What is the Empirical Formula for this?

Watch this thread
zattyzatzat
Badges: 9
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#1
Report Thread starter 5 years ago
#1
A compound, T, contains 41.0 % potassium, 33.7 % sulfur and 25.3 % oxygen by mass. What is the empirical formula of T?[Ar: K, 39; S, 32; O, 16]
0
reply
alow
Badges: 19
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#2
Report 5 years ago
#2
What have you tried?
0
reply
zattyzatzat
Badges: 9
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#3
Report Thread starter 5 years ago
#3
(Original post by alow)
What have you tried?
I used the method which i was taught in school. Divide the percentage of mass of each element by it's atomic number. Then divide that the results with the smallest outcome. And then that is the amount of atoms for each element. so:

K:41/39 = 1.0512
S:33.7/32 = 1.0531
O:25.3/ = 1.5812

K:1.0512 / 1.0512 = 1
S:1.0531 / 1.0512 = 1
O:25.3 / 1.0512 = 1.5

Previously when i tried doing empirical formula questions like these. I could do it. But I know this is wrong because of the 1.5 value for oxygen.
0
reply
alow
Badges: 19
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#4
Report 5 years ago
#4
(Original post by zattyzatzat)
I used the method which i was taught in school. Divide the percentage of mass of each element by it's atomic number. Then divide that the results with the smallest outcome. And then that is the amount of atoms for each element. so:

K:41/39 = 1.0512
S:33.7/32 = 1.0531
O:25.3/ = 1.5812

K:1.0512 / 1.0512 = 1
S:1.0531 / 1.0512 = 1
O:25.3 / 1.0512 = 1.5

Previously when i tried doing empirical formula questions like these. I could do it. But I know this is wrong because of the 1.5 value for oxygen.
http://www.chemspider.com/Chemical-Structure.55421.html
0
reply
Plantagenet Crown
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#5
Report 5 years ago
#5
(Original post by zattyzatzat)
I used the method which i was taught in school. Divide the percentage of mass of each element by it's atomic number. Then divide that the results with the smallest outcome. And then that is the amount of atoms for each element. so:

K:41/39 = 1.0512
S:33.7/32 = 1.0531
O:25.3/ = 1.5812

K:1.0512 / 1.0512 = 1
S:1.0531 / 1.0512 = 1
O:25.3 / 1.0512 = 1.5

Previously when i tried doing empirical formula questions like these. I could do it. But I know this is wrong because of the 1.5 value for oxygen.
If you get half numbers then you multiply all of them by the smallest number that will take them all to whole numbers:

K: 1 X 2 = 2
S: 1 X 2 = 2
O: 1.5 X 2 = 3
1
reply
charco
Badges: 18
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#6
Report 5 years ago
#6
(Original post by Plantagenet Crown)
If you get half numbers then you multiply all of them by the smallest number that will take them all to whole numbers:

K: 1 X 2 = 2
S: 1 X 2 = 2
O: 1.5 X 2 = 3
"divide"
0
reply
Plantagenet Crown
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#7
Report 5 years ago
#7
(Original post by charco)
"divide"
Once you have done the calculations and are down to the final numbers you most certainly can multiply to get rid of non integers. After all, multiplying 1,1,1.5 by 2 is exactly the same as dividing 1,1,1.5 by 0.5 and the multiplication is often easier to see.

https://www.chem.tamu.edu/class/majo.../empirical.htm
0
reply
charco
Badges: 18
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#8
Report 5 years ago
#8
(Original post by Plantagenet Crown)
Once you have done the calculations and are down to the final numbers you most certainly can multiply to get rid of non integers. After all, multiplying 1,1,1.5 by 2 is exactly the same as dividing 1,1,1.5 by 0.5 and the multiplication is often easier to see.

https://www.chem.tamu.edu/class/majo.../empirical.htm
carbon 81.82%
hydrogen 18.18%

C: 81.82/12 = 6.82
H: 18.18/1 = 18.18

What are you going to multiply by?
0
reply
charco
Badges: 18
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#9
Report 5 years ago
#9
(Original post by Plantagenet Crown)
Once you have done the calculations and are down to the final numbers you most certainly can multiply to get rid of non integers. After all, multiplying 1,1,1.5 by 2 is exactly the same as dividing 1,1,1.5 by 0.5 and the multiplication is often easier to see.

https://www.chem.tamu.edu/class/majo.../empirical.htm
Phosphorus: 43.66%
Oxygen: 56.34%

P: 43.66/31 = 1.41
O: 56.34/16 = 3.52

What are you going to multiply by?
0
reply
Plantagenet Crown
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#10
Report 5 years ago
#10
(Original post by charco)
Phosphorus: 43.66%
Oxygen: 56.34%

P: 43.66/31 = 1.41
O: 56.34/16 = 3.52

What are you going to multiply by?
(Original post by charco)
carbon 81.82%
hydrogen 18.18%

C: 81.82/12 = 6.82
H: 18.18/1 = 18.18

What are you going to multiply by?
I was referring to the part of the calculation when you get the FINAL numbers, in the example given by the OP 1, 1, 1.5, not immediately after dividing the percentage by the atomic mass.
0
reply
macpatgh-Sheldon
Badges: 20
Rep:
? You'll earn badges for being active around the site. Rep gems come when your posts are rated by other community members.
#11
Report 5 years ago
#11
With this Oxygen and Phosphorus one, you can see straight away that both 1.41 and 3.52 are just about multiples of 0.7.

1.4/0.7 = 2
3.5/0.7 = 5

Also you know that O has 6 electrons in outer shell, so can gain or share two and P has 5, so can lose or share 5 (or 3).

So you can work out that the answer is P2O5
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Year 12s - where are you at with making decisions about university?

I’ve chosen my course and my university (9)
36%
I’ve chosen my course and shortlisted some universities (8)
32%
I’ve chosen my course, but not any universities (2)
8%
I’ve chosen my university, but not my course (1)
4%
I’ve shortlisted some universities, but not my course (1)
4%
I’m starting to consider my university options (3)
12%
I haven’t started thinking about university yet (0)
0%
I’m not planning on going to university (1)
4%

Watched Threads

View All
Latest
My Feed