Empirical formular
22.3g of oxide of lead was reduced with a gas x to produce 20.7g of lead. Find the empirical formular of the oxide. If the R.M.M of the oxide is 223gmole-1. What is the molar mass of the oxide
Reply 1
20
Original post
by From salman22.3g of oxide of lead was reduced with a gas x to produce 20.7g of lead. Find the empirical formular of the oxide. If the R.M.M of the oxide is 223gmole-1. What is the molar mass of the oxide
What have you tried so far?
When working out empirical formulae, you need to work out the amount (in mol) of the various components of the compound.
amount = mass / relative mass.
All you therefore need to know is the masses and relative masses of the lead and the oxygen and then simplify that ratio to their lowest whole numbers.
Reply 2
Original post
by From salman22.3g of oxide of lead was reduced with a gas x to produce 20.7g of lead. Find the empirical formular of the oxide. If the R.M.M of the oxide is 223gmole-1. What is the molar mass of the oxide
Okay, so to figure out the mass of gas X, we take the starting mass of the lead oxide and subtract the mass of the lead we got at the end:
Gas X = 22.3 g - 20.7 g = 1.6 g
To get the formula for the lead oxide, we first figure out how much oxygen we have:
Oxygen Mass = 22.3 g - 20.7 g = 1.6 g
Then, we find out how many moles of lead and oxygen there are:
Lead moles = 20.7 g / 207.2 g/mol = 0.1 mol
Oxygen moles = 1.6 g / 16 g/mol = 0.1 mol
The ratio of lead to oxygen moles is 0.1:0.1 (which simplifies to 1:1), so the lead oxide formula is PbO.
Now, let's check PbO's molar mass:
PbO Molar mass = (1 x 207.2 g/mol) + (1 x 16 g/mol) = 223.2 g/mol
That's pretty close to the molar mass they gave us (223 g/mol), so PbO is surely right.
The lead oxide formula is PbO.
Bye,
Sandro
Reply 3
Original post
by NitrotolueneOkay, so to figure out the mass of gas X, we take the starting mass of the lead oxide and subtract the mass of the lead we got at the end:
Gas X = 22.3 g - 20.7 g = 1.6 g
To get the formula for the lead oxide, we first figure out how much oxygen we have:
Oxygen Mass = 22.3 g - 20.7 g = 1.6 g
Then, we find out how many moles of lead and oxygen there are:
Lead moles = 20.7 g / 207.2 g/mol = 0.1 mol
Oxygen moles = 1.6 g / 16 g/mol = 0.1 mol
The ratio of lead to oxygen moles is 0.1:0.1 (which simplifies to 1:1), so the lead oxide formula is PbO.
Now, let's check PbO's molar mass:
PbO Molar mass = (1 x 207.2 g/mol) + (1 x 16 g/mol) = 223.2 g/mol
That's pretty close to the molar mass they gave us (223 g/mol), so PbO is surely right.
The lead oxide formula is PbO.
Bye,
Sandro
Gas X = 22.3 g - 20.7 g = 1.6 g
To get the formula for the lead oxide, we first figure out how much oxygen we have:
Oxygen Mass = 22.3 g - 20.7 g = 1.6 g
Then, we find out how many moles of lead and oxygen there are:
Lead moles = 20.7 g / 207.2 g/mol = 0.1 mol
Oxygen moles = 1.6 g / 16 g/mol = 0.1 mol
The ratio of lead to oxygen moles is 0.1:0.1 (which simplifies to 1:1), so the lead oxide formula is PbO.
Now, let's check PbO's molar mass:
PbO Molar mass = (1 x 207.2 g/mol) + (1 x 16 g/mol) = 223.2 g/mol
That's pretty close to the molar mass they gave us (223 g/mol), so PbO is surely right.
The lead oxide formula is PbO.
Bye,
Sandro
Whilst I agree PbO is the correct answer, attempting to work out the mass of gas X is unnecessary and your approach to doing so is not correct. If you were making some lead compound where gas X was incorporated into the structure, it would have been appropriate.
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