Chemistry problem (unit conversion) help?

Question

I have some trouble converting squared or cube units and stuff.

Here's the problem:

The amount of mercury in a lake is 0.4 micrograms Hg/mL. What is the total mass in kg of mercury in the lake? (lake has surface area of 100 mi^2, and an average depth of 20 ft.)

Answer 1

First, you need to find the volume of the lake in mL. This is done by using the definition of volume. Volume = Area x Height. To keep units simple, convert miles to feet. Since one mile is equal to 5280 feet, 100 miles squared multiplied by ((5280x5280)/1) = 27878840000 ft^2. Multiply this by 20 ft to get your volume, which = 5.57568 x 10^10 ft^3.

Since 1 cubic foot = 28 316.8466 milliliter, you find that your volume of the lake x 28 316.8466 milliliter is equal to your volume in mL. This equals 1.58 x 10^15 mL.

Take that volume and multiply it by how much mercury in the lake.

0.4 ug Hg/mL (1.58 x 10^15 mL) = 6.32 x 10^14 ug of Hg (Since mL canceled out).

Now convert Micrograms to Kilograms. Since 10^9 ug is in one kg...

6.32 x 10^14 ug (1kg/10^9 ug) = 6.32 x 10^5 kg.

Hope this helps you understand.

UPDATE (For your concern):

Cubes and Squares are dealt as if you are doing the same during the conversion. For example...

1 m = 100 cm.

1 m^2 = 100 x 100 cm.

1 m^3 = 100 x 100 x 100 cm. Etc.

So for the above problems, since 1 mile has 5280 ft in it, that means 1 mile^2 has 5280x5280 ft^2. Then you go onto your conversion of multiplying the 100 to that number to get your ft^2.

Answer 2

There is process called unit factor method or factor label method for unit conversions that is taught in basically every chemistry program and every engineering program today.  I highly recommend you learn this method.  

I'll solve your problem first, then provide information about how to handle other unit conversion problems at the end....

here we go...

***** volume of lake... ****

volume = cross sectional area x depth = 100 mi² x 20 ft x (5280 ft / mi)²  = 5.58 x 10^10 ft³

**** conversion from ft³ water to kg of Hg ****

(5.58 x 10^10 ft³) x (12 in / ft)³ x (2.54 cm / in)³ x (1 ml / 1 cm³) x (0.40 µg Hg / ml) x (1 g Hg / 10^6 µg Hg) x (1 kg / 10³ g) = 6.32 x 10^5 kg

**********************

"unit factor method"

**********************

rule 1) units on top and bottom of a fraction cancel.

examples:

in / in = 1

ft / ft = 1

cm / cm cancels.

sec / sec cancel

sec² / sec² cancel

m³ / m³ cancels

got that down?

****

rule 2) any number x 1 = that number

examples:

4 x 1 = 4

4 in x 1 = 4 in

5280 ft x 1 = 5280 ft

9.8 m / s² x 1 = 9.8 m / s²

easy right?

****

rule 3) any equality can be rearranged to = 1

examples:

2.54 cm = 1 in

if we divide both sides by 2.54 we get

2.54 cm / 2.54 cm = 1 in / 2.54 cm

1 = 1 in / 2.54 cm

12 in = 1 ft

12 in / 12 in = 1 ft / 12 in

1 = 1 ft / 12 in

also... since 1/1 = 1....

1 / 1 = 1 / (1 ft / 12 in)

1 = 12 in / 1 ft

meaning for any equality of the form a = b, .....a / b = b / a = 1

a/b and b/a are called unit factors because they = unity and we will be using them in rules 4 and 5 to factor (or more precisely "change" units)

ok? this may be a bit more complicated than the first two rules but if you play around with it for a while I'm sure you will get it.

****

rule 4) units can be changed by multiplying by an appropriate "unit factor".

this essentially combines rules 1, 2, and 3

example....

10 in = ? cm

10 in x 1 = 10 in right?

and 2.54 cm / 1 in = 1 right?

substituting

10 in x (2.54 cm / 1 in ) = 10 x 2.54 x in x cm / in = 25.4 cm

because inches cancel....

3 ft = ? in

3 ft x 1 = 3 ft x (12 in / ft) = 36 in

here is an important trick... if the units you want to cancel are on the top, put the matching units for the "unit factor" on the bottom.. and vice versa. if on the bottom, then put the matching ones one the top....

this will definitely take practice...

****

rule 5) exponents....

since 1 to any power = 1... any "unit factor" raised to any power = 1

examples:

1² = 1

(2.54 cm / in) = 1

(2.54 cm / in)² = 1² = 1

similiarly

(2.54 cm / in)³ = 1³ = 1

how you use this is like this...

10 in³ = ? cm³

10 in³ x 1 = 10 in³ x (2.54 cm / in)³

= 10 in³ x (2.54 cm)³ / (1 in)³

= 10 in³ x 16.4 cm³ / 1 in³

= 164 cm³

and that's all there is to it...

