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Chapter 3: Problem 88

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 compound is \(P_5N_5Cl_{10}\).

Step by step solution

01

Calculate the moles of each element in 100 grams of the compound.

Since we are given the percentage composition of each element, we can assume that we have 100 grams of the compound, and based on the given percentages, we have: - \(26.7 g\) of P - \(12.1 g\) of N - \(61.2 g\) of Cl. Now, let's calculate the number of moles for each element: - Moles of P: \(\dfrac{26.7}{30.97} = 0.862\) - Moles of N: \(\dfrac{12.1}{14.01} = 0.864\) - Moles of Cl: \(\dfrac{61.2}{35.45} = 1.727\)
02

Determine the empirical formula.

To find the empirical formula, we need to find the ratio of moles between these elements. We can do this by dividing the number of moles of each element by the smallest number among them: - Moles ratio of P: \(\dfrac{0.862}{0.862} = 1\) - Moles ratio of N: \(\dfrac{0.864}{0.862} \approx 1\) - Moles ratio of Cl: \(\dfrac{1.727}{0.862} \approx 2\) Based on the mole ratios, the empirical formula is \(PNCl_2\).
03

Calculate the molar mass of the empirical formula.

To find the molar mass of the empirical formula, we add up the molar masses of the elements in the empirical formula: Molar mass of \(PNCl_2\) = molar mass of P + molar mass of N + 2 (molar mass of Cl) = \(30.97 + 14.01 + 2 \times 35.45 = 116.88\,g/mol\)
04

Determine the molecular formula using the molar mass.

We are given that the molar mass of the compound is \(580\,g/mol\). To find the molecular formula, we need to find the ratio of the molar mass of the compound to the molar mass of the empirical formula: Molar mass ratio = \(\dfrac{580}{116.88} \approx 5\) Since the ratio is approximately 5, the molecular formula will be 5 times the empirical formula: Molecular Formula = \(5 \times (PNCl_2) = P_5N_5Cl_{10}\)
05

Final Result

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

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Most popular questions from this chapter

Consider the following reaction: $$ 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ If a container were to have 10 molecules of \(\mathrm{O}_{2}\) and 10 molecules of \(\mathrm{NH}_{3}\) initially, how many total molecules (reactants plus products) would be present in the container after this reaction goes to completion?

Vitamin A has a molar mass of \(286.4 \mathrm{~g} / \mathrm{mol}\) and a general molecular formula of \(\mathrm{C}_{x} \mathrm{H}_{2} \mathrm{E}\), where \(\mathrm{E}\) is an unknown element. If vitamin \(\mathrm{A}\) is \(83.86 \% \mathrm{C}\) and \(10.56 \% \mathrm{H}\) by mass, what is the molecular formula of vitamin \(\mathrm{A}\) ?

An iron ore sample contains \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) plus other impurities. \(\mathrm{A}\) \(752-\mathrm{g}\) sample of impure iron ore is heated with excess carbon, producing \(453 \mathrm{~g}\) of pure iron by the following reaction: $$ \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{C}(s) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}(g) $$ What is the mass percent of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) in the impure iron ore sample? Assume that \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) is the only source of iron and that the reaction is \(100 \%\) efficient.

A common demonstration in chemistry courses involves adding a tiny speck of manganese(IV) oxide to a concentrated hydrogen peroxide \(\left(\mathrm{H}_{2} \mathrm{O}_{2}\right)\) solution. Hydrogen peroxide decomposes quite spectacularly under these conditions to produce oxygen gas and steam (water vapor). Manganese(IV) oxide is a catalyst for the decomposition of hydrogen peroxide and is not consumed in the reaction. Write the balanced equation for the decomposition reaction of hydrogen peroxide.

Zinc and magnesium metal each reacts with hydrochloric acid to make chloride salts of the respective metals, and hydrogen gas. A \(10.00-\mathrm{g}\) mixture of zinc and magnesium produces \(0.5171 \mathrm{~g}\) of hydrogen gas upon being mixed with an excess of hydrochloric acid. Determine the percent magnesium by mass in the original mixture.
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