help with chem.subtract? samples

Chemistry
Tutor: None Selected Time limit: 1 Day

PHIL weighed a 150-mL beaker (67.34 g) and a piece of filter paper (0.34 g).  He added his unknown sample to the beaker and reweighed it (97.12 g).

PHIL separated the sand from the unknown sample using a funnel.  He washed it and then dried the sample on the filter paper.  The resulting mass of the sand on the filter paper was 5.81 g.

What was the mass of sand that PHIL recovered from his sample? Would i subtract 5.81 from sample?


Oct 6th, 2015

Thank you for the opportunity to help you with your question!

beaker is 67.34g. Beaker+ unknown sample is 97.12g. 

filter paper is 0.34g

sand+filter paper is 5.81g

the mass of the sand alone that was recovered from the unknown sample is 5.81-0.34=5.47g

the mass of the unknown sample is 97.12-67.34=29.78g 

so if you want the percent of sand in the unknown sample as well, you would calculate: 5.47/29.78 x 100 = 18.37% of the unknown sample was sand 

hope that helps

I am very good in general chemistry, whatever problems you have I can upload photos of my work for you and I have very neat handwriting. I will bid on more of your questions in the future. 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Oct 6th, 2015

so i  separate the crystals from the remaining solution.  He cleaned and dried the crystals and put them in the 150-mL beaker. When he weighed it, the mass was 77.02 g.


KNO 3 i have no is .32 mg what would be the % KNO 3 recovered? I just wanted to see if i am doing it right

Oct 7th, 2015

not .32, 9/68 g

Oct 7th, 2015

9.68 g sry

Oct 7th, 2015

Ok, I will reply in about five minutes, let me review your question 

Oct 7th, 2015

So you recovered 9.68g of crystals from your solution. Your calculation was correct. However, I do not know the composition of the crystals or the amount of KNO3 you started out with, so I cannot tell you the percent composition. What I can tell you is that the percent KNO3 recovered is (mass of KNO3 recovered)/(mass of original mixture/solute) * 100. For example, if you recovered 9.68g of KNO3 from an unknown solute that was originally 29.78g, the percent KNO3 recovered would be (9.68/29.78) *100=3.51%

Oct 7th, 2015

i got 32.5%

Oct 8th, 2015

Alright, if you upload your work I can check your solution if you like. 

Oct 8th, 2015

I am just saying (9.68/29.78) x 100 % = 32.5% not 3.51%, correct?

Oct 8th, 2015

Yes, that is correct, sorry for the calculation error. 

Oct 8th, 2015

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