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Chapter 5: 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 P₅N₅Cl₁₀.

Step by step solution

01

Calculate moles of constituent elements from their given percentages

To determine the empirical formula, we first need to convert these given percentages into moles for each element. We can assume we have 100g of the compound, which will make our task easier. The percentage amounts will then become grams: 1. P: 26.7% → 26.7 g 2. N: 12.1% → 12.1 g 3. Cl: 61.2% → 61.2 g Now, we need to convert these masses into moles using their respective molar masses. The molar masses for P, N, and Cl are approximately 31 g/mol, 14 g/mol, and 35.5 g/mol, respectively.
02

Convert grams into moles

1. Moles of P: \( \frac{26.7 \ \mathrm{g}}{31 \ \mathrm{g/mol}} \approx 0.861 \ \mathrm{mol} \) 2. Moles of N: \( \frac{12.1 \ \mathrm{g}}{14 \ \mathrm{g/mol}} \approx 0.864 \ \mathrm{mol} \) 3. Moles of Cl: \( \frac{61.2 \ \mathrm{g}}{35.5 \ \mathrm{g/mol}} \approx 1.724 \ \mathrm{mol} \)
03

Determine the mole ratio

Next, divide all the mole values by the smallest mole value to get the mole ratio. Mole ratio (P:N:Cl) = (0.861:0.864:1.724) ÷ 0.861 ≈ (1:1:2) So, the empirical formula for this compound will be P₁N₁Cl₂, which has a molar mass of approximately: \(31 \ \mathrm{g/mol} + 14\ \mathrm{g/mol} + 2 \times 35.5 \ \mathrm{g/mol} = 116 \ \mathrm{g/mol}\)
04

Determine the molecular formula

Now, we need to find the factor(system)(integer) we need to multiply the empirical formula to get the molecular formula. To do this, divide the given molar mass of the compound (580 g/mol) by the molar mass of the empirical formula (116 g/mol): \( \frac{580 \ \mathrm{g/mol}}{116 \ \mathrm{g/mol}} = 5 \) The molecular formula will then be obtained by multiplying the empirical formula with the factor, 5: Molecular Formula = P₅N₅Cl₁₀ So, the molecular formula of the compound is P₅N₅Cl₁₀.

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Most popular questions from this chapter

Hydrogen peroxide is used as a cleansing agent in the treatment of cuts and abrasions for several reasons. It is an oxidizing agent that can directly kill many microorganisms; it decomposes on contact with blood, releasing elemental oxygen gas (which inhibits the growth of anaerobic microorganisms); and it foams on contact with blood, which provides a cleansing action. In the laboratory, small quantities of hydrogen peroxide can be prepared by the action of an acid on an alkaline earth metal peroxide, such as barium peroxide: $$\mathrm{BaO}_{2}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{BaCl}_{2}(a q)$$ What mass of hydrogen peroxide should result when \(1.50 \mathrm{g}\) barium peroxide is treated with \(25.0 \mathrm{mL}\) hydrochloric acid solution containing \(0.0272 \mathrm{g}\) HCl per mL? What mass of which reagent is left unreacted?

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 compound contains only carbon, hydrogen, and oxygen. Combustion of \(10.68 \mathrm{mg}\) of the compound yields \(16.01 \mathrm{mg}\) \(\mathrm{CO}_{2}\) and \(4.37 \mathrm{mg} \mathrm{H}_{2} \mathrm{O} .\) The molar mass of the compound is \(176.1 \mathrm{g} / \mathrm{mol} .\) What are the empirical and molecular formulas of the compound?

Over the years, the thermite reaction has been used for welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors. The reaction is $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s) \longrightarrow 2 \mathrm{Fe}(l)+\mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ What masses of iron(III) oxide and aluminum must be used to produce \(15.0 \mathrm{g}\) iron? What is the maximum mass of aluminum oxide that could be produced?

A \(2.077-\mathrm{g}\) sample of an element, which has an atomic mass between 40 and \(55,\) reacts with oxygen to form 3.708 g of an oxide. Determine the formula of the oxide (and identify the element).
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