**************************************...

if you've made it this far..... here's your problem explained

**************************************...

volume = cross sectional area x depth = 100 mi² x 20 ft x (5280 ft / mi)²  = 5.58 x 10^10 ft³

since 5280 ft  = 1 mi

5280 ft / 1 mi = 1

(5280 ft / 1 mi) ² = 1

by multiplying 100 mi² x 20 ft x (5280 ft / 1 mi)² , I changed the units from mi² x ft to ft² x ft = ft³...  do you see that?

next....

(5.58 x 10^10 ft³) x (12 in / ft)³ x (2.54 cm / in)³ x (1 ml / 1 cm³) x (0.40 µg Hg / ml) x (1 g Hg / 10^6 µg Hg) x (1 kg / 10³ g) = 6.32 x 10^5 kg

since

12 in = 1 ft → (12in / 1 ft)³ = 1

2.54 cm = 1 in → (2.54 cm / 1 in)³ = 1

1 ml = 1 cm³ → (1 ml / cm³) = 1

(0.40 µg Hg / ml) = 1

1 g = 10^6 µg → (1 g Hg / 10^6 µg Hg) = 1

1 kg = 10³ g → (1 kg / 10³ g) = 1

so in the above conversion, I keep multiplying by different forms of "1" and progressively changing the units...

***************

final word

***************

although my answer may look complicated at first...  notice the simplicity compared to the other answers.  and also notice the difference in results....

volume of lake.....

100 mi² x 20 ft x (5280 ft / mi)²  = 5.58 x 10^10 ft³  

mass in kg....

(5.58 x 10^10 ft³) x (12 in / ft)³ x (2.54 cm / in)³ x (1 ml / 1 cm³) x (0.40 µg Hg / ml) x (1 g Hg / 10^6 µg Hg) x (1 kg / 10³ g) = 6.32 x 10^5 kg

relatively easy to follow right?  each () =  a conversion factor..... each conversion factor = 1 right ?

now let's compare to the other answers...

**** chem man ****

this was actually quite well done except that he misread the problem... V = A x depth  and area was given to be 100 mi²  chem man took A to be 100 mi x 100 mi...

if instead, he did this....

A = 100 mi² x (1609 m / 1 mi)² x (100 cm / m)² = 2.59 x 10^12 cm²

V = V = A x d = (2.59 x 10^12 cm²) x 610. cm = 1.58 x 10^15 cm^3

etc....

then the answer would have been mass Hg = 6.32 x 10^5 kg

*****  norrie *****

= 1,579,776,000,000,000mL is correct....

0.4μg/ml / 1,000 = 0.0004μg/L is correct but why did you do this?

1,579,776,000,000,000 mL x 0.0004μg/L does NOT =

= 631,910,400,000μg Hg....

it does = 631,910,400,000μg Hg x ml / L.... the ml / L does NOT cancel...

if you did this instead....

1,579,776,000,000,000 mL x 0.4μg/ml = 6.32 x 10^14  ÃƒÂŽÃ‚¼g = 6.32 x 10^8 g = 6.32 x 10^5 kg

then your answer would agree with the rest of ours...

***** for Steve Y and rmjrenne *******

Steve has the basic idea in that (5280ft x 5280 ft / 1 mile x 1 mile) is the conversion factor.. however (5280 ft / mi)²  is generally how this is written for simplicity

rmjrenee.... you converted from 100 mi²  ÃƒÂ¢Ã‚†Â’ 10 mi x 10mi →

16 km x 16 km → 256 km²  ÃƒÂ¢Ã‚†Â’ 16 x 10^5 cm x 16 x 10^5 cm →2.59 x 10^12 cm².... that worked out ok but...

100 mi² x (1.6 km / mi)² x (10^6 cm / km)² = 2.59 x 10^12 cm²

is easier to follow isn't it?

******* an update for Norrie *******

Thank you for critiquing my answer.  This method, that I keep trying to "teach" in my answers to "how do you do this conversion" questions is in fact  the method taught for unit conversions in Universities for at least the past 30 years.   I know this because I have studied this... have taught it.... and use it all the time.  And yes, it always is difficult at first as you suggest, but then it becomes easier as you use it...

So why is this method taught?  here are some examples.... let's start by comparing your method to mine...looking at just the first part...

yours...

100mi² = 10mi x 10mi.

1mi = 5,280ft/mi.

10mi = 52,800ft

52,800ft x 52,800ft x 20ft = 55,756,800,000ft³

mine

100 mi² x 20 ft x (5280 ft / mi)²  = 5.58 x 10^10 ft³

ok?  notice the (5280 ft / mi)² converts the mi² to ft² ?   mi² is on the top and bottom of the fraction and therefore cancels... leaving units of ft² x ft.  you may not see the advantage yet... so let's continue....

Now what would your answer look like if your units were cubic?

100mi³  = ³√10 mi x ³√10 mi x ³√10 mi

or perhaps if you think about it for awhile....

100mi³  = 10mi x 10mi x 1 mi

1mi = 5,280ft/mi.

