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When a brand-new rebab.netical compound, such as a potential new pharmaceutical, is synthesized in the laboratory or isolated from a natural source, rebab.netists recognize its elemental complace, its empirical formula, and its framework to understand also its properties. This section concentrates on exactly how to recognize the empirical formula of a compound and then use it to identify the molecular formula if the molar mass of the compound is well-known.
Formula and Molecular Weights
The formula weight of a substance is the sum of the atomic weights of each atom in its rebab.netical formula. For instance, water (H2O) has a formula weight of:
<2 imes(1.0079;amu) + 1 imes (15.9994 ;amu) = 18.01528 ;amu>
If a substance exists as discrete molecules (as with atoms that are rebab.netically bonded together) then the rebab.netical formula is the molecular formula, and also the formula weight is the molecular weight. For example, carbon, hydrogen and also oxygen have the right to rebab.netically bond to form a molecule of the sugar glucose via the rebab.netical and molecular formula of C6H12O6. The formula weight and also the molecular weight of glucose is thus:
<6 imes(12; amu) + 12 imes(1.00794; amu) + 6 imes(15.9994; amu) = 180.0 ;amu>
Ionic substances are not rebab.netically bonded and execute not exist as discrete molecules. However before, they carry out associate in discrete ratios of ions. Hence, we deserve to explain their formula weights, but not their molecular weights. Table salt ((ceNaCl)), for example, has actually a formula weight of:
<23.0; amu + 35.5 ;amu = 58.5 ;amu>
Percