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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鈧匩鈧匔l鈧佲個.

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% 鈫 26.7 g 2. N: 12.1% 鈫 12.1 g 3. Cl: 61.2% 鈫 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) 梅 0.861 鈮 (1:1:2) So, the empirical formula for this compound will be P鈧丯鈧丆l鈧, 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 = P鈧匩鈧匔l鈧佲個 So, the molecular formula of the compound is P鈧匩鈧匔l鈧佲個.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Empirical Formula
In chemistry, the empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It does not show the exact number of atoms, but rather the simplest form that indicates their relative proportions.

To determine the empirical formula, you typically start by converting the mass of each element in the compound to moles. As in our exercise, assuming that the total mass is 100 grams makes the conversion straightforward since percentages can be directly taken as grams. For example:
  • P: 26.7% translates to 26.7 grams.
  • N: 12.1% translates to 12.1 grams.
  • Cl: 61.2% translates to 61.2 grams.
Next, using the molar mass of each element, convert these masses into moles. This helps in identifying the ratio of elements. Once the moles are calculated, the smallest number of moles is used to divide each to find the simple ratio. Thus, deriving the empirical formula requires understanding these simple steps to showcase the ratio correctly.
Moles Calculation
For chemists, calculating moles is a foundational skill. It involves converting the mass of an element into moles using its molar mass, which is the mass of one mole of that element.

In our context, when given a percentage composition, you first think of it as a mass in grams if the whole is 100 grams. Then you use the molar mass of each element to convert the mass to moles. Let's walk through how this is done:
  • P: Convert 26.7 g to moles using 31 g/mol (molar mass of P).
  • N: Convert 12.1 g to moles using 14 g/mol (molar mass of N).
  • Cl: Convert 61.2 g to moles using 35.5 g/mol (molar mass of Cl).
The calculation considers this formula: \[ \text{Moles} = \frac{\text{Mass in grams}}{\text{Molar mass in g/mol}} \] For instance, the moles of phosphorus would be calculated as \( \frac{26.7}{31} \approx 0.861 \text{ mol} \). This same approach applies to the other elements, helping to reveal the simplest mole ratios.
Molar Mass
The molar mass is a crucial component in determining and working with the molecular and empirical formulas. It represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). Each element has its specific molar mass, based on the atomic masses listed in the periodic table.

In the example at hand, the molar masses are:
  • Phosphorus (P): 31 g/mol
  • Nitrogen (N): 14 g/mol
  • Chlorine (Cl): 35.5 g/mol
Grouping these together to find the empirical formula鈥檚 molar mass, you calculate the combined contribution of each element: \[ 31 \ \text{g/mol} + 14 \ ext{g/mol} + 2 \times 35.5 \ ext{g/mol} = 116 \ ext{g/mol} \] This tells us the weight of one mole of the empirical formula, in this case, PNCl鈧. To find the molecular formula, this value is compared with the given compound's molar mass allowing us to scale up as necessary.
Percentage Composition
Percentage composition provides insight into the relative amount of each element within a compound. It鈥檚 a way of describing the elements' concentrations as percentages by mass鈥攁 crucial step in working out molecular formulas.

Using the data from our exercise:
  • Phosphorus: 26.7%
  • Nitrogen: 12.1%
  • Chlorine: 61.2%
These percentages mean that in 100 grams of the compound, there would be 26.7 grams of phosphorus, 12.1 grams of nitrogen, and 61.2 grams of chlorine.
This method simplifies the conversion of percentages directly to grams when considering a 100g sample size, allowing straightforward calculations into moles. Understanding percentage composition helps in visualizing compound makeup, and this estimation is key for both empirical and molecular formula calculations.

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Most popular questions from this chapter

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 following unbalanced reaction: $$\mathrm{P}_{4}(s)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{PF}_{3}(g)$$ What mass of \(\mathrm{F}_{2}\) is needed to produce \(120 . \mathrm{g}\) of \(\mathrm{PF}_{3}\) if the reaction has a \(78.1 \%\) yield?

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