Finding mass of a percent composition with dimen. analysis?
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#1
March 3rd, 2006, 01:56
 jaded Junior Member Join Date: Sep 2005 Posts: 3
Finding mass of a percent composition with dimen. analysis?

For some reason I'm blocking these problems. My instructor did one on the board that was similar, but with no explanation. It's a dimensional analysis solution.

If someone could start setting them up and explain why, it would really help me solve them. Thanks in advance. My own "notes" are below #1, indented.

1) A bracelet that has a volume of 15.40mL is made of an alloy that is 60.0% Gold by mass and has a density of 14.3g/mL. How many ounces (mass ounces) of Gold are in the bracelet?
--For example, it seems as though all you would have to do for this problem is take the volume and the density of the bracelet, solve for the mass and then take 60% of that number and convert grams to ounces. His method was solely using dimensional analysis though...so what am I missing?

2) A type of bronze is 80.0% by weight Copper and 20.0% Tin. The bronze has a density of 8.74 g/cubic cm. A sculptor is preparing to cast a bronze figure that requires 0.350L of bronze. How many mass ounces of Copper must be used to prepare the bronze to use in the figure?
#2
March 29th, 2006, 08:19
 Terry Junior Member Join Date: Mar 2006 Posts: 6

Dimensional analysis is really cool and handy when you get used to it. To me it is baciscally using the property of 1.
You can multiply any thing by one and not change its value....Like duh..
But when you tag numbers with units(dimensions) it opens up a bit for you.
Now stuff like 1 foot = 12 inches can be seen as
1 foot/12 inches = 1.
12 inches / 1 foot = 1.

but you have an extra bookkeeping duty to keep track of the units and the the numbers.

____________________________________

So then
you have .3 feet and you want to know how many inches.

0.3 foot * [ 12 inches / 1 foot ] = ??? inches

or

(0.3*12)/1 for numbers and (foot*inches)/foot for dimensions.

3.6 inches {...note foot/foot cancels out...}

Viola!

3.6 inches.

Simple case here, but it can be dragged out on and on and usually is.
__________________________________

You could have a problem with a set of numbers like

=(1000 * 50.2 *34)/ 24 * 43 * 500
=3.3

and a set of units like

=(Liters of Y * gms of X * moles of Y ) / (Liters of Y * gms of X * L of Y)
units cancel out and you are left with
=moles of Y / Liters of Y
= Molar Y

or

overall
3.3 Molar Y

They can even be longer and I will save your sufferring by not going any further with it.
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But it boils down to bookkeeping your numbers and units....
Konwing what dimensions you have
Knowing what units you want to get to.
Using relationships that equal 1 with the dimensions you want to cancel or get to.

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Do the same with your problem

You have ... 15.4 mL Bracelet.
You want .... ??? oz of Gold.

Relationships that equal 1 (note these are given here and only apply to this problem...Some constants and stufF can be applied to other problem,
but do not always make that assumption)

60.0% Gold by mass....so think of 60% as parts per hundred or

60 gms Gold / 100 gms Bracelet = 1
and
14.3 gm Bracelet / mL Bracelet = 1

Work the prob

15.4 mL Bracelet * (14.3 gm Bracelet / mL Bracelet) * (60 gms Gold / 100 gms Bracelet)
ml Bracelet cancels out
gm Bracelet cancels out

you are left with gms Gold and some numbers to multiply.
but you need to convert gms to ozs, so you need a conversion table for that and you should be home free.

I know this sounds complicated and maybe in the way I walked through it made it that way, but it really is a beatiful way to walk through problems
and you can use it in very many other disciplines as well. Learn it...it will make your life a whole lot easier down the road.

Terry

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