Question

Calculate the volume that 4.5 \(kg\) of ethylene gas \(\left(C_{2} H_{4}\right)\) will occupy at STP.

Short answer

At STP, 4.5 kg of ethylene gas will occupy a volume of approximately 3,593.29 L.

Step-by-step solution

(Step 1: Determine the molar mass of ethylene gas)

. The first step is to determine the molar mass of ethylene gas (\(C_{2} H_{4}\)). The molar mass of carbon (C) is 12.01 g/mol, and the molar mass of hydrogen (H) is 1.01 g/mol. The molecule has 2 carbon atoms and 4 hydrogen atoms. Therefore, the molar mass of ethylene is: Molar mass of \(C_{2} H_{4}\) = (Number of carbon atoms × molar mass of carbon) + (Number of hydrogen atoms × molar mass of hydrogen) = (2 × 12.01 g/mol) + (4 × 1.01 g/mol) = 28.04 g/mol

(Step 2: Determine the number of moles of ethylene gas)

. Next, we need to convert the given mass of ethylene gas (4.5 kg) into the number of moles. To do this, we will use the molar mass of ethylene gas (28.04 g/mol) that we found in step 1. First, convert the mass of ethylene from kg to grams: 4.5 kg = 4,500 g Now, determine the number of moles (n) using the formula: n = Mass of ethylene / Molar mass of ethylene n = 4,500 g / 28.04 g/mol = 160.62 moles

(Step 3: Apply the Ideal Gas Law and STP conditions to calculate volume)

. Using the Ideal Gas Law formula, PV = nRT, and the STP conditions, we will find the volume of ethylene gas. At STP, the pressure is 1 atm and the temperature is 273.15 K. The ideal gas constant, R, is 0.0821 L atm / (mol K). Now, rearrange the Ideal Gas Law to solve for volume: V = nRT / P Plugging in the values, we get: V = (160.62 moles × 0.0821 L atm / (mol K) × 273.15 K) / 1 atm V = 3,593.29 L

(Step 4: State the final answer)

. At STP, 4.5 kg of ethylene gas will occupy a volume of approximately 3,593.29 L.

Key concepts

Understanding Molar Mass

Molar mass is a fundamental concept in chemistry. It represents the mass of one mole of a substance, measured in grams per mole (g/mol). Each element in the periodic table has its own molar mass based on its atomic weight.
For molecules like ethylene (C_{2}H_{4}), the molar mass is found by summing the molar masses of all atoms in the molecule.

• The molar mass of carbon (C) is 12.01 g/mol.
• The molar mass of hydrogen (H) is 1.01 g/mol.

In ethylene, there are two carbon atoms and four hydrogen atoms. Thus, the molar mass calculation is: \[(2 \times 12.01 \text{ g/mol}) + (4 \times 1.01 \text{ g/mol}) = 28.04 \text{ g/mol}\]This means one mole of ethylene weighs 28.04 grams. Understanding molar mass helps us connect macroscopic quantities, like grams, to macroscopic quantities like moles, which describe the amount of substance.

Calculating Moles from Mass

Once you have the molar mass, you can easily convert a known mass of a substance to moles. The mole is a central unit in chemistry that helps to quantify the amount of substance.
Here is the simple formula you need:\[n = \frac{\text{Mass in grams}}{\text{Molar mass (g/mol)}}\]In our example, we converted 4.5 kg of ethylene gas into grams (4,500 g) and used the molar mass of 28.04 g/mol.

• Substitute the values into the formula:

\[n = \frac{4,500 \text{ g}}{28.04 \text{ g/mol}} = 160.62 \text{ moles}\]Calculating moles is a key step before applying the Ideal Gas Law, especially when transitioning from mass to volume of a gas.

Exploring STP Conditions

Standard Temperature and Pressure (STP) provides a set of predefined conditions to enable consistent and comparable results. Under STP, the pressure is set to 1 atmosphere (atm) and the temperature to 273.15 Kelvin (K).
These conditions are pivotal in problems involving gases because they provide a baseline for calculations.
The Ideal Gas Law states that the volume of a gas (V) is determined by the number of moles (n), the ideal gas constant (R = 0.0821 L atm / (mol K)), and the temperature (T) under a certain pressure (P):\[PV = nRT\]This can be rearranged to find volume:\[V = \frac{nRT}{P}\]For ethylene gas under STP conditions and using the moles (160.62 mol) calculated earlier, we substitute:\[V = \frac{(160.62 \text{ moles}) \times 0.0821 \text{ L atm / (mol K)} \times 273.15 \text{ K}}{1 \text{ atm}}\]Resulting in a total volume of about 3,593.29 L. This understanding is crucial for predicting and applying the behavior of gases systematically in chemical reactions and processes.