Confused about Chemical Q Decay Calculation?

[helpm3pl3ase]

I have a question Iam stuck on. I was wondering if i did the right calculations. Please let me know:

Through decay, Chemical Q loses one-third of its mass every 7 years.
Twenty-five years ago, a container of Chemical Q was buried. The EPA has
just dug it up and found 1,000 grams remaining. How much Chemical Q was
originally buried?

1000=(Q)(1/3)^(x/7) when x = 25

1000=(Q)(1/3)^(25/7)

Q = 1000/(3)^(3/7)/81

Q = 50583?

I am not sure if I did this correct??

 

[berkeman]

If it loses 1/3 of its mass every 7 years, then 2/3 of its mass is left at the end of 7 years, and (2/3)^2 at the end of 14 years, etc. I think you used 1/3 where you should have used 2/3...?

 

[helpm3pl3ase]

hmm I don't really understand that?? The question called for one third. Iam not understanding wher eyou get 2/3

 

[daniel_i_l]

If it loses 1/3 then you have 2/3 left. so every 7 years the new amount would equal 2/3 the old amount. that's why the equation is
(Q)(2/3)^(x/7) - you're doing (Q)(2/3 * 2/3 * 2/3 * ... x/7 times)
in other words, to take away 1/3 each time you multiply by 2/3

 

[helpm3pl3ase]

O alright.. I get it now. Thank you

So everything thing else is correct, but instead of 1/3 use 2/3?

 

[HallsofIvy]

Yes, use 2/3 instead of 1/3.