Gold Atom Count in O-Ring: 10.4x10^21

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SUMMARY

The calculation of gold atoms in a golf O-ring, formed from a 100-mm length of 1.5-mm-diameter wire, reveals a significant error in unit conversion. The volume was initially calculated as 176.7 mm² instead of the correct volume in mm³. The density of gold is 19.28 g/cm³, and the atomic mass unit (amu) of gold is 196.97 grams. After correcting the volume to 0.1767 cm³, the accurate calculation yields approximately 10.4x10^21 atoms, confirming the expected result.

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shreddinglicks
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Homework Statement


A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring.

Homework Equations


Volume of cylinder = PI*h*r^2

The Attempt at a Solution


I got the volume of the cylinder:
PI*100*.75^2 = 176.7

I used the formula m = D*V:

density gold = 19.28 g/cm^3
amu gold = 196.97 grams

19.28*176.7 = 3406.776

I then multiplied by avogadro's number over amu.

3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms

The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.
 
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shreddinglicks said:

Homework Statement


A golf O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed from a 100-mm length of 1.5-mm-diameter wire. Calculate the number of gold atoms in the O-ring.

Homework Equations


Volume of cylinder = PI*h*r^2

The Attempt at a Solution


I got the volume of the cylinder:
PI*100*.75^2 = 176.7

I used the formula m = D*V:

density gold = 19.28 g/cm^3
amu gold = 196.97 grams

19.28*176.7 = 3406.776

I then multiplied by avogadro's number over amu.

3406.776 *(6.02x10^23 / 196.97) = 1.04x10^25 atoms

The answer should be 10.4x10^21 atoms. I'm thinking I messed up some unit conversion. I am unsure of where.

What are the units on your volume? What are the units on your density?
 
The density is g/cm^3

the volume is mm^2

I see the issue, how do I convert this? Is there a way to convert so I can cancel out a unit?
 
shreddinglicks said:
The density is g/cm^3

the volume is mm^2

I see the issue, how do I convert this? Is there a way to convert so I can cancel out a unit?

Volume should be cubic something. mm^2 is an area.

To convert a volume in mm^3 to cm^3 think about the following:

Conversion factor * X mm^3 = Y cm^3

The conversion factor needs to have units of cm^3 on the top and mm^3 on the bottom. What about [( 1 cm/ 10 mm)]^3 ? I.e. can you use the conversion between cm and mm to come up with a conversion factor between mm^3 and cm^3 ?
 
I see, so I have [176.7 mm^3] * [1cm/10mm]^3

basically

176.7 / 10^3 = .1767
 
Numbers are OK, but in the final formula you again ignored units. That's the simplest way of making mistakes, as you have already seen. Don't do that.
 
.1767cm^3
 
Now use this number to answer the problem.
 

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