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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 PClâ‚‚N.

Step by step solution

01

Calculate moles of each element

First, we need to calculate the moles of each element present in the compound. The given percentages of the compound are: - P: 26.7% - N: 12.1% - Cl: 61.2% Since we are given a molar mass of 580 g/mol, we can use these percentages to calculate the grams of each element present. - Mass of P: (\(0.267 \times 580\text{g/mol}\)) = 154.86 g/mol - Mass of N: (\(0.121 \times 580\text{g/mol}\)) = 70.18 g/mol - Mass of Cl: (\(0.612 \times 580\text{g/mol}\)) = 354.96 g/mol Next, divide the mass of each element by its respective molar mass to determine the moles: - Moles of P: (\(\frac{154.86\text{g/mol}}{30.97\text{g/mol}}\)) = 5.0 mol - Moles of N: (\(\frac{70.18\text{g/mol}}{14.01\text{g/mol}}\)) = 5.0 mol - Moles of Cl: (\(\frac{354.96\text{g/mol}}{35.45\text{g/mol}}\)) = 10.0 mol
02

Determine the mole ratios

To find the mole ratio of the elements in the compound, we'll divide the moles of each element by the smallest number of moles: - P: \(\frac{5.0}{5.0}\) = 1 - N: \(\frac{5.0}{5.0}\) = 1 - Cl: \(\frac{10.0}{5.0}\) = 2 Thus, the mole ratio of P:N:Cl in the compound is 1:1:2.
03

Convert the mole ratio into the molecular formula

Following the mole ratio obtained in the previous step, the molecular formula for the given compound is P1N1Cl2, or simply PClâ‚‚N.

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Key Concepts

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

Understanding Mole Ratio
In chemistry, the mole ratio is a fundamental concept that helps determine the proportions of elements in a compound. When a compound's composition is given, as in the percentage of each element, the first step is to convert these percentages into a usable form by calculating the moles of each element. This calculation involves using the molar mass of each element, which allows us to convert grams of the element into moles.
The mole ratio is then found by comparing the amounts of each element in moles. The goal is to express these quantities in their simplest whole number ratios, often by dividing each by the smallest mole value present among them. For instance, if you have 5 moles of phosphorus (P), 5 moles of nitrogen (N), and 10 moles of chlorine (Cl), you divide all moles by 5, the smallest value, resulting in a simple whole number ratio of 1:1:2 for P:N:Cl.
  • This ratio is crucial as it determines the subscript numbers in the empirical formula.
  • The empirical formula reflects the simplest ratio of the compound's elements.
Defining Molar Mass
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole ( \( ext{g/mol} \)). It is an essential concept, especially when relating mass to moles, allowing us to calculate how much of a substance is present for a given mass.
Understanding molar mass begins with the periodic table. Each element has an atomic mass unit, which is numerically similar to its molar mass. For example, phosphorus (P) has a molar mass of 30.97 g/mol. This means 30.97 grams of phosphorus is equivalent to one mole of phosphorus atoms.
  • Molar mass allows conversion from mass to moles, a crucial step in determining mole ratios and molecular formulas.
  • In the problem, the molar mass of 580 g/mol is a value representing the entire compound and helps verify the molecular formula's accuracy.
Percentage Composition in Chemistry
Percentage composition indicates the relative mass or percentage mass of each element in a compound. It is an informative representation of a compound's formula, showing how much each element contributes to the compound's total mass.
To find the percentage composition, one calculates the mass fraction of each element in the compound and then expresses it as a percentage. For example, if a compound contains 26.7% phosphorus, it means that in a sample weighing 100 grams, 26.7 grams would be phosphorus. This is often the starting point in problems involving molecular or empirical formula determination.
  • Percentage composition is used to guide calculations of individual element masses within a given compound.
  • These calculations lead directly to the next step: converting masses to moles.
Moles Calculation Methodology
Calculating moles involves determining how many units of a substance are present, using the relation between mass and molar mass. The formula for calculating moles is given by:
\[ ext{Moles} = \frac{ ext{Mass in grams}}{ ext{Molar mass in g/mol}} \]
This calculation is the foundation for advancing to other determinations, such as mole ratios and eventually finding a compound's molecular formula.
In practice, these steps involve:
  • Determining the mass of each element based on the percentage composition provided.
  • Utilizing the known molar mass for each element from the periodic table to calculate the mole values.
  • Using these mole values to compute the mole ratio.
For instance, by knowing the mass of phosphorus, nitrogen, and chlorine in the compound as deduced from the given percentages, and dividing these masses by their respective molar masses, we can find out the moles of each element present.

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Most popular questions from this chapter

Consider the following unbalanced equation: $$ \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{CaSO}_{4}(s)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) $$ What masses of calcium sulfate and phosphoric acid can be produced from the reaction of 1.0 \(\mathrm{kg}\) calcium phosphate with 1.0 \(\mathrm{kg}\) concentrated sulfuric acid \(\left(98 \% \mathrm{H}_{2} \mathrm{SO}_{4} \text { by mass)? }\right.\)

Consider the following unbalanced chemical equation for the combustion of pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right) :\) $$ \mathrm{C}_{5} \mathrm{H}_{12}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) $$ If 20.4 g of pentane are burned in excess oxygen, what mass of water can be produced, assuming 100\(\%\) yield?

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

According to the law of conservation of mass, mass cannot be gained or destroyed in a chemical reaction. Why can’t you simply add the masses of two reactants to determine the total mass of product?

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?

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