This document discusses density, specific gravity, and specific volume calculations. It defines density as mass per unit volume and provides examples of calculating density. Specific gravity is defined as the ratio of the weight of a substance to the weight of an equal volume of water. Methods for calculating specific gravity using weights and volumes are shown. Specific volume is the ratio of the volume of a substance to the volume of an equal weight of water. Examples are provided for calculating specific gravity and specific volume.

1. Lecture 7
Density, specific gravity, and specific volume calculations
Laith Al-Asadi

2. Density
 Density (d) is mass per unit volume of a substance. It is
usually expressed as grams per cubic centimeter (g/cc).
 Because the gram is defined as the mass of 1 cc of water
at 4C, the density of water is 1 g/cc.
 the United States Pharmacopeia1 states that 1 mL may be
used as the equivalent of 1 cc, the density of water may
be expressed as 1 g/mL.
 In contrast, one milliliter of mercury weighs 13.6 g;
hence, its density is 13.6 g/mL.

3. Density
 Density may be calculated by dividing mass by volume,
that is:
Density =
𝑴𝒂𝒔𝒔
𝑽𝒐𝒍𝒖𝒎𝒆
 Thus, if 10 mL of sulfuric acid weighs 18 g, its density is:
Density =
18 (𝑔)
10 (𝑚𝐿)
= 1.8 grams per milliliter

4. Example 1
 A piece of copper metal weighs 53.6 g, and has a volume
of 6 mL. Calculate its density?
Solution :
Density =
𝑴𝒂𝒔𝒔
𝑽𝒐𝒍𝒖𝒎𝒆
=
53.6 𝑔
6 𝑚𝑙
= 8.933 g/ml answer.

5. Example 2
 If 250 mL of alcohol weighs 203 g, what is its density?
Solution:
Density =
𝑴𝒂𝒔𝒔
𝑽𝒐𝒍𝒖𝒎𝒆
=
203 𝑔
250 𝑚𝑙
= 0.812 g/ml answer.

6. Specific gravity
 Specific gravity (sp gr) is a ratio, expressed decimally, of
the weight of a substance to the weight of an equal
volume of a substance chosen as a standard, both
substances at the same temperature or the temperature
of each being known.
 Water is used as the standard for the specific gravities of
liquids and solids; the most useful standard for gases is
hydrogen.

7. Specific gravity
 Specific gravity may be calculated by dividing the weight of a given
substance by the weight of an equal volume of water, that is:
Specific gravity =
Weight of substance
Weight of equal volume of water
Thus, if 10 mL of sulfuric acid weighs 18 g, and 10 mL of water, under
similar conditions, weighs 10 g, the specific gravity of the acid is:
Specific gravity =
18 (𝑔)
10 (g)
= 1.8
• Substances that have a specific gravity less than 1 are lighter than
water.
• Substances that have a specific gravity greater than 1 are heavier than
water.

8. Specific gravity
 Specific gravities may be expressed decimally to as many
places as the accuracy of their determination warrants.
 In pharmaceutical work, this expression may be to two,
three, or four decimal places.
 Because substances expand or contract at different rates
when their temperatures change, accurate work
necessitates allowing carefully for variations in the
specific gravity of a substance.
 In the United States Pharmacopeia, the standard
temperature for specific gravities is 25 °C, except for that
of alcohol, which is 15.56 °C by government regulation.

9. Example 1
 If 150 mL of a sorbitol solution weigh 170 g, what is
its specific gravity?
Solution:
Specific gravity =
Weight of substance
Weight of equal volume of water
=
170 𝑔
150 𝑔
= 1.1333

10. Example 2
 If 2 fl. oz. of glycerol weighs 74.1 g, what is its specific gravity?
Solution:
2 fl. oz. = 2 × 29.57 ml = 59.14 ml
Wt of equal volume of water = 59.14 g
Specific gravity =
Weight of substance
Weight of equal volume of water
=
74.1 𝑔
59.14 𝑔
= 1.253 answer.

11. Pycnometer or Specific Gravity
Bottle
 A pycnometer is a special glass bottle used to determine
specific gravity .
 Pycnometersare generally available for laboratory use in
volumes ranging from 1 mL to 50 mL.
 Pycnometers have fitted glass stoppers with a capillary
opening to allow trapped air and excess fluid to escape.
 Some pycnometers have thermometers affixed, because
temperature is a factor in specific gravity determinations.
 In using a pycnometer, it is first weighed empty and then
weighed again when filled to capacity with water.
 The weight of the water is calculated by difference.

