You are watching: Mass of one molecule of sucrose

When a new chathamtownfc.netical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, chathamtownfc.netists determine its elemental composition, its empirical formula, and its structure to understand its properties. This section focuses on how to determine the empirical formula of a compound and then use it to determine the molecular formula if the molar mass of the compound is known.

## Formula and Molecular Weights

*The formula weight of a substance is the sum of the atomic weights of each atom in its chathamtownfc.netical formula*. For example, water (H2O) has a formula weight of:

\<2\times(1.0079\;amu) + 1 \times (15.9994 \;amu) = 18.01528 \;amu\>

If a substance exists as discrete molecules (as with atoms that are *chathamtownfc.netically bonded* together) then the *chathamtownfc.netical formula* **is** the *molecular formula*, and the *formula weight* **is** the *molecular weight*. For example, carbon, hydrogen and oxygen can chathamtownfc.netically bond to form a molecule of the sugar **glucose** with the chathamtownfc.netical and molecular formula of C6H12O6. The formula weight and the molecular weight of glucose is thus:

\<6\times(12\; amu) + 12\times(1.00794\; amu) + 6\times(15.9994\; amu) = 180.0 \;amu\>

Ionic substances are not chathamtownfc.netically bonded and do not exist as discrete molecules. However, they do associate in discrete ratios of ions. Thus, we can describe their formula weights, but not their *molecular weights*. Table salt (\(\ce{NaCl}\)), for example, has a formula weight of:

\<23.0\; amu + 35.5 \;amu = 58.5 \;amu\>

## Percentage Composition from Formulas

In some types of analyses of it is important to know the ** percentage by mass** of each type of element in a compound. The law of definite proportions states that a chathamtownfc.netical compound always contains the same proportion of elements by mass; that is, the percent composition—the percentage of each element present in a pure substance—is

**constant**(although there are exceptions to this law). Take for example methane (\(CH_4\)) with a Formula and molecular weight:

\<1\times (12.011 \;amu) + 4 \times (1.008) = 16.043 \;amu\>

the relative (mass) percentages of carbon and hydrogen are

\<\%C = \dfrac{1 \times (12.011\; amu)}{16.043 amu} = 0.749 = 74.9\%\>

\<\%H = \dfrac{4 \times (1.008 \;amu)}{16.043\; amu} = 0.251 = 25.1\%\>

A more complex example is sucrose (table sugar), which is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g of sucrose always contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen. First the molecular formula of sucrose (C12H22O11) is used to calculate the mass percentage of the component elements; the mass percentage can then be used to determine an **empirical formula**.

According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole of sucrose molecules therefore contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. This information can be used to calculate the mass of each element in 1 mol of sucrose, which gives the molar mass of sucrose. These masses can then be used to calculate the percent composition of sucrose. To three decimal places, the calculations are the following:

\< \text {mass of C/mol of sucrose} = 12 \, mol \, C \times {12.011 \, g \, C \over 1 \, mol \, C} = 144.132 \, g \, C \label{3.1.1a}\>

\< \text {mass of H/mol of sucrose} = 22 \, mol \, H \times {1.008 \, g \, H \over 1 \, mol \, H} = 22.176 \, g \, H \label{3.1.1b}\>

\< \text {mass of O/mol of sucrose} = 11 \, mol \, O \times {15.999 \, g \, O \over 1 \, mol \, O} = 175.989 \, g \, O \label{3.1.1c}\>

Thus 1 mol of sucrose has a mass of 342.297 g; note that more than half of the mass (175.989 g) is oxygen, and almost half of the mass (144.132 g) is carbon.

The mass percentage of each element in sucrose is the mass of the element present in 1 mol of sucrose divided by the molar mass of sucrose, multiplied by 100 to give a percentage. The result is shown to two decimal places:

\< \text {mass % C in Sucrose} = {\text {mass of C/mol sucrose} \over \text {molar mass of sucrose} } \times 100 = {144.132 \, g \, C \over 342.297 \, g/mol } \times 100 = 42.11 \% \>

\< \text {mass % H in Sucrose} = {\text {mass of H/mol sucrose} \over \text {molar mass of sucrose} } \times 100 = {22.176 \, g \, H \over 342.297 \, g/mol } \times 100 = 6.48 \% \>

\< \text {mass % O in Sucrose} = {\text {mass of O/mol sucrose} \over \text {molar mass of sucrose} } \times 100 = {175.989 \, g \, O \over 342.297 \, g/mol } \times 100 = 51.41 \% \>

This can be checked by verifying that the sum of the percentages of all the elements in the compound is 100%:

\< 42.11\% + 6.48\% + 51.41\% = 100.00\%\>

If the sum is not 100%, an error has been made in calculations. (Rounding to the correct number of decimal places can, however, cause the total to be slightly different from 100%.) Thus 100.00 g of sucrose contains 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to two decimal places, the percent composition of sucrose is indeed 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen.

Figure \(\PageIndex{1}\): Percent and absolute composition of sucroseIt is also possible to calculate mass percentages using atomic masses and molecular masses, with atomic mass units. Because the answer is a ratio, expressed as a percentage, the units of mass cancel whether they are grams (using molar masses) or atomic mass units (using atomic and molecular masses).

Example \(\PageIndex{1}\): NutraSweet

Aspartame is the artificial sweetener sold as NutraSweet and Equal. Its molecular formula is \(\ce{C14H18N2O5}\).

Molecular Structure of Aspartame. (CC BY-NC-SA 3.0; anonymous) Calculate the mass percentage of each element in aspartame. Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.**Given**: molecular formula and mass of sample

**Asked for**: mass percentage of all elements and mass of one element in sample

**Strategy**:

**Solution**:

a.

**A **We calculate the mass of each element in 1 mol of aspartame and the molar mass of aspartame, here to three decimal places:

\< 14 \,C (14 \, mol \, C)(12.011 \, g/mol \, C) = 168.154 \, g \nonumber\>

\< 18 \,H (18 \, mol \, H)(1.008 \, g/mol \, H) = 18.114 \, g\nonumber\>

\< 2 \,N (2 \, mol \, N)(14.007 \, g/mol \, N) = 28.014 \, g\nonumber\>

\< +5 \,O (5 \, mol \, O)(15.999 \, g/mol \, O) = 79.995 \, g\nonumber\>

\

Thus more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).

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**B** To calculate the mass percentage of each element, we divide the mass of each element in the compound by the molar mass of aspartame and then multiply by 100 to obtain percentages, here reported to two decimal places:

\< mass \% \, C = {168.154 \, g \, C \over 294.277 \, g \, aspartame } \times 100 = 57.14 \% C\nonumber\>

\< mass \% \, H = {18.114 \, g \, H \over 294.277 \, g \, aspartame } \times 100 = 6.16 \% H\nonumber\>

\< mass \% \, N = {28.014 \, g \, N \over 294.277 \, g \, aspartame } \times 100 = 9.52 \% \nonumber\>

\< mass \% \, O = {79.995 \, g \, O \over 294.277 \, g \, aspartame } \times 100 = 27.18 \%\nonumber \>

As a check, we can add the percentages together:

\< 57.14\% + 6.16\% + 9.52\% + 27.18\% = 100.00\% \nonumber\>

If you obtain a total that differs from 100% by more than about ±1%, there must be an error somewhere in the calculation.