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]kavu - 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.sargent1015 - 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?CHRIS25 - 8-5-2012 at 07:13
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.
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.CHRIS25 - 8-5-2012 at 13:08
Thankyou Adamsium, a little bit of reading I think you put me in the right direction.
