To recognize the empirical formula that a compound from its composition by mass. To have the molecular formula the a compound from the empirical formula.

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When a brand-new rebab.netical compound, such together a potential new pharmaceutical, is synthesized in the laboratory or isolated indigenous a herbal source, rebab.netists identify its elemental composition, that empirical formula, and its framework to know its properties. This section focuses on how to determine the empirical formula of a compound and also then use it to identify the molecule formula if the molar massive of the link is known.


Formula and Molecular Weights

The formula weight that a substance is the amount of the atomic weights of every atom in that rebab.netical formula. Because that example, water (H2O) has actually a formula load of:

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

If a problem exists as discrete molecule (as v atoms that are rebab.netically bonded together) climate the rebab.netical formula is the molecular formula, and the formula weight is the molecular weight. Because that example, carbon, hydrogen and also oxygen deserve to rebab.netically bond to type a molecule of the sugar glucose through the rebab.netical and molecular formula the C6H12O6. The formula weight and also the molecular load of glucose is thus:

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

Ionic substances room not rebab.netically bonded and also do not exist together discrete molecules. However, they do associate in discrete ratios the ions. Thus, us can describe their formula weights, however not their molecular weights. Table salt (\(\ceNaCl\)), for example, has a formula load of:

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


Percentage composition from Formulas

In some varieties of analyses of it is necessary to recognize the percentage through mass that each type of element in a compound. The law of identify proportions says that a rebab.netical compound always contains the exact same proportion of aspects by mass; that is, the percent composition—the percent of each element present in a pure substance—is constant (although there space exceptions come this law). Take for instance methane (\(CH_4\)) through a Formula and molecular weight:

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

the family member (mass) percentages the carbon and hydrogen are

\<\%C = \dfrac1 \times (12.011\; amu)16.043 amu = 0.749 = 74.9\%\>

\<\%H = \dfrac4 \times (1.008 \;amu)16.043\; amu = 0.251 = 25.1\%\>

A more complicated example is sucrose (table sugar), i beg your pardon is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This means that 100.00 g that sucrose always contains 42.11 g that carbon, 6.48 g the hydrogen, and 51.41 g that oxygen. First the molecule formula that sucrose (C12H22O11) is offered to calculation the mass percentage of the ingredient elements; the mass percentage deserve to then be offered to recognize an empirical formula.

According come its molecule formula, every molecule of sucrose has 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole that sucrose molecules thus contains 12 mol of carbon atoms, 22 mol of hydrogen atoms, and also 11 mol that oxygen atoms. This information deserve to be supplied to calculation the massive of each aspect in 1 mol the sucrose, which gives the molar fixed of sucrose. This masses have the right to then be supplied to calculation the percent composition of sucrose. To three decimal places, the calculations room the following:

\< \text mass that C/mol the sucrose = 12 \, mol \, C \times 12.011 \, g \, C \over 1 \, mol \, C = 144.132 \, g \, C \label3.1.1a\>

\< \text mass of H/mol the sucrose = 22 \, mol \, H \times 1.008 \, g \, H \over 1 \, mol \, H = 22.176 \, g \, H \label3.1.1b\>

\< \text mass of O/mol that sucrose = 11 \, mol \, O \times 15.999 \, g \, O \over 1 \, mol \, O = 175.989 \, g \, O \label3.1.1c\>

Thus 1 mol of sucrose has actually a massive of 342.297 g; keep in mind that more than half of the fixed (175.989 g) is oxygen, and also almost fifty percent of the massive (144.132 g) is carbon.

The mass portion of each element in sucrose is the mass of the aspect present in 1 mol that sucrose separated by the molar massive of sucrose, multiply by 100 to offer a percentage. The result is presented to two decimal places:

\< \text mass % C in Sucrose = \text mass that 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 fixed of sucrose \times 100 = 22.176 \, g \, H \over 342.297 \, g/mol \times 100 = 6.48 \% \>

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

This deserve to be confirm by verifying that the amount of the percentages of every the aspects in the link is 100%:

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

If the sum is no 100%, an error has been make in calculations. (Rounding come the correct variety of decimal places can, however, reason the full to be slightly various from 100%.) therefore 100.00 g of sucrose consists of 42.11 g the carbon, 6.48 g the hydrogen, and also 51.41 g the oxygen; to 2 decimal places, the percent composition of sucrose is certainly 42.11% carbon, 6.48% hydrogen, and also 51.41% oxygen.

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Figure \(\PageIndex1\): Percent and absolute ingredient of sucrose

It is also feasible to calculation mass percentages making use of atomic masses and molecular masses, v atomic mass units. Since the prize is a ratio, expressed as a percentage, the units of massive cancel whether they space grams (using molar masses) or atomic mass units (using atomic and molecular masses).


Example \(\PageIndex1\): NutraSweet

Aspartame is the man-made sweetener marketed as NutraSweet and Equal. Its molecule formula is \(\ceC14H18N2O5\).

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Molecular framework of Aspartame. (CC BY-NC-SA 3.0; anonymous) calculation the mass percentage of each element in aspartame. Calculation the fixed of carbon in a 1.00 g packet the Equal, assuming that is pure aspartame.

Given: molecular formula and also mass of sample

Asked for: mass portion of all elements and mass that one element in sample

Strategy:

usage atomic masses indigenous the routine table to calculation the molar fixed of aspartame. Divide the mass of each element by the molar mass of aspartame; climate multiply by 100 to attain percentages. To uncover the massive of an aspect contained in a given mass of aspartame, multiply the fixed of aspartame by the mass percent of the element, expressed together a decimal.

Solution:

a.

A We calculate the fixed of each facet in 1 mol of aspartame and the molar fixed of aspartame, here to three decimal places:

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

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

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

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

\

Thus an ext than half the fixed of 1 mol of aspartame (294.277 g) is carbon (168.154 g).

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B To calculation the mass percent of every element, we divide the fixed of each element in the compound by the molar mass of aspartame and also then multiply by 100 to obtain percentages, below reported to two decimal places:

\< massive \% \, C = 168.154 \, g \, C \over 294.277 \, g \, aspartame \times 100 = 57.14 \% C\>

\< fixed \% \, H = 18.114 \, g \, H \over 294.277 \, g \, aspartame \times 100 = 6.16 \% H\>

\< massive \% \, N = 28.014 \, g \, N \over 294.277 \, g \, aspartame \times 100 = 9.52 \% \>

\< fixed \% \, O = 79.995 \, g \, O \over 294.277 \, g \, aspartame \times 100 = 27.18 \% \>

As a check, we can add the percentages together:

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

If you attain a full that differs from 100% by much more than about ±1%, there need to be an error what in the calculation.