12. Example 1
 A pycnometer weighs 21.62 g. Filled with water, it weighs
46.71 g; filled with an- liquid, it weighs 43.28 g. Calculate
the specific gravity of the liquid?
Solution:
Specific gravity =
Weight of substance
Weight of equal volume of water
Weight of equal volume of water = 43.28g + 46.71g – 21.62g
= 68.28 g
Specific gravity =
43.28 𝑔
68.28 𝑔
= 0.633

13. Example 2
 A 50 mL pycnometer is found to weigh 120 g when empty,
171 g when filled with water, and 160g when filled with
an unknown liquid. Calculate the specific gravity of the
unknown liquid.
Solution
Weight of water: 171 g - 120 g = 51 g
Weight of unknown liquid: 160 g - 120 g = 40 g
 Specific gravity =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑒𝑞𝑢𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
Specific
 gravity of unknown liquid =
40 𝑔
51 𝑔
40 = 0.78 answer.

14. Homework
 A modified Ringer’s Irrigation has the following
formula:
Sodium chloride 8.6 g
Potassium chloride 0.3 g
Calcium chloride 0.33 g
PEG 3350 60 g
Water for injection ad 1000 mL
Assuming that 980 mL of water is used in preparing the
irrigation, calculate its specific gravity?

15. Displacement or Plummet Method
 Calculating the specific gravity of a liquid determined by
the displacement or plummet method is based on
Archimedes’ principle, which states that a body immersed
in a liquid displaces an amount of the liquid equal to its
own volume and suffers an apparent loss in weight equal to
the weight of the displaced liquid.
 We can weigh a plummet when suspended in water and
when suspended in a liquid the specific gravity of which we
want to determine; and by subtracting these weights from
the weight of the plummet in air, we get the weights of
equal volumes of the liquids needed in our calculation.

16. Example:
 A glass plummet weighs 12.64 g in air, 8.57 g when
immersed in water, and 9.12 g when immersed in an oil.
Calculate the specific gravity of the oil.
Solution:
12.64 g - 9.12 g = 3.52 g of displaced oil
12.64 g - 8.57 g = 4.07 g of displaced water
Specific gravity of oil =
3.52 (𝑔)
4.07 (𝑔)
= 0.865 answer.

17. Calculating Weight, Knowing the
Volume and Specific Gravity
 Based on the explanation in the previous paragraphs, we
can derive the following equation:
Grams = Milliliters × Specific gravity
 Although it is both obvious and true that one cannot
multiply milliliters by specific gravity and have a product in
grams, the equation ‘‘works’’ because the volume of the
liquid in question is assumed to be the same volume as
water for which milliliters equal grams. So, in essence,
the true equation would be:
 Grams (other liquid) = Grams (of equal volume of
water) × Specific gravity (other liquid)

18. Examples: 1
 What is the weight, in grams, of 3620 mL of alcohol with a
specific gravity of 0.820?
Solution:
3620 mL of water weigh = 3620 g
3620 g × 0.820 = 2968 g, answer.

19. Example 2
What is the weight, in grams, of 2 fl. oz. of a liquid having a
specific gravity of 1.118?
In this type of problem, it is best to convert the given
volume to its metric equivalent first and then solve the
problem in the metric system.
Solution:
2 × 29.57 mL = 59.14 mL
59.14 mL of water weigh = 59.14 g
59.14 g × 1.118 = 66.12 g answer.

20. Calculating Specific Volume
 Specific volume, in pharmaceutical practice, is usually
defined as an abstract number representing the ratio,
expressed decimally, of the volume of a substance to the
volume of an equal weight of another substance taken as
a standard, both having the same temperature. Water is
the standard.
Specific volume =
𝒕𝒉𝒆 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒂 𝒈𝒊𝒗𝒆𝒏 𝒎𝒂𝒔𝒔
𝒕𝒉𝒆 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒂𝒏 𝒆𝒒𝒖𝒂𝒍 𝒘𝒆𝒊𝒈𝒉𝒕 𝒐𝒇 𝒘𝒂𝒕𝒆𝒓

21.  if 25 g of glycerin measures 20 mL and 25 g of water
measures 25 mL under the same conditions, the
specific volume of the glycerin is:
Solution:
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 25 𝑔 𝑜𝑓 𝑔𝑙𝑦𝑐𝑒𝑟𝑖𝑛
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 25 𝑔 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
=
20 𝑚𝑙
25𝑚𝑙
= 0.8

22. Example
Calculate the specific volume of a syrup, 91.0 mL of
which weighs 107.16 g.
Solution:
107.16 g of water measures 107.16 mL
Specific volume of syrup =
91 𝑚𝑙
107.16 𝑚𝑙
= 0.849, answer.