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The compound \(\mathrm{XYZ}_{3}\) has a molar mass of \(100.09 \mathrm{~g}\) and a percent composition (by mass) of \(40.04 \% \mathrm{X}, 12.00 \% \mathrm{Y}\), and \(47.96 \% \mathrm{Z}\). What is the formula of the compound?

Short Answer

Expert verified
The molecular formula of the compound is \(\mathrm{XYZ}_3\).

Step by step solution

01

Determine the Mass of Each Element in 100 g of Compound

Assume a 100 g sample of the compound. Thus, the masses of the elements would be:- X: 40.04 g- Y: 12.00 g- Z: 47.96 g
02

Calculate the Moles of Each Element

Use the molar masses of each element to find the number of moles:- Moles of X = \( \frac{40.04}{\text{Molar mass of X}} \)- Moles of Y = \( \frac{12.00}{\text{Molar mass of Y}} \)- Moles of Z = \( \frac{47.96}{\text{Molar mass of Z}} \)
03

Find the Simplest Whole Number Ratio

Divide the number of moles of each element by the smallest number of moles calculated. This will provide the simplest whole number ratio of atoms.
04

Determine the Empirical Formula

Use the whole number ratios to determine the empirical formula of the compound. This will be in the form of \(\mathrm{X}_a\mathrm{Y}_b\mathrm{Z}_c\).
05

Convert to the Molecular Formula

If necessary, use the molar mass of the compound (100.09 g/mol) and the empirical formula mass to find the molecular formula. Multiply the subscripts in the empirical formula by the ratio of the compound's molar mass to the empirical formula mass.

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Key Concepts

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

Percent Composition
Understanding percent composition is crucial for determining the chemical formula of a compound. Percent composition by mass refers to the percentage of each element in a compound relative to the total mass of the compound. For example, in the compound \(\text{XYZ}_3\), if 40.04% is \'X\', 12.00% is \'Y\', and 47.96% is \'Z\', these figures represent the mass contribution of each element to the compound. This initial data helps us determine how many grams of each element are present in a 100 g sample, simplifying the following calculations.
Molar Mass
Molar mass is the mass of one mole of a given substance, typically measured in grams per mole (g/mol). In this exercise, knowing the molar mass of each element (X, Y, and Z) is essential for converting mass (grams) to moles. For instance, the molar masses allow us to calculate the number of moles of each element by dividing the mass of each element by its molar mass: \( \text{Moles of X} = \frac{40.04 \text{ g}}{\text{Molar mass of X}} \). This conversion from mass to moles is a necessary step in determining the empirical formula.
Chemical Formula Determination
Chemical formula determination involves finding the simplest whole number ratio of atoms in a compound, known as the empirical formula. To do this, we start by calculating the moles of each element. We then divide each mole value by the smallest number of moles calculated among the elements to get a simple ratio. For example, if we calculated 2 moles of X, 1 mole of Y, and 3 moles of Z, and 1 mole is the smallest calculated amount, then the ratio is 2:1:3. These ratios give us the empirical formula (e.g., \( \text{X}_2 \text{Y}_1 \text{Z}_3 \)). The molecular formula is then found by comparing the empirical formula mass with the given molar mass of the compound. If needed, this ratio is used to scale up the empirical formula to match the given molar mass.
Mole Concept
The mole concept is central to understanding chemical reactions and stoichiometry. A mole defines a specific number of entities, usually 6.022 x 10^23 atoms or molecules (Avogadro's number). In the context of the exercise, it helps in converting mass percentages into the actual number of moles of each element. This is achieved by dividing the given mass of each element by its molar mass. Converting mass values into moles using the molar mass makes it easier to derive the simplest whole number ratio of atoms in the compound, which is key to determining both the empirical and molecular formulas.

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Most popular questions from this chapter

The compound \(\mathrm{X}_{2}\left(\mathrm{YZ}_{3}\right)_{3}\) has a molar mass of \(282.23 \mathrm{~g}\) and a percent composition (by mass) of \(19.12 \% \mathrm{X}, 29.86 \% \mathrm{Y}\), and \(51.02 \% \mathrm{Z}\). What is the formula of the compound?

Determine the molar masses of these compounds: (a) \(\mathrm{NaOH}\) (f) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\) (b) \(\mathrm{Ag}_{2} \mathrm{CO}_{3}\) (g) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (c) \(\mathrm{Cr}_{2} \mathrm{O}_{3}\) (h) \(\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6}\) (d) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}\) (i) \(\mathrm{BaCl}_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) (e) \(\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}\)

Given 1.00-g samples of each of the compounds \(\mathrm{CO}_{2}, \mathrm{O}_{2}\), \(\mathrm{H}_{2} \mathrm{O}\), and \(\mathrm{CH}_{3} \mathrm{OH}\), (a) which sample will contain the largest number of molecules? (b) which sample will contain the largest number of atoms? Show proof for your answers.

Calcium tartrate is used as a preservative for certain foods and as an antacid. It contains \(25.5 \% \mathrm{C}, 2.1 \% \mathrm{H}, 21.3 \% \mathrm{Ca}\), and \(51.0 \% \mathrm{O}\). What is the emnirical formula for calcium tartrate?

Cosmone is a molecule used by fragrance manufacturers to provide a rich and elegant musky essence to many perfumes. Cosmone has the molecular formula \(\mathrm{C}_{15} \mathrm{H}_{26} \mathrm{O}\). (a) Calculate the molar mass of cosmone. (b) Calculate the mass of \(3.82\) moles of cosmone. (c) Calculate the number of molecules of cosmone in a sample containing \(8.36 \times 10^{-4} \mathrm{~mol}\) cosmone. (d) Calculate the number of moles of carbon in a \(8.35-\mathrm{mol}\) sample of cosmone. (e) Calculate the mass of oxygen in a 4.29-g sample of cosmone. (f) Calculate the mass of a sample of cosmone that contains \(6.58 \times 10^{19}\) atoms of hydrogen. (g) Calculate the mass of one molecule of cosmone. (h) Calculate the number of atoms of carbon in \(8.00-g\) of cosmone.

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