How many moles of Ca(NO3)2 contain 2.95 x 10^24 oxygen atoms?
I set up the conversions like this:
2.95 x 10^24 / 6.022 x 10^23 to get # of moles Oxygen. Then, I divided that answer by 8, the ratio of oxygen needed to 'make' one unit of calcium nitrite. I get the wrong answer. What am I doing wrong?
each molecule of Ca(NO3)2 contains 6 atoms of oxygen, not 8
To calculate the number of moles of Ca(NO3)2 that contain 2.95 x 10^24 oxygen atoms, you need to consider the stoichiometry of the compound.
First, let's look at the balanced equation for the formation of Ca(NO3)2:
2 Ca + 2 HNO3 ā Ca(NO3)2 + H2
From this equation, we can see that 2 moles of calcium nitrate (Ca(NO3)2) are formed from 2 moles of nitrate ions (NO3-) and 2 moles of calcium ions (Ca2+). Therefore, the ratio of nitrate ions to calcium nitrate is 2:1.
Now, let's determine the number of moles of nitrate ions:
2.95 x 10^24 oxygen atoms x (1 mol NO3- / 1 mol O ) x (2 mol NO3- / 1 mol Ca(NO3)2)
Using Avogadro's number, 6.022 x 10^23, the calculation becomes:
2.95 x 10^24 / (6.022 x 10^23) x (2 / 1) = 9.7684 moles of nitrate ions (NO3-)
Since the ratio of nitrate ions to calcium nitrate is 2:1, the number of moles of calcium nitrate will be half that:
9.7684 moles / 2 = 4.8842 moles of Ca(NO3)2
So, 4.8842 moles of Ca(NO3)2 contain 2.95 x 10^24 oxygen atoms.
In order to correctly solve this problem, we need to use the concept of molar ratios and the Avogadro's number. Let's break down the steps to find the number of moles of Ca(NO3)2 containing 2.95 x 10^24 oxygen atoms:
1. Start by identifying the molar ratio of oxygen atoms to calcium nitrite (Ca(NO3)2) in the compound. From the formula, we can see that there are two oxygen atoms per Ca(NO3)2 molecule.
2. Use Avogadro's number, 6.022 x 10^23, to convert the number of oxygen atoms to moles of oxygen. Divide 2.95 x 10^24 oxygen atoms by 6.022 x 10^23 (1 mole) to find the moles of oxygen.
Now, here is where the error occurred:
3. Instead of dividing your answer by 8, you need to divide it by 2. This is because we established earlier that there are two oxygen atoms per Ca(NO3)2 molecule. The ratio of oxygen atoms to Ca(NO3)2 molecules is 2:1.
So, the correct calculation would be to divide the moles of oxygen obtained in step 2 by 2 (not 8).
Let's do the calculation correctly:
Moles of oxygen = (2.95 x 10^24) / (6.022 x 10^23) = 4.894 moles (approximately, rounded to three decimal places)
Dividing this answer by 2 (the ratio of oxygen atoms to Ca(NO3)2 molecules), we get:
Moles of Ca(NO3)2 = 4.894 moles / 2 = 2.447 moles (approximately, rounded to three decimal places)
Therefore, 2.95 x 10^24 oxygen atoms are present in approximately 2.447 moles of Ca(NO3)2.