To recognize the empirical formula of a compound from its composition by mass. To derive the molecular formula that a link from its empirical formula.

You are watching: Difference between formula weight and molecular weight


When a new benidormclubdeportivo.orgical compound, such as a potential brand-new pharmaceutical, is synthesized in the laboratory or isolated native a herbal source, benidormclubdeportivo.orgists identify its element composition, the empirical formula, and also its structure to know its properties. This section concentrates on exactly how to identify the empirical formula of a compound and also then use it to identify the molecule formula if the molar fixed of the link is known.


Formula and Molecular Weights

The formula weight that a substance is the amount of the atom weights of every atom in that benidormclubdeportivo.orgical 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 problem exists as discrete molecule (as through atoms that room benidormclubdeportivo.orgically bonded together) climate the benidormclubdeportivo.orgical formula is the molecular formula, and also the formula weight is the molecular weight. For example, carbon, hydrogen and oxygen deserve to benidormclubdeportivo.orgically bond to form a molecule that the street glucose v the benidormclubdeportivo.orgical and also molecular formula that 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 are not benidormclubdeportivo.orgically bonded and also do no exist together discrete molecules. However, they do associate in discrete ratios the ions. Thus, we can describe their formula weights, but not your molecular weights. Table salt (\(\ceNaCl\)), because that example, has a formula weight of:

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


Percentage ingredient from Formulas

In some varieties of analyses of that is crucial to understand the percentage through mass of each kind of element in a compound. The legislation of definite proportions states that a benidormclubdeportivo.orgical compound always contains the exact same proportion of facets by mass; the is, the percent composition—the portion of each aspect present in a pure substance—is constant (although there room exceptions come 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 also 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 m sorry is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This way that 100.00 g of sucrose always contains 42.11 g the carbon, 6.48 g of hydrogen, and 51.41 g the oxygen. An initial the molecule formula of sucrose (C12H22O11) is supplied to calculate the mass percent of the ingredient elements; the massive percentage can then be provided to determine an empirical formula.

According come its molecular formula, each molecule that sucrose consists of 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. A mole the sucrose molecules thus contains 12 mol the carbon atoms, 22 mol of hydrogen atoms, and also 11 mol that oxygen atoms. This information can be offered to calculation the mass of each aspect in 1 mol the sucrose, which gives the molar mass of sucrose. These masses can then be supplied to calculation the percent composition of sucrose. To three decimal places, the calculations are the following:

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

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

\< \text mass that 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 that sucrose has a massive of 342.297 g; keep in mind that much more than half of the fixed (175.989 g) is oxygen, and also almost fifty percent of the fixed (144.132 g) is carbon.

The mass percent of each element in sucrose is the mass of the element present in 1 mol of sucrose separated by the molar massive of sucrose, multiplied by 100 to give a percentage. The an outcome is displayed to 2 decimal places:

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

\< \text mass % H in Sucrose = \text mass that 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 of O/mol sucrose \over \text molar massive 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 sum of the percentages of every the facets in the compound is 100%:

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

If the sum is no 100%, one error has actually been do in calculations. (Rounding come the correct number of decimal places can, however, reason the complete to be slightly different from 100%.) hence 100.00 g that sucrose includes 42.11 g the carbon, 6.48 g of hydrogen, and 51.41 g that oxygen; to 2 decimal places, the percent composition of sucrose is without doubt 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen.

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

It is also feasible to calculate mass percentages using atomic masses and molecular masses, through atomic massive units. Because the price is a ratio, expressed together a percentage, the units of fixed cancel even if it is they space grams (using molar masses) or atom mass systems (using atomic and also molecular masses).


Example \(\PageIndex1\): NutraSweet

Aspartame is the artificial sweetener offered as NutraSweet and also Equal. Its molecule formula is \(\ceC14H18N2O5\).

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Molecular structure of Aspartame. (CC BY-NC-SA 3.0; anonymous) calculation the mass percent of each facet in aspartame. Calculate the mass of carbon in a 1.00 g packet of Equal, assuming it is pure aspartame.

Given: molecule formula and also mass of sample

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

Strategy:

usage atomic masses native the routine table to calculate the molar massive of aspartame. Divide the fixed of each element by the molar fixed of aspartame; then multiply through 100 to attain percentages. To uncover the mass of an facet contained in a given mass the aspartame, main point the massive of aspartame by the mass portion of the element, expressed together a decimal.

Solution:

a.

A We calculation the massive of each element in 1 mol of aspartame and the molar fixed of aspartame, below 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 fifty percent the massive of 1 mol of aspartame (294.277 g) is carbon (168.154 g).

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B To calculation the mass percentage of each element, we division the mass of each facet in the link by the molar mass of aspartame and also then multiply by 100 to attain percentages, below reported to two decimal places:

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

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

\< mass \% \, 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 achieve a full that differs from 100% by more than around ±1%, there have to be an error somewhere in the calculation.