International Cookbooks and Ingredient Mass vs. Volume - ChemPRIME

International Cookbooks and Ingredient Mass vs. Volume

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If you read recipes from any country outside the US, you'll note that quantities of ingredients (especially solids) are normally given in mass units (grams or kilograms) rather than volume units (cups, tablespoons, or milliliters). There are several reasons for measuring masses of ingredients:

A digital kitchen scale, or in Dutch, Digi-keukenweegschaal

  • Convenience. It's easy to put a bowl on a scale, tare it, measure in the ingredient, tare the scale again, measure in the next ingredient, and so on.
  • Unambiguous. The definition of a "cup" is not precise. There's a US legal cup (240 ml, 8.12 US customary fluidounces), a US customary cup (236.6 ml, 8 US customary fluidounces), a metric cup (250 ml, 8.45 US customary fluidounces), and an Imperial cup (284 ml).
  • Accuracy. If a recipe calls for a cup of flour, this can be a wide range of masses. It depends on humidity, which makes flour expand in volume. [1] It depends on whether the flour is "Scooped", "dip and sweeped", "Spooned" or "Sifted", [2], on the shape of the cup [3], and it depends on the type of flour,[4] but all references do not agree: [5][6]. Salt densities also range widely; 1 tsp of ordinary granulated table salt is equivalent to 1-1/4 tsp of Morton's Coarse Kosher Salt, and 2 tsp of Crystal Kosher Salt.[7]

Flour Type Mass of
1 cup (g)
Deaf Smith 130 g 0.55
U.K. Self Raising 111 0.47
US All-purpose 99 0.42
buckwheat 170 0.72
cake flour 90 0.38
semolina flour 175 0.74
wheat bread 99 0.42
whole wheat 130 0.55
flour, rye 90 0.38

Bakeries, therefore, always measure dry ingredients by weight. But US home cooks can still use volume measurements with international recipes, if they use the appropriate "conversion factors". Of course, the best thing to do is use a kitchen scale to measure masses directly.

Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 or lb/ft3. Now we will see how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the volume of a given mass of flour? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.

When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another (for example, convert a volume of flour to a mass of flour. We might also want to know, if we need 12 cups of flour, how much it will cost if the price is given per pound of flour. This requires conversion of a volume to a mass, then to a cost.

Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, but the conversion factor is based on a mathematical function (D = m / V) or mathematical equation that relates parameters. An example involving the familiar quantities mass and volume will be used to illustrate the way conversion factors are employed.

Suppose a cookie recipe calls for 1 cup of caster (or castor) sugar (called "superfine", 0.35 mm sugar in the US, originally sprinkled from a "castor"), [8] and we want to calculate the cost of the recipe, anticipating scale up in a bakery. The price of superfine sugar is about $3.00 for 5 lb, and so we need to know the mass rather than the volume. If we didn't trust web converters to apply to this specific sugar, we could add it to a rectangular container and weigh it to determine the density. Suppose we have a rectangular container which measures 3.04 cm × 8.14 cm × 17.3 cm. We can easily calculate that its volume is 428 cm3. If the sugar weighs 347 g, it has a density of 347 g / 428 cm3 = 0.811 g/cm3. But how much is 1 cup worth? To find out, we can manipulate the equation which defines density, ρ = m / V. If we multiply both sides by V, we obtain

\text{V}\times \rho =\frac{\text{m}}{\text{V}}\times \text{V}=\text{m}

\text{Mass}=\text{m}=\text{V}~\times~\rho        (1)

or mass = volume × density

One cup is 236.6 cm3, and taking the density of caster sugar to be 0.811 g/cm3 we can now calculate

\text{Mass}=\text{m}=\text{V}~\times~\rho =\text{236.6 cm}^{3}\times \frac{\text{0}\text{.811 g}}{\text{1 cm}^{3}}=\text{192 g}

We can use a unity factor to get the weight in pounds:

192 g x 1 lb / 453.6 g = .423 lb,

and finally calculate the cost with another conversion factor:

0.423 lb x ($3.00 / 5.00 lb) = $0.254, or a little over 25 cents.

The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. (1) are multiplied by 1/ρ, we have

  & \frac{\text{1}}{\rho }\times \text{m}=\text{V}\rho \times \frac{\text{1}}{\rho }=\text{V} \\ 
 & \text{         V}=\text{m}\times \frac{\text{1}}{\rho } \\ 
\end{align}       (2)

So if a German cookie recipe calls for 250 g of caster sugar, we can calculate the volume in cups:

\text{V}~=~\text{250 g}~\times~\frac{\text{1 cm}^3}{\text{0.811 g}} = 308 cm3, and

308 cm3 x (1 cup / 236.6 cm3) = 1.30 cups.

Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?

