Question regarding concentration units

Click For Summary
SUMMARY

The discussion centers on calculating the final concentration of a substance after adding 800 µL into a liter of a sample, with a density of 1020 kg/m³ and a concentration of 60% w/w. The correct final concentration is determined to be 490 mg/L based on the calculations provided. The participants clarify that maintaining consistent units is crucial for accurate results, and they discuss the implications of using a density of 1.020 kg/m³, which is considered unusually low for a liquid. The conversation emphasizes the importance of understanding density and concentration units in chemical calculations.

PREREQUISITES
  • Understanding of concentration units (mg/L, µg/L)
  • Basic knowledge of density (kg/m³, gm/cc)
  • Familiarity with weight/weight percentage (w/w)
  • Ability to perform unit conversions (mL to L, µL to mL)
NEXT STEPS
  • Research the implications of density variations on concentration calculations
  • Learn about weight/volume and weight/weight concentration calculations
  • Study the significance of unit consistency in scientific measurements
  • Explore common density values for various substances in laboratory settings
USEFUL FOR

Chemists, laboratory technicians, and students involved in experimental design and concentration calculations will benefit from this discussion.

Lassy
Messages
4
Reaction score
0
Hi all,

I've done some experiments and I'm trying to work out the correct unit of my final concentration. The experiment goes thus far:
I added 800 uL of a substance into a litre of my sample (800uL/L).
This substance has a density of 1020 kg/m3 and is 60% w/w.
I'm trying to work out the amount of substance per litre of sample.
After my calculations I have arrived at a number of 490 but I will appreciate some clarification on this as well as if the correct final unit of concentration will be ug/L or mg/L.

Thanks in advance.
 
Physics news on Phys.org
800 ##\mu L## = 0.8 cc

1020 kg/m3 = 1.02 gm/cc

(0.8)(0.6)(1.02) = 0.49 gm = 490 mg
 
Chestermiller said:
800 ##\mu L## = 0.8 cc

1020 kg/m3 = 1.02 gm/cc

(0.8)(0.6)(1.02) = 0.49 gm = 490 mg

Thanks a lot for your reply, truly appreciated.

I have one more question, what if the density was 1.020 kg/m3?

Will the final concentration still be in mg/L or ug/L?
 
I moved the thread to our homework section.

As long as you keep your units consistent, that problem never comes up.

1.020 kg / m3 * 1m3/(1000 L) * 800 µL * 0.6 = ..
 
Lassy said:
Thanks a lot for your reply, truly appreciated.

I have one more question, what if the density was 1.020 kg/m3?

Will the final concentration still be in mg/L or ug/L?
That's a pretty low density
 
Chestermiller said:
That's a pretty low density

Yes, but that's what is quoted on the data sheet for the substance.

This is the calculation I did:

Mass = D x V = 1.02mg / mL x 0.8mL / 1000 mL = 0.000816 mg/mL = 0.816 mg/L = 816 ug/L
816 ug/L x 60% = 490 ug/L
 
Lassy said:
Yes, but that's what is quoted on the data sheet for the substance.

This is the calculation I did:

Mass = D x V = 1.02mg / mL x 0.8mL / 1000 mL = 0.000816 mg/mL = 0.816 mg/L = 816 ug/L
816 ug/L x 60% = 490 ug/L
So, where did the 1020 kg/m3 come from? This is 1.02 gm/cc.
 
1.020 kg/m3 is a density lower than air. There is no liquid with that density.

Could the source be from some foreign country? Compared to English, some other languages have a swapped use of "," and "." for numbers (e.g. 12345.6 can be written as 12.345,6), so 1.020 could mean 1020.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 56 ·
2
Replies
56
Views
4K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 23 ·
Replies
23
Views
5K
Replies
2
Views
3K
Replies
1
Views
2K
Replies
2
Views
2K