Part 1: Each molecule of NO2, nitrogen dioxide, consists of 1 atom of N and 2 atoms of O. If you have 0.155 g. of NO2, how many molecules of NO2 do you have ?
0.155gx46.0 = moles
moles x 6.02 E23 = answer
Part 1 Answer: 2030000000000000000000
Part 2: How many atoms of N do you have ?
3 total parts, 1 part is N
So take first answer x 1
Part 2 Answer: 2030000000000000000000
Part 3: How many atoms of O do you have?
3 total parts, 2 parts are O
So take first answer x 2
Part 3: 4060000000000000000000
Part 4: What is the % by mass of N?
Part 5: What is the % by mass of O?
I need help on Part 4 & 5, but you'll need the previous information to do the problem. Thank you if you can help.
Solved, Nevermind.
See my note at your original post about the number of significant figures.
Also, in this post note the typo on the first problem. moles = 0.155/46.0 and not 0.155 x 46.0
Sure, I can help you with Parts 4 and 5. In order to calculate the percent by mass of an element, you need to know the mass of that element and the total mass of the compound.
For Part 4 (the percent by mass of N):
1. Calculate the molar mass of N:
- N has a molar mass of 14.01 g/mol.
2. Calculate the mass of N in the compound NO2:
- The molecular formula of NO2 consists of 1 atom of N and 2 atoms of O.
- We already know that you have 0.155 g of NO2.
- To find the mass of N, you can multiply the molar mass of N by the number of moles of N, which can be calculated by dividing the mass of NO2 by its molar mass.
mass of N = (0.155 g) x (1 mol NO2 / 46.0 g) x (1 mol N / 1 mol NO2) x (14.01 g / 1 mol N)
3. Calculate the percent by mass of N:
- To find the percent by mass of N, you need to divide the mass of N by the total mass of the compound and multiply by 100%.
percent by mass of N = (mass of N / mass of NO2) x 100%
You can follow a similar process for Part 5 (the percent by mass of O), but instead, you would calculate the mass of O in the compound NO2 and divide it by the total mass of NO2, then multiply by 100%. The molar mass of O is 16.00 g/mol.
I hope this helps! Let me know if you have any further questions.