Question

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

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

Step-by-step solution

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)

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 \(moles_{P} = \frac{26.7 g}{30.97 g/mol} = 0.862 mol\) \(moles_{N} = \frac{12.1 g}{14.01 g/mol} = 0.863 mol\) \(moles_{Cl} = \frac{61.2 g}{35.45 g/mol} = 1.726 mol\)

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).

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: \(Molar\,mass \,of\, PNCl2 = 30.97 + 14.01 + 2(35.45) = 115.88 g/mol\)

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.

Key concepts

Empirical Formula

The empirical formula represents the simplest whole-number ratio of elements in a compound. It's derived from the percent composition, which details the proportion of each element. By assuming a sample mass of 100 grams, the percentage directly translates into grams.
For example, if a compound contains 26.7% phosphorus, 12.1% nitrogen, and 61.2% chlorine, we can assume:

• 26.7 grams of phosphorus
• 12.1 grams of nitrogen
• 61.2 grams of chlorine

Next, convert these amounts into moles using the molar masses of each element. Divide the number of moles of each element by the smallest mole value to get the simplest integer ratios. These ratios will form the empirical formula of the compound. For the given exercise, this resulted in the empirical formula of PNCl₂.

Mole Ratio

The mole ratio is crucial for determining the empirical formula. It involves comparing the amount (in moles) of each element in a compound. First, compute the number of moles of each element by dividing the mass of the element by its molar mass.
For instance, with phosphorus at 26.7 grams, you compute the moles through \( \frac{26.7g}{30.97 g/mol} \) which simplifies to 0.862 moles.
Similarly, determine the moles for nitrogen and chlorine. Once all mole values are established, identify the smallest value among them. The smallest value serves as the denominator for each mole quantity in the ratio.
In our example, the mole values are:

• Phosphorus: 0.862 mol
• Nitrogen: 0.863 mol
• Chlorine: 1.726 mol

Dividing each by the smallest value (0.862) gives the simplest whole numbers: 1:1:2 ratio, confirming the empirical formula PNCl₂.

Molar Mass Calculation

Molar mass calculations ensure the empirical formula corresponds correctly to the molecular formula of the compound. Begin by determining the molar mass of the empirical formula using the molar masses of its constituent elements.
For PNCl₂, calculate as follows: \( 30.97 \) (for P) + \( 14.01 \) (for N) + \( 2 \times 35.45 \) (for 2 Cl atoms) = 115.88 g/mol. This combined weight is the molar mass of the empirical formula.
With the given molar mass of the compound (580 g/mol in the exercise), it’s vital to ascertain how many times the empirical formula fits into the actual molecular composition. By dividing the molar mass of the compound by the empirical formula’s molar mass, \( \frac{580 g/mol}{115.88 g/mol} \) yields a factor of 5. This factor indicates how many empirical units meld to form the compound's molecular structure, allowing us to conclude the complete molecular formula.

Percentage Composition

Percentage composition assists in defining the makeup of a compound, displaying what percentage of the total mass each individual element holds.
It’s calculated based on the atomic mass of the atoms within the formula. To simplify, presume a total mass of 100 grams for the compound since percentages directly translate into grams under this assumption.
This exercise highlights the necessity to transition from percentage to mass, subsequently to moles, leading to determining the elemental ratios. With values such as 26.7% phosphorus translating to 26.7g in grams, similar transitions occur for other elements.
These initial steps are fundamental in leading to the empirical formula and further to the molecular formula. By making the connections from mass to moles and ratios, one can understand how the element distribution contributes towards the compound's full characterization.