A 1.58-g sample of \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{X}_{3}(g)\) has a volume of \(297 \mathrm{~mL}\) at \(769 \mathrm{~mm} \mathrm{Hg}\) and \(35^{\circ} \mathrm{C}\). Identify the element \(\mathrm{X}\).

Short Answer

Expert verified
Answer: The element X in the compound C2H3X3 is sulfur (S).

Step by step solution

01

Convert the given data into suitable units

We are given the temperature in Celsius, volume in milliliters, and pressure in mmHg. We need to convert these values into Kelvin, liters, and atmospheres respectively for working with the ideal gas equation. Temperature, T = \(35^{\circ}\mathrm{C} = 35 + 273.15 = 308.15\mathrm{K}\) Volume, V = \(297\mathrm{mL} = 297/1000 = 0.297\mathrm{L}\) Pressure, P = \(769\mathrm{mmHg} = 769/760 = 1.0118\mathrm{atm}\)
02

Calculate the moles of gas using the Ideal Gas Law

The Ideal Gas Law equation is given by: PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the gas constant and T is the temperature. Here, R = 0.0821 \(\frac{\mathrm{L.atm}}{\mathrm{mol.K}}\) We will solve the Ideal Gas Law for n: n = \(\frac{PV}{RT}\) Now, substitute the known values and calculate the moles of gas: n = \(\frac{1.0118\mathrm{atm} * 0.297\mathrm{L}}{0.0821\frac{\mathrm{L.atm}}{\mathrm{mol.K}} * 308.15\mathrm{K}} = 0.01205\mathrm{mol}\)
03

Calculate the Molar Mass of the compound

To find the molar mass, we can use the formula: Molar mass = \(\frac{\mathrm{Mass}}{\mathrm{Moles}}\) We are given the mass of the sample as 1.58 grams. Molar mass = \(\frac{1.58\mathrm{g}}{0.01205\mathrm{mol}} = 131.12\mathrm{g/mol}\)
04

Identify the Element X

The molecular formula of the compound is \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{X}_{3}\). We know that the molar mass of carbon (C) is 12.01 g/mol and the molar mass of hydrogen (H) is 1.01 g/mol. Calculating the molar mass of C2H3: (2 * 12.01) + (3 * 1.01) = 29.05 g/mol The molar mass of three atoms of element X can be calculated as: Molar mass of X3 = Molar mass of the compound - Molar mass of C2H3 = 131.12 - 29.05 = 102.07 g/mol Now we can find the molar mass of a single atom of element X: Molar mass of X = \(\frac{102.07\mathrm{g/mol}}{3} = 34.02\mathrm{g/mol}\) Looking at the periodic table, we find that the element with a molar mass close to 34.02 g/mol is sulfur (S), which has a molar mass of 32.07 g/mol. There could be a small discrepancy in our calculation or the given data, but the closest element is sulfur. So, the element X is sulfur, and the compound is \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{S}_{3}\).

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Key Concepts

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

Molar Mass Calculation
Understanding how to calculate molar mass is essential in the field of chemistry, especially when working with gas laws. Molar mass represents the mass of one mole of a substance, usually expressed in grams per mole (g/mol). Calculating it involves summing the atomic masses of all atoms in a molecule.

For example, for a compound with the molecular formula \( \mathrm{C}_2 \mathrm{H}_3 \mathrm{X}_3 \), you first need to calculate the molar mass of the known atoms—carbon (\(\mathrm{C}\)) and hydrogen (\(\mathrm{H}\)). Each carbon atom has an approximate atomic mass of 12.01 g/mol and each hydrogen atom has an approximate atomic mass of 1.01 g/mol. With two carbons and three hydrogens, the calculated molar mass for the \(\mathrm{C}_2 \mathrm{H}_3\) part is \((2 \times 12.01) + (3 \times 1.01) = 29.05 \mathrm{g/mol}\).

