Sciencemadness Discussion Board
Not logged in [Login ]
Go To Bottom

Printable Version  
Author: Subject: Specific Gravity clarification asked
CHRIS25
National Hazard
****




Posts: 951
Registered: 6-4-2012
Location: Ireland
Member Is Offline

Mood: No Mood

[*] posted on 8-5-2012 at 05:15
Specific Gravity clarification asked


I know google has so much to say about thjis topic but trying to dissect the simplest of info that I need was becoming a bit of a headache.

This is what I am looking for: When I measure 1 mL of water that is same as 1gram. Easy. But if a liquid has a specific gravity of 0.7 for example and I am weighing this on my scales would I then multiply 0.7mL by 1000mL which = 700, and this would be the number of grams of that liquid that I need to use. have I worked all this out correctly please?

Sorry I forgot to say that the equation was asking for 1litre of liquid, but I measure out 700 grams and this would be fairly accurate, at least this is what I am understanding from all the information overload.

[Edited on 8-5-2012 by CHRIS25]




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
View user's profile View All Posts By User
kavu
Hazard to Others
***




Posts: 207
Registered: 11-9-2011
Location: Scandinavia
Member Is Offline

Mood: To understand is to synthesize

[*] posted on 8-5-2012 at 05:27


A sure way of checking reasoning is to use some algebra! For material with density ρ mass m and volume V are related by ρ = m/V. Solving for mass m you get m = ρV. If you want to have one liter of ρ = 0.7 g/mL liquid, you need to weigh out m = ρV = 0.7 g/mL * 1000 mL = 700 g. Your reasoning is thus correct. Just plug in the values and presto! Another way of thinking about this would be to say that ρ = 0.7 g/mL means that for every mL you have the stuff weighs 0,7 grams. For 10 mL 7 g, for 100 mL 70 g and for 1 L 700 g.



View user's profile View All Posts By User
sargent1015
Hazard to Others
***




Posts: 315
Registered: 30-4-2012
Location: WI
Member Is Offline

Mood: Relaxed

[*] posted on 8-5-2012 at 05:28


Yeah, but measuring out 1 liter is just as easy, if not easier. Is there a point to weighing it out?



The Home Chemist Book web page and PDF. Help if you want to make Home Chemist history! http://www.bromicacid.com/bookprogress.htm
View user's profile View All Posts By User
CHRIS25
National Hazard
****




Posts: 951
Registered: 6-4-2012
Location: Ireland
Member Is Offline

Mood: No Mood

[*] posted on 8-5-2012 at 07:13


Quote: Originally posted by sargent1015  
Yeah, but measuring out 1 liter is just as easy, if not easier. Is there a point to weighing it out?
Thanks Kavu. And Sargent, I anticipated one bright smart chemist was going to comment on the logically obvious - apart from just wanting to understand gravity mass I also came across a couple of times when measuring the liquid was not possible in the middle of a reaction and I need to put the whole beaker on the scales to double check something. Also it is handy when some data from certain sites measure things in PARTS as opposed to grams or mils. In those instances I wish to keep consistency and therefore weigh the liquids and powders all in grams. Hope this satisfies your curiosity.

kind regards




‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
View user's profile View All Posts By User
adamsium
Hazard to Others
***




Posts: 180
Registered: 9-4-2012
Location: \ƚooɿ\
Member Is Offline

Mood: uprooting

[*] posted on 8-5-2012 at 08:53


Quote: Originally posted by CHRIS25  
would I then multiply 0.7mL by 1000mL which = 700, and this would be the number of grams of that liquid that I need to use. have I worked all this out correctly please?
[Edited on 8-5-2012 by CHRIS25]


One thing that might help you a bit with this and many other things is to read up a bit on dimensional analysis. I'm not sure if that was just a typo, but you'll notice that 0.7mL x 1000mL = 700mL2, and you clearly aren't looking for a value with units of millilitres squared. As kavu indicated, density is defined as mass per volume, or ρ = m/V ('ρ' is the Greek letter rho, lower case, and is used to represent the dimension of density). Since you want to find mass (m) from density (ρ ) and volume (V), you can rearrange the equation for density to m = ρV by multiplying both sides by V. Mass has (in this case) units of grams (the SI unit, however, is the kilogram), density has units of mass per volume (or, with the units you are using, grams per millilitre -- g/mL) and volume has units of millilitres (again, the dimension derived directly from SI units would be the m3, though the litre, or dm3 - decimetre cubed - exactly equal to 1L - is commonly used. This is simply a cube with sides of 10cm (10cm = 1dm (1/10th of a metre, hence "deci-"))). It's important that, whatever units you use, you are consistent. From this, you can see that you would get X = Yg/mL x ZmL, so we have X = YZ((gxmL)/mL), and cancelling the mL in the YZ term, we have X = YZg, with appropriate units (i.e. a mass).

This is quite simple in this application, but is actually quite important, particularly for more complex calculations, and is a great way to rationalise a method for calculating a desired quantity (e.g. finding moles from molarity and volume C = mol/V, so, mol = CV .... mol/L = mol/L, so, mol = (mol/L)L gives mol = mol, etc), as well as for simply ensuring that you (probably) haven't made a silly error in your calculation, the logic being that if the units have been carried through correctly to give the correct dimension at the end of the calculation, the values associated with them likely have been, also.
View user's profile View All Posts By User
CHRIS25
National Hazard
****




Posts: 951
Registered: 6-4-2012
Location: Ireland
Member Is Offline

Mood: No Mood

[*] posted on 8-5-2012 at 13:08


Thankyou Adamsium, a little bit of reading I think you put me in the right direction.



‘Calcination… is such a Separation of Bodies by Fire, as makes ‘em easily reducible into Powder; and for that reason ‘tis call’d by some Chymical Pulverization.’ (John Friend, Chymical Lectures London, 1712)

Right is right, even if everyone is against it, and wrong is wrong, even if everyone is for it. (William Penn 1644-1718)

The very nature of Random, Chance development precludes the existence of Order - strange that our organic and inorganic world is so well defined by precision and law. (me)
View user's profile View All Posts By User

  Go To Top