What is the mass of 0.455 L of butane gas, C4H10, at STP?
Well, the mass of butane gas can be calculated using the formula:
mass = density × volume
At STP (Standard Temperature and Pressure), the volume of 1 mole of any gas is 22.4 L. So, we can calculate the number of moles of butane gas using the given volume:
moles = volume / 22.4 L
moles = 0.455 L / 22.4 L
moles ≈ 0.02033
Now, to calculate the mass, we need the molar mass of butane, which is:
C4H10:
(4 × molar mass of carbon) + (10 × molar mass of hydrogen)
So, let's calculate it:
(4 × 12.01 g/mol) + (10 × 1.01 g/mol) = 58.12 g/mol
Finally, we can calculate the mass:
mass = moles × molar mass
mass = 0.02033 mol × 58.12 g/mol
mass ≈ 1.18 g
So, the mass of 0.455 L of butane gas at STP is approximately 1.18 grams. Hope this helps, and remember, never trust an atom... they make up everything!
To find the mass of butane gas, C4H10, at STP (Standard Temperature and Pressure), we will follow these steps:
Step 1: Determine the molar mass of butane.
The molecular formula of butane, C4H10, indicates that it contains four carbon atoms (C) and ten hydrogen atoms (H).
The atomic mass of carbon (C) is approximately 12.01 g/mol, and the atomic mass of hydrogen (H) is approximately 1.01 g/mol.
To calculate the molar mass of butane (C4H10), we multiply the atomic mass of each element by the number of atoms and sum them up:
Molar mass of C4H10 = (4 * atomic mass of C) + (10 * atomic mass of H)
Step 2: Convert the given volume of gas to the number of moles using the ideal gas law.
Under STP conditions, one mole of gas occupies a volume of 22.4 L.
To convert the given volume (0.455 L) to moles, we divide the volume by the molar volume:
Number of moles = Volume of gas / Molar volume
Step 3: Calculate the mass of butane using the number of moles and molar mass.
To find the mass of butane, we can multiply the number of moles by the molar mass of butane:
Mass of butane = Number of moles * Molar mass
Let's calculate the mass of 0.455 L of butane gas at STP:
Step 1: Molar mass of C4H10 = (4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol
Step 2: Number of moles = 0.455 L / 22.4 L/mol = 0.0203 mol
Step 3: Mass of butane = 0.0203 mol * 58.12 g/mol ≈ 1.18 g
Therefore, the mass of 0.455 L of butane gas at STP is approximately 1.18 grams.
To determine the mass of butane gas (C4H10) at STP (standard temperature and pressure), we need to use the ideal gas law.
The ideal gas law equation is given by:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in Kelvin)
At STP, the pressure is 1 atm, and the temperature is 273 K. We are given the volume as 0.455 L.
First, we need to calculate the number of moles (n) of butane gas.
Rearranging the ideal gas law equation, we have:
n = PV / RT
Substituting the given values, we get:
n = (1 atm * 0.455 L) / (0.0821 L·atm/mol·K * 273 K)
Simplifying, we find:
n = 0.019 moles
Next, we find the molar mass of butane gas (C4H10).
The molar mass is calculated by adding up the atomic masses of each element in the chemical formula.
C (carbon) has an atomic mass of 12.01 g/mol, and H (hydrogen) has an atomic mass of 1.01 g/mol.
The molar mass of butane (C4H10) is:
(4 * 12.01 g/mol) + (10 * 1.01 g/mol) = 58.12 g/mol
Finally, we can calculate the mass of butane gas using the equation:
Mass = number of moles * molar mass
Substituting the values, we have:
Mass = 0.019 moles * 58.12 g/mol
Calculating the result, we find:
Mass = 1.10 g
Therefore, the mass of 0.455 L of butane gas at STP is 1.10 grams.
1 mol butane at STP has a volume of 22.4 L. Calculate the number of mols from that. Then 1 mol butane has a mass of (4*12) +))10*1); therefore, you number of mols will have a mass of ?
Post your work if you get stuck.