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Problem Set 3 Answer Key .pdf


Original filename: Problem Set 3 Answer Key.pdf
Author: Robert Bauer

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Chemistry Lesson: Steemit Edition Problem Set 3

1.) How many neutrons are present in 54
24๐ถ๐‘Ÿ ? (refer to Chemistry Lesson: Part 3 if you do not
remember how to determine that)

There are 30 neutrons in that Chromium atom. Remember molecular mass is that of protons + neutrons,
and the atomic number is the number of protons. To calculate number of neutrons we subtract the
atomic number from the atomic mass (54 - 24 = 30).

2.) What is the empirical formula for a compound with the following composition: 54.5 % C, 13.7 %
H, and 31.8 % N? (Bonus: the molecular mass of this compound is ~88 g/mol, the first person to
post the correct molecular formula and name of the compound in the comments of Chemistry
Lesson: Part 4 will get 5 SBD as a reward).
1 ๐‘š๐‘œ๐‘™
54.5 % ๐ถ โˆ— 100๐‘” = 54.5 ๐‘” ๐ถ โˆ—
= 4.54 ๐‘€๐‘œ๐‘™ ๐ถ
12.01 ๐‘” ๐ถ
1 ๐‘š๐‘œ๐‘™
13.7 % ๐ป โˆ— 100๐‘” = 13.7 ๐‘” ๐ป โˆ—
= 13.59 ๐‘š๐‘œ๐‘™ ๐ป
1.008 ๐‘” ๐ถ
1 ๐‘š๐‘œ๐‘™
31.8 % ๐‘ โˆ— 100๐‘” = 31.8 ๐‘” ๐‘ โˆ—
= 2.26 ๐‘š๐‘œ๐‘™ ๐‘
14.01 ๐‘” ๐ถ
Nitrogen is the smallest # of moles so we divide each number by 2.26. 4.54/2.26 = 2C.
13.59/2.26 = 6 H, 2.26/2.26 = 1 N. So our empirical formula is C2H6N.

If you want the bonusโ€ฆ youโ€™ve got to figure that one out on your own.
3.) Balance the following chemical reaction: Al + H2SO4 ๏ƒ  Al2(SO4)3 + H2
This is the reaction of aluminum with concentrated sulfuric acid, if you are curious what this
reaction looks like see the following YouTube video (itโ€™s NOT a crazy one):
(https://www.youtube.com/watch?v=vR2ivraWTyU)
2Al + H2SO4 ๏ƒ  Al2(SO4)3 + H2
2Al + 3H2SO4 ๏ƒ  Al2(SO4)3 + H2
3Al + 3H2SO4 ๏ƒ  Al2(SO4)3 + 3H2
Balanced Aluminum because it was the easiest first. Then balanced SO4, there were 3 on the
product side (Compound)number of units so added 3 to the H2SO4 to make that equal. Finally
balanced the hydrogens. We had 6 on the reactants side, so added the coefficient of 3 to the H2
to put 6 on the product side.


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