An Important Caveat

A mistake sometimes made by beginning students is to confuse density with concentration, which also may have units of g/cm3. For example, sugar syrups can be described by their density in grams of syrup per cm3 or their concentration in grams of sugar per cm3 of syrup. By dimensional analysis, it looks like density and concentration can be used interchangeably. To see the error, we must understand the meaning of the function

C = \frac{m}{V}.

In this case, V refers to the volume of a solution, which contains both a solute (the sugar) and solvent (the water), but m is the mass of just the solute (sugar), not the whole solution.

Given a concentration of a syrup is 25 g sugar in 100 cm3 of syrup, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the volume of sugar in 20 g of the syrup by using the concentration, as follows:

 \text{20 g syrup}~\times~\frac{100 \text{cm}^{3}\text{syrup}}{\text{25 g sugar}}     = 80 cm3 sugar

It is only possible to calculate the volume of sugar if the density of the syrup is known, so that the volume of syrup represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function. If the density is 1.11 g/cm3, 20.0 g syrup / 1.11 g syrup/cm3 syrup= 18.0 cm3 syrup.

The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.


Honey with density of 1.44 g/ml is sometimes substituted for sugar in a recipe. As a sugar solution with a concentration of about 80 g / 100 cm3 or 0.80 g/cm3 it supplies both sugar and water. Assuming that the sugars in honey have properties similar to caster sugar, what is the volume equivalent of caster sugar (D = 0.811 g / cm3 in 100 g of honey?


The volume of 100 g of honey is

V = m / D = 100 g /1.44 g cm-3 = 69.4 cm3.

The mass of sugar in this volume is

m = V x C = 69.4 cm3 x 0.80 g / cm3 = 55.6 g.

The volume of sugar = m / D = 55.6 g / 0.811 g / cm3 = 68.5 cm3

This is about 0.3 cup. The 100 g of honey would provide 100 g - 55.6 g = 44.4 g of water (44.4 ml or about 0.2 cup).

Note that we cannot calculate the volume of sugar by

 \frac {\frac{1.44 g}{cm^3} x 100 cm^3}{\frac {0.811 g}{cm^3}} = 177 cm3

even though this equation is dimensionally correct, it would be impossible for 100 g of honey to contain 177 cm3 of sugar.

Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.

EXAMPLE 2 Find the volume occupied by a 500 g sample of salt, with a (quite variable) density of about 1.10 g/cm3.

Solution Using Eq. (2),

\text{Volume = }V\text{ = }m\text{ }\times \text{ }\frac{\text{1}}{\rho }\text{ = 500}\times ~ \frac{\text{1 cm}^{\text{3}}}{\text{1}\text{.10 g}}\text{ = 455}\text{cm}^{\text{3}}

(Note that taking the reciprocal of \tfrac{\text{1}\text{.10 g}}{\text{1 cm}^{3}} simply inverts the fraction ― 1 cm3 goes on top, and 1.10 g goes on the bottom.)

If we had multiplied by the density instead of the reciprocal of the density, the units of the result would immediately show our error:

\text{V}=\text{500 g}~\times~ \frac{\text{1.10 g}}{\text{1 cm}^{3}}~=~\text{550 g}^{2}/{\text{cm}^{3}}\; (no cancellation!)

It is clear that square grams per cubic centimeter are not the units we want.

The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because the mathematical formula defining density relates it to mass and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. (1) and (2)], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate converstion factors by unit cancellation, as the following example shows:

EXAMPLE 3 Compare the masses of 1 cup (236.6 mL) of water and honey, with a density of 1.44 g/cm3.


We know that mass is related to volume through density, and that the volume in each case is 236.6 cm3.


m = V x conversion factor

Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel. For the honey,

\text{m}~=~\text{V}~\times ~\rho ~=~\text{236.6 g}~\times~ \text{1}\text{.44 g/cm}^3=\text{341 g}

Using a conversion factor is very similar to using a unity factor — we know the conversion factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of the relationship [ie. the definition of density as defined by Eqs. (1) and (2) includes the relationships between density, mass, and volume], not because it is has a value of one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.

A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:

\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume   or   }m\overset{\rho }{\longleftrightarrow}V\text{ }

This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written

\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity}

As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.

EXAMPLE 4 Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).

Solution The road map

V\xrightarrow{\rho }m\text{ }

tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:

Mass = m = 47.3 cm3 × \frac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}}

Since the volume units are different, we need a unity factor to get them to cancel:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \frac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}

We now have the mass in pounds, but we want it in grams, so another unity factor is needed:

m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\frac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\frac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{.9 g}

In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique.

Web Sources:


  1. Weight-per-volume
  3. Wolke, R.L. "What Einstein Told His Cook", W.W. Norton & Co., NY, 2002, p. 293
  7. Wolke, R.L. "What Einstein Told His Cook", W.W. Norton & Co., NY, 2002, p. 57
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