To calculate the mass percent of NaClO3 in the original sample, we need to determine the mass of NaClO3 and the total mass of the sample.
First, let's calculate the moles of oxygen gas (O2) collected over water using the ideal gas law:
PV = nRT
Where:
P = pressure of oxygen gas (converted to atm)
V = volume of oxygen gas (converted to liters)
n = number of moles of oxygen gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (converted to Kelvin)
Converting the given values:
P = 734 torr - 19.8 torr (vapor pressure of water) = 714.2 torr
V = 57.2 mL = 0.0572 L
T = 22°C + 273.15 = 295.15 K
Converting pressure to atm:
714.2 torr / 760 torr/atm = 0.94 atm
Now we can calculate the number of moles of oxygen gas:
n = PV / RT
n = (0.94 atm) * (0.0572 L) / (0.0821 L.atm/mol.K * 295.15 K)
Calculate n:
n ≈ 0.00231 mol
Since sodium chlorate (NaClO3) decomposes to produce oxygen gas at a 2:3 molar ratio, the number of moles of NaClO3 is:
(2/3) * n ≈ (2/3) * 0.00231 mol
Calculate moles of NaClO3:
0.00154 mol
Now we need to determine the molecular weight of NaClO3. The atomic masses are: Na = 22.99 g/mol, Cl = 35.45 g/mol, and O = 16.00 g/mol.
Molecular weight of NaClO3:
Na = 22.99 g/mol
Cl = 35.45 g/mol
O = 16.00 g/mol
Molecular weight of NaClO3:
22.99 g/mol + (35.45 g/mol) + (3 * 16.00 g/mol)
Molecular weight of NaClO3:
106.44 g/mol
Finally, we can calculate the mass of NaClO3 in the original sample:
mass of NaClO3 = moles of NaClO3 × molecular weight of NaClO3
mass of NaClO3 ≈ 0.00154 mol × 106.44 g/mol
mass of NaClO3 ≈ 0.164 g
Given that the original sample weighed 0.8915 g, the mass percent of NaClO3 is:
mass percent of NaClO3 = (mass of NaClO3 / mass of sample) × 100
mass percent of NaClO3 = (0.164 g / 0.8915 g) × 100
mass percent of NaClO3 ≈ 18.36%
Therefore, the mass percent of NaClO3 in the original sample is approximately 18.36%.