A 0.528 gram sample of a metal, M, reacts completely with sulfuric acid according to: M+H2SO4 -> MSO4+H2
A volume of 205 mL of hydrogen is collected over water; the water level in the collecting vessel is the same as the outside level. Atmospheric pressure is 756.0 Torr and the temperature is 25°C. Calculate the atomic mass of the metal.
? g/mol
PV = nRT and solve for n = number of mols of H2. In the process one should reduce the total P by vapor pressure of H2O at 25C; however, that information is not included in the problem.
Then n mols H2 = same mols M, then
grams = mols/molar mass. You know grams and mols, solve for molar (atomic) mass M.
Is the answer 3.24x10^-3 g/mol?
To calculate the atomic mass of the metal (M), we need to use the ideal gas law equation to find the number of moles of hydrogen gas (H2) produced in the reaction.
First, let's convert the volume of hydrogen gas collected over water to the volume of dry hydrogen gas. We can assume that the vapor pressure of water at 25°C is 23.8 Torr.
Subtracting the vapor pressure of water from the atmospheric pressure gives us the pressure of dry hydrogen gas.
Pressure of dry H2 = Atmospheric pressure - Vapor pressure of water
= 756.0 Torr - 23.8 Torr
= 732.2 Torr
Next, let's convert the pressure from Torr to atm by dividing it by 760 Torr/atm.
Pressure of dry H2 = 732.2 Torr / 760 Torr/atm
= 0.962 atm
Now, let's convert the volume of dry hydrogen gas to moles using the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 L·atm/mol·K), and T is the temperature in Kelvin.
Converting the temperature from Celsius to Kelvin:
Temperature (K) = Temperature (°C) + 273.15
= 25°C + 273.15
= 298.15 K
Now we can calculate the number of moles of hydrogen gas (n) produced:
n = PV / RT
= (0.962 atm) * (205 mL / 1000 L) / (0.0821 L·atm/mol·K * 298.15 K)
= 0.00924 moles
Since 1 mole of metal (M) reacts with 1 mole of hydrogen gas (H2) in the balanced equation, the molar ratio is 1:1.
Now we can calculate the molar mass of the metal (M):
Molar mass of M = Mass of M / Moles of M
= 0.528 g / 0.00924 moles
= 57.07 g/mol
Therefore, the atomic mass of the metal (M) is approximately 57.07 g/mol.
To calculate the atomic mass of the metal, we first need to determine the number of moles of hydrogen gas produced in the reaction. We can use the ideal gas law, which is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvins.
Before we use the ideal gas law, we need to correct the pressure of hydrogen to account for the presence of water vapor. The pressure of the water vapor can be found using a vapor pressure table for water at different temperatures. At 25°C, the vapor pressure of water is 23.76 Torr.
So, the corrected pressure of hydrogen gas is:
Corrected Pressure = Total Pressure - Vapor Pressure of Water
= 756.0 Torr - 23.76 Torr
= 732.24 Torr
Next, we need to convert the volume of hydrogen gas from milliliters to liters:
Volume of Hydrogen Gas = 205 mL * (1 L / 1000 mL)
= 0.205 L
Now, we can rearrange the ideal gas law equation to solve for the number of moles of hydrogen gas:
n = PV / RT
Substituting the values into the equation:
n = (732.24 Torr) * (0.205 L) / (0.0821 L·atm/mol·K * 298 K)
= 6.03 x 10^-3 mol
Since the balanced chemical equation indicates that 1 mole of metal, M, reacts with 1 mole of hydrogen gas to produce 1 mole of MSO4, the number of moles of metal is also 6.03 x 10^-3 mol.
Finally, we can calculate the atomic mass of the metal using the formula:
Atomic Mass = Mass of Metal / Moles of Metal
Mass of Metal = 0.528 g
Atomic Mass = 0.528 g / (6.03 x 10^-3 mol)
≈ 87.6 g/mol
Therefore, the atomic mass of the metal is approximately 87.6 g/mol.