Question #37fe4

Let's say that the sample that is too big to fit in the graduated cylinder has a volume #V# #"mL"#. After you weight the sample, you find that it has a mass #M# #"g"#.

By definition, the density of the mineral, #rho#, is equal to the ratio that exists between the mass of any sample of this mineral and the volume it occupies.

So for the initial sample, you have--I'll skip the units for simplicity

#rho = M/V" "" "color(darkorange)("(*)")#

Now, cut a piece of this sample that is small enough to fit in a graduated cylinder. Place this small piece on a scale and record its mass, let's say #m# #"g"#.

Use the graduated cylinder to determine its volume.

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Let's say that this small piece has a volume #v# #"mL"#.

Since the density of the mineral must be the same regardless of the mass of the sample and the volume it occupies, you can say that--I'll skip the units for simplicity

#rho = m/v#

You can now use equation #color(darkorange)("(*)")# to say that

#M/V = m/v#

Rearrange to find #V#, the volume of the initial sample

#M/V = m/v implies V = (Mcolor(red)(cancel(color(black)("g"))))/(mcolor(red)(cancel(color(black)("g")))) * vcolor(white)(.)"mL"#

#V = (M/m * v)color(white)(.)"mL"#

And there you have it, the volume of the initial sample as a function of its mass #M# and the mass #m# and volume #v# of a smaller piece of this mineral.