A mixture of powdered carbon and an oxide of lead was heated producing lead metal and carbon dioxide.when 13.70g of the lead oxide was reduced to lead by carbon,960cm^3 of carbon dioxide measured at r.t.p was formed.
Calculate the empirical formular of the lead oxide???
There may be shorter ways to do this but here is what I would do.
PbxOy + C ==> yCO2 + xPb
Use PV = nRT and solve for mols CO2 at rtp.
Convert mols CO2 to g oxygen.
g Pb = 13.70-g O = g Pb
Then mols Pb = grams pb/atomic mass Pb = ?
mols O atoms = grams O/atomc mass O = ?
Now find the ratio of the elements to each other with the smallest number being no less than 1.00. The easy way to do that is to divide both numbers by the smaller number of mols. That way you get 1.00 for the smallest ratio. Then round to whole numbers but be careful that you don't get heavy handed in rounding; i.e., it's ok to round 1.9 to 2 and 2.1 to 2 but not to round 1.4 to 1 or 1.6 to 2. If you get a number like that, then multiply both small numbers by 2, then 3, then 4, then 5, etc until you get numbers close enough to whole numbers that they can be rounded. That will be the empirical formula. Post your work if you get stuck and I can help you through it.
To determine the empirical formula of the lead oxide, we first need to find the moles of lead oxide and carbon dioxide produced.
Step 1: Calculate the moles of carbon dioxide produced.
Using the Ideal Gas Law, we can convert the volume of carbon dioxide to moles.
PV = nRT, where:
P = pressure (r.t.p = 1 atm)
V = volume (960 cm^3)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (r.t.p = 273 K)
Rearranging the equation to solve for moles:
n = PV / RT
n = (1 atm * 960 cm^3) / (0.0821 L.atm/mol.K * 273 K)
n ≈ 38.11 mol
Step 2: Determine the moles of lead oxide.
We'll assume that the powdered carbon is in excess, so all the lead oxide is used up.
To find the moles of lead oxide, we can use its molar mass.
Let's assume the lead oxide has a molar mass of M grams/mol.
Using the molar mass and mass of lead oxide, we can calculate moles:
moles of lead oxide = mass of lead oxide / molar mass
moles of lead oxide = 13.70 g / M g/mol
Step 3: Equate the moles of lead oxide and carbon dioxide.
Since the reaction is balanced, the moles of lead oxide and carbon dioxide should be equal.
moles of lead oxide = moles of carbon dioxide
13.70 g / M g/mol = 38.11 mol
Step 4: Solve for the molar mass and empirical formula.
Let's solve for M to find the molar mass of the lead oxide:
13.70 g / (38.11 mol) = M g/mol
M ≈ 0.359 g/mol
Now that we have the molar mass, we can calculate the empirical formula by assuming a convenient molar mass of 1 g/mol.
Empirical formula: round the subscripts of the elements to the nearest whole number.
For the lead oxide, the empirical formula would be:
Pb2O4
To determine the empirical formula of the lead oxide, you first need to find the number of moles of each element present in the given data. The balanced chemical equation for the reaction is:
PbO + C → Pb + CO2
From the equation, the stoichiometric ratio between PbO and CO2 is 1:1. This means that for every mole of PbO, one mole of CO2 is produced.
First, let's find the number of moles of CO2 formed:
Using the ideal gas law equation PV = nRT, we can convert the volume of carbon dioxide measured at r.t.p (960 cm^3) to moles.
1. Convert the volume to liters:
Volume = 960 cm^3 = 960/1000 = 0.960 L
2. The molar volume of a gas at r.t.p is approximately 22.4 liters/mol.
Moles of CO2 = Volume (in liters) / Molar volume (22.4 L/mol)
Moles of CO2 = 0.960 / 22.4 = 0.0429 mol
Since the stoichiometric ratio between PbO and CO2 is 1:1, this also represents the number of moles of PbO present.
Next, let's determine the number of moles of Pb in 13.70 g of PbO:
The molar mass of PbO is 207.2 g/mol.
Moles of PbO = Mass of substance / Molar mass
Moles of PbO = 13.70 g / 207.2 g/mol = 0.0661 mol
Now, we need to find the empirical formula by dividing the number of moles of each element by the smallest number of moles.
The number of moles of Pb is 0.0429 mol, and the number of moles of O is also 0.0429 mol.
Dividing by the smallest number of moles (0.0429 mol), we get:
Pb: 0.0429 mol / 0.0429 mol = 1
O: 0.0429 mol / 0.0429 mol = 1
Therefore, the empirical formula of the lead oxide is PbO.