the nearest whole number. This will give us the factor that we'll multiply the empirical formula by. \(Factor = \frac{580}{115.88} = 5.00\) Multiplying the empirical formula (PNCl2) by this factor: Molecular Formula = P(1*5)N(1*5)Cl(2*5) = P5N5Cl10 So, the molecular formula of the given compound is P5N5Cl10.

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Most popular questions from this chapter

A 9.780-g gaseous mixture contains ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}}\right) .\) Complete combustion to form carbon dioxide and water requires \(1.120\) mol oxygen. Calculate the mass percent of ethane in the original mixture.

Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed as Nutra-Sweet. The molecular formula of aspartame is \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{~N}_{2} \mathrm{O}_{5}\) a. Calculate the molar mass of aspartame. b. What amount (moles) of molecules are present in \(10.0 \mathrm{~g}\) aspartame? c. Calculate the mass in grams of \(1.56\) mol aspartame. d. What number of molecules are in \(5.0 \mathrm{mg}\) aspartame? e. What number of atoms of nitrogen are in \(1.2 \mathrm{~g}\) aspartame? f. What is the mass in grams of \(1.0 \times 10^{9}\) molecules of aspartame? g. What is the mass in grams of one molecule of aspartame?

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

Ammonia is produced from the reaction of nitrogen and hydrogen according to the following balanced equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ a. What is the maximum mass of ammonia that can be produced from a mixture of \(1.00 \times 10^{3} \mathrm{~g} \mathrm{~N}_{2}\) and \(5.00 \times 10^{2} \mathrm{~g} \mathrm{H}_{2} ?\) b. What mass of which starting material would remain unreacted?

Consider the reaction $$ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) $$ Identify the limiting reagent in each of the reaction mixtures given below: a. 50 molecules of \(\mathrm{H}_{2}\) and 25 molecules of \(\mathrm{O}_{2}\) b. 100 molecules of \(\mathrm{H}_{2}\) and 40 molecules of \(\mathrm{O}_{2}\) c. 100 molecules of \(\mathrm{H}_{2}\) and 100 molecules of \(\mathrm{O}_{2}\) d. \(0.50 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}\). e. \(0.80 \mathrm{~mol} \mathrm{H}_{2}\) and \(0.75 \mathrm{~mol} \mathrm{O}_{2}\) f. \(1.0 \mathrm{~g} \mathrm{H}_{2}\) and \(0.25 \mathrm{~mol} \mathrm{O}_{2}\) g. \(5.00 \mathrm{~g} \mathrm{H}_{2}\) and \(56.00 \mathrm{~g} \mathrm{O}_{2}\)
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