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?

Short Answer

Expert verified
Solve for X, Y, Z molar masses assuming approximate values and test on total molar mass 282.23 g/mol.

Step by step solution

01

Find the mass of each element in 100 grams of compound

Given the percent compositions: \(19.12\text{ % X, } 29.86\text{ % Y, } \text{and } 51.02\text{ % Z}\). This means in 100 grams of compound, there are: \(19.12 \text{ g of X, } 29.86 \text{ g of Y, and } 51.02 \text{ g of Z}\).
02

Convert masses to moles

Assume the molar masses are: \(X \text{: M}_X \text{ g/mol, Y: M}_Y \text{ g/mol, Z: M}_Z \text{ g/mol}\). Using the formula: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\), calculate the number of moles of each element: \( \text{moles of X} = \frac{19.12 \text{ g}}{M_X} \text{moles of Y} = \frac{29.86 \text{ g}}{M_Y} \text{moles of Z} = \frac{51.02 \text{ g}}{M_Z} \).
03

Define the moles in terms of the compound formula

For the compound \(\text{X}_2\text{(YZ}_3\text{)}_3\): Let the moles of the formula be \(2a\) for X, \(3a\) for Y, and \(9a\) for Z, based on the structure. So: \(19.12 = 2a \times M_X \29.86 = 3a \times M_Y \51.02 = 9a \times M_Z\).
04

Solve for the molar masses of X, Y, and Z

Now, we solve these equations assuming \(a\) is a constant. For example: \(M_X = \frac{19.12}{2a}, M_Y = \frac{29.86}{3a}, M_Z = \frac{51.02}{9a}\). Utilize the total molar mass of the compound \(282.23 \text{ g/mol}\): \(2a \times M_X + 3a \times M_Y + 9a \times M_Z = 282.23\).
05

Observe and apply

Because of their relationship in the compound and the option to scale, we find a logical value for \(a\) which will satisfy our equation, acknowledging typical molar masses from the periodic table.
06

Solve numerically for known molar masses

Given common elements and logical values for \(M_X, M_Y, M_Z\), calculate \(a\) accordingly and verify that it fits the given molar mass and percentages appropriately. Usually determined elements algebraically.

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Key Concepts

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

Percent Composition by Mass
Percent composition by mass is an important concept in chemistry. It tells us the percentage of each element in a compound by weight. It’s calculated using the formula:

\(\text{Percent Composition} = \frac{\text{mass of element in 1 mole of the compound}}{\text{molar mass of compound}} \times 100\)

Here's how you can understand this better with the compound \(\text{X}_{2}\text{(YZ}_{3}\text{)}_{3}\). Given the percent compositions: 19.12% X, 29.86% Y, and 51.02% Z, in 100 grams of the compound, these percentages translate directly to the mass of each element: 19.12 g of X, 29.86 g of Y, and 51.02 g of Z.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance (usually in g/mol). To find the molar mass of a compound, you sum up the molar masses of all atoms in its formula.

Given the compound \(\text{X}_{2}\text{(YZ}_{3}\text{)}_{3}\), the total molar mass is 282.23 g/mol.

Therefore, the problem is worked out by assuming molar masses for X, Y, and Z. We use their masses (from percent composition) and convert these to moles by the formula:

\(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\).

For example, moles of X are calculated as:

\(\frac{19.12 \text{ g}}{M_{X}}\)

And similarly for Y and Z. By substituting these back into the compound’s structure and solving, we can find the approximate molar masses of X, Y, and Z.
Mole Concept
The mole concept is fundamental in chemistry. A mole represents 6.022 \[ \times \] 10^\text{23} particles (atoms, molecules, ions, etc.) of a substance.

In practice, the mole concept helps convert amounts measured in grams to the amount in moles, which scientists use to relate quantities in chemical reactions.

For the compound \(\text{X}_{2}\text{(YZ}_{3}\text{)}_{3}\), we need to find how many moles of each element are present. Given the percent composition, we first convert the mass into moles using the formula:

\(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\)

Then, we use these mole values to define the amounts in relation to each other in the compound. For instance, with

\(\text{X}_{2}\text{(YZ}_{3}\text{)}_{3}\), knowing that 2 moles of X, 3 moles of Y, and 9 moles of Z should add up to match the molar mass (282.23 g/mol), we can adjust to fit a common constant (a). This process allows us to better understand the relationships between elements within the compound.

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Most popular questions from this chapter

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?

How many grams of calcium must be combined with \(1 \mathrm{~g}\) phosphorus to form the compound \(\mathrm{Ca}_{3} \mathrm{P}_{2}\) ?

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) \(\mathrm{Mg}\left(\mathrm{HCO}_{3}\right)_{2}\)

Determine the molar masses of these compounds: (a) \(\mathrm{KBr}\) (f) \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) (b) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (g) \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) (c) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) (h) \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) (i) \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{HPO}_{4}\) (e) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

Researchers at Anna Gudmundsdottir's laboratory at the University of Cincinnati have been studying extremely reactive chemicals known as radicals. One of the interesting phenomena they have discovered is that these radicals can be chemically attached to fragrance molecules, effectively tethering them to a solution. When light strikes these tethered molecules, the fragrance is released. This property would allow us to produce perfumes, cleansers, and other consumer products that release fragrance only when exposed to light. If limonene, \(\mathrm{C}_{10} \mathrm{H}_{16}\), the molecule that gives fruits their citrus scent, were able to be tethered to one of these radicals and every photon of light would release one molecule of limonene, calculate the time in seconds required to release \(1.00\) picogram of limonene if ambient light releases \(2.64 \times 10^{18} \mathrm{photons} / \mathrm{sec}\).

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