Once the molar mass of the other constituent element (X) is needed, the overall molar mass of the compound calculated from the given mass and the number of moles (from the Ideal Gas Law) is used. Subtract the known molar mass of carbon and hydrogen from the total molar mass to find the combined molar mass of the X atoms. Divide this by the number of X atoms present to find the molar mass of a single X atom. This final step allows the identification of the unknown element by comparing the calculated molar mass to known atomic masses.
Gas Constant
In the context of the Ideal Gas Law \( PV = nRT \), the gas constant (\(\mathrm{R}\)) is a fundamental value that relates the pressure, volume, temperature, and moles of a gas. Its value is determined experimentally and changes depending on the units used for pressure, volume, and temperature. For instance, when pressure is measured in atmospheres, volume in liters, and temperature in Kelvin, \(\mathrm{R}\) is approximated as 0.0821 \( \mathrm{L} \cdot \mathrm{atm} / (\mathrm{mol} \cdot \mathrm{K}) \).

It's crucial to use the appropriate gas constant value that corresponds to the units. Misaligning units may lead to incorrect results. Additionally, it’s significant to note that the Ideal Gas Law assumes a 'perfect' gas, where gas particles do not interact and occupy no volume. While real gases do not perfectly align with this model, the Ideal Gas Law offers a close approximation for many conditions, especially at low pressures and high temperatures, where gas molecules are less interactive.
Stoichiometry
Stoichiometry is the quantitative relationship between the elements and compounds as they undergo chemical reactions. It's grounded in the law of conservation of mass, which states that matter cannot be created or destroyed. In stoichiometry, balanced chemical equations guide us in determining the amounts of reactants and products.

When connected with gas laws, stoichiometry can determine the amounts of gases involved in reactions, including those formed and consumed. It relies on the molar ratios established in balanced equations and the molar volumes of gases under given conditions. For example, using the Ideal Gas Law, if you know the volume, pressure, and temperature of a gas, you can calculate the number of moles and, through stoichiometry, relate it to masses or volumes of other substances in a reaction. This approach is vital in the lab and industry, enabling the accurate prediction and measurement of substances needed for chemical processes.

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Most popular questions from this chapter

A tube \(5.0 \mathrm{ft}\) long is evacuated. Samples of \(\mathrm{NH}_{3}\) and \(\mathrm{HCl}\), at the same temperature and pressure, are introduced simultaneously through tiny openings at opposite ends of the tube. When the two gases meet, a white ring of \(\mathrm{NH}_{4} \mathrm{Cl}(s)\) forms. How far from the end at which ammonia was introduced will the ring form?

I A sample of air is originally at \(32^{\circ} \mathrm{C}\). If \(P\) and \(n\) are kept constant, to what temperature must the air be cooled to (a) decrease its volume by \(25 \%\) ? (b) decrease its volume to \(25 \%\) of its original volume?

The air is said to be "thinner" in higher altitudes than at sea level. Compare the density of air at sea level where the barometric pressure is \(755 \mathrm{~mm} \mathrm{Hg}\) and the temperature is \(0^{\circ} \mathrm{C}\) with the density of air on top of \(\mathrm{Mt}\). Everest at the same temperature. The barometric pressure at that altitude is \(210 \mathrm{~mm} \mathrm{Hg}\) (3 significant figures). Take the molar mass of air to be \(29.0 \mathrm{~g} / \mathrm{mol}\).

A Porsche \(928 \mathrm{~S} 4\) engine has a cylinder volume of \(618 \mathrm{~cm}^{3}\). The cylinder is full of air at \(75^{\circ} \mathrm{C}\) and \(1.00\) atm. (a) How many moles of oxygen are in the cylinder? (Mole percent of oxygen in air \(=21.0\).) (b) Assume that the hydrocarbons in gasoline have an average molar mass of \(1.0 \times 10^{2} \mathrm{~g} / \mathrm{mol}\) and react with oxygen in a \(1: 12 \mathrm{~mole}\) ratio. How many grams of gasoline should be injected into the cylinder to react with the oxygen?

The normal boiling points of \(\mathrm{CO}\) and \(\mathrm{SO}_{2}\) are \(-192^{\circ} \mathrm{C}\) and \(-10^{\circ} \mathrm{C}\), respectively. (a) At \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), which gas would you expect to have a molar volume closest to the ideal value? (b) If you wanted to reduce the deviation from ideal gas behavior, in what direction would you change the temperature? The pressure?

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