A compound of the element A and oxygen has a mole ratio of A:O=2:3. If 8.0 grams of the oxide contains 2.4 grams of the oxygen,
a. What is the atomic weight of A?
b. What is the weight of one mole of the oxide?
c. What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen? What is the % yield, if 38 grams of the oxide was produced?
A2O3 is the compound. If 8.0 g of A2O3 contains 2.4 g oxygen, then mass of A must be 8.0-2.4 = 5.6. If we were trying to determine the formula we would convert grams to moles and find the ratio of the moles to each other. In this case, we already have the mole ratio; we want to determine the molar mass.
moles A = 5.6g/molar mass of x
moles O = 2.4/15 = 0.15
If we want the ratio to be 2:3 and we know O is 0.15 mole, what is the moles A? That must be 0.667*0.15 = 0.1.
Then moles = 5.6/x
0.1 = 5.6/x
x = 56 for the atomic mass.
mass of 1 mole is the molar mass. 2*atomic mass Z + 3*atomic mass O = molar mass.
For c part.
Write and balance the equation between A and O (but you can use Ba for A if you wish).
Convert 28.0 g Ba to moles. moles = grams/molar mass.
Using the coefficients in the balanced equation, convert moles Ba to moles of the oxide.
The g oxide = moles x molar mass. This is the theoretical yield.
Finally, percent yield = (actual yield/theoretical yield)*100 = ?
thanks for the help :)
To solve this problem, we need to use the concept of moles and the mole ratio between element A and oxygen. Let's go through each part of the question step by step:
a. What is the atomic weight of A?
To find the atomic weight of A, we need to calculate the molar mass of A using the given information. The mole ratio of A to oxygen is 2:3. This means that for every 2 moles of A, we have 3 moles of oxygen.
Since we know the mass of the oxygen (2.4 grams) and the molar mass of oxygen (16 g/mol), we can calculate the number of moles of oxygen:
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 2.4 grams / 16 g/mol
Now, using the mole ratio, we can calculate the number of moles of A:
moles of A = (2/3) * moles of oxygen
Since we know the mass of A (8.0 grams), we can calculate the molar mass of A:
molar mass of A = mass of A / moles of A
b. What is the weight of one mole of the oxide?
To find the weight of one mole of the oxide, we need to consider the mole ratio between A and oxygen given in the question (2:3) and the molar mass of A and oxygen.
First, we calculate the molar mass of the oxide using the given mass of oxygen (2.4 grams) and the molar mass of oxygen (16 g/mol):
moles of oxygen = mass of oxygen / molar mass of oxygen
moles of oxygen = 2.4 grams / 16 g/mol
Then, we determine the moles of A using the mole ratio of A to oxygen (2:3):
moles of A = (2/3) * moles of oxygen
Finally, we calculate the molar mass of A using the given mass of the oxide (8.0 grams) and the moles of A:
molar mass of A = mass of A / moles of A
c. What theoretical weight of the oxide is formed when 28 grams of A is heated in excess oxygen? What is the % yield, if 38 grams of the oxide was produced?
To find the theoretical weight of the oxide when 28 grams of A is heated in excess oxygen, we need to calculate the moles of A and the moles of oxygen using the mole ratio.
First, we calculate the moles of A using the molar mass of A (calculated in part a):
moles of A = mass of A / molar mass of A
Using the mole ratio, we find the moles of oxygen:
moles of oxygen = (3/2) * moles of A
Now, we calculate the molar mass of oxygen using the given mass of the oxide (8.0 grams):
moles of oxygen in the oxide = mass of oxygen in the oxide / molar mass of oxygen
moles of oxygen in the oxide = 2.4 grams / 16 g/mol
To determine the theoretical weight of the oxide, we use the molar mass of the oxide:
weight of the oxide = moles of oxygen in the oxide * molar mass of the oxide
Finally, to find the percent yield, we divide the actual weight of the oxide produced (38 grams) by the theoretical weight of the oxide (calculated above) and multiply by 100:
% yield = (actual weight of the oxide / theoretical weight of the oxide) * 100