10mi = 52,800ft

52,800ft x 52,800ft x 5,280ft ft = 1.47 x 10^13 ft³

versus

100 mi³ x  (5280 ft / mi)³  = 1.47 x 10^13 ft³

which is simplier?

or what if the cube roots were more difficult.. let's say convert 157 mi³ to ft³

your way...

157 mi³  = ³√157mi x ³√157mi x ³√157mi

1mi = 5,280ft/mi.

³√157mi = 5.3947 x 5,280ft/mi = 28484 ft

28484 ft x 28484 ft x 28484 ft= 2.31 x 10^13 ft³

my way

157 mi³ x  (5280 ft / mi)³  = 2.31 x 10^13 ft³

or how about this...

157 ft³ /s to m³ / hr

your method

157 ft³ = ³√157ft x ³√157ft x ³√157ft

3.2808 ft = 1 m.

³√157ft / 3.2808ft / 1 m  = 1.644 m

1.644 m x 1.644 m x 1.644 m= 4.446 m³

1 hr = 3600 sec

4.446 m³ /s = 4.446 m³/ s/ 1 hr /3600 sec = 1.60 x 10^4 m³/hr

my method...

157 ft³ /s x (1 m / 3.2808 ft)³ x (3600 s/hr) = 1.60 x 10^4 m³/hr

you may also notice that the  

"³√157ft / 3.2808ft / 1 m" and the "4.446 m³/ s/ 1 hr /3600 sec"  steps are very complicated... do you multiply or divide?   you don't have that problem with my method. right?

So what happens if the units are even more complicated?   g cm²/s² /ft² to lbsf / in² for example?  or to lbsm / in² and so on....

All that said..your original answer was incorrect.  The reason you made a mistake was because your method does not clearly carry units through and you cannot watch them cancel.  Which is the whole point of the unit factor method.  I gave you a thumbs down because your answer was incorrect and I gave you an  explanation as to where you went wrong.   I could certainly give you a  thumbs down without the explanation.  Would that be better?

One last thing.   This is yahoo answers.  People ask questions.  People answer questions.   There is a rating system for the answers.   Thumbs up if you like the others answers, thumbs down if you don't.   You gave an incorrect answer and you have really have no reason to take offense for a thumbs down now do you.  So instead of getting upset why don't you focus a bit of that energy on trying to understand or learn what the points others are making? I certainly understand that you have spent many years in industry and have much  valuable knowledge.   But so have I.  Take my advice and read this answer again and try to pick up this method.

Answer 3

Volume of the lake in mL (or cm^3):

V = A x d (surface area times depth)

100 miles x (1609 m / 1 mile) x (100 cm / 1 m) = 1.609 x 10^7 cm

20 ft x (12 inches / 1 ft) x (2.54 cm / 1 inch) = 610. cm

V = A x d = (1.609 x 10^7 cm)^2 x 610. cm = 1.58 x 10^17 cm^3

1 mL = 1 cm^3

1.58 x 10^17 mL x (0.4 x 10^-6 g / 1 mL) = 6.32 x 10^10 g

1000 g = 1 kg

6.32 x 10^10 g x (1 kg / 1000 g) = 6.32 x 10^7 kg

Answer 4

100mi² = 10mi x 10mi.

1mi = 5,280ft/mi.

10mi = 52,800ft

52,800ft x 52,800ft x 20ft = 55,756,800,000ft³ = 1,579,776,000m³

= 1,579,776,000,000,000mL.

1,579,776,000,000,000mL x 0.4μg/mL.

= 631,910,400,000,000μg / 1,000,000μg/g = 631,910.4kg

m w: Notice what simplicity in your answer ?

I've never seen such a convoluted calculation. Why don't you leave other answerers alone ?. I can see my error and have corrected it.

Answer 5

Looks like you have to convert British otr standard measure to metric.

Convert the miles to kilometres first. The lake is 10 miles by 10 miles (think of it as a square since it is 100 square miles in area).

1 mile = 1.6 km, so 10 miles = 16 km

the area of the lake is therefore 16 X 16 = 256 km^2

Now convert km^2 to cm^2 (since 1 cm^3 = 1 mL)

There are 100 cm in a metre and 1000 metres in a km

so there are 10^5 cm in a km (check it so you understand why...)

That means the lake is 16km X 16 km = (16 X 10^5) X (16 X 10^5) = 256 X 10^10 cm^2 in area (with me so far?)

Now for the depth:

1 inch = 2.54 cm and there are 12 inches in 1 foot

The lake therefore has an average depth of 20 ft X 12 in/ft X 2.54 cm/in

or 599.6 cm

So the volume of the lake must therefore be

256 X 10^10 cm^2 X 599.6 cm = 153497.6 X 10^10 cm^3

or approximately 1.54 X 10^15 mL

Each mL contains 0.4 micrograms, or 4 X 10^-7 grams of Hg

so the total amount of Hg in the lake water is

4 X 10^-7 g/mL X 1.54 X 10^15mL

or approximately 6 X 10^8 g of Hg

That would be equivalent to 6 X 10^5 kilograms of Hg in the water of this lake... you could walk on this stuff!