Calculating the percentage of silica

  • Thread starter Thread starter xiphoid
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on calculating the percentage of silica in a partially dried clay mineral sample. The original sample contained 12% water and 45% silica, leading to a non-water component of 88%. After drying, the sample retains the same amount of silica and other components, but the water content decreases to 8%. Participants suggest using a hypothetical 100 g sample to simplify calculations, emphasizing the need to establish a relationship between the water content before and after drying to determine the new percentage of silica.

PREREQUISITES
  • Understanding of percentage calculations
  • Basic knowledge of clay mineral composition
  • Familiarity with mass balance concepts
  • Ability to set up and solve equations
NEXT STEPS
  • Learn how to perform mass balance calculations in mineralogy
  • Study percentage composition calculations in chemistry
  • Explore the effects of drying on mineral properties
  • Investigate the role of water in clay mineral behavior
USEFUL FOR

Students in chemistry or geology, educators teaching mineral composition, and anyone involved in materials science or clay mineralogy.

xiphoid
Messages
57
Reaction score
0

Homework Statement


A partially dried clay mineral contains 8% of water, the original sample contained 12% water and 45% silica. Then, the percentage of silica in the partially dried sample is nearly?


Homework Equations


Don't know if this is would help in the calculation,
in the original sample, 12% water is present, hence non water components are consisting of 88%


The Attempt at a Solution


Attempted one but was out of range of the concept to be used.
 
Physics news on Phys.org
hi xiphoid! :smile:

hint: suppose the original sample was 100 g (12 g water, 45 g silica, 43 g other stuff) :wink:
 
I had that stuff in my mind but couldn't manage to calculate further
tiny-tim said:
hi xiphoid! :smile:

hint: suppose the original sample was 100 g (12 g water, 45 g silica, 43 g other stuff) :wink:
 
xiphoid said:
I had that stuff in my mind but couldn't manage to calculate further

ok, after the sample has dried, the water is less but everything else is the same

so call it x g water, 45 g silica, 43 g other stuff, …

from the question, what equation can you find for x ? :wink:
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
7K
Replies
3
Views
2K
Replies
4
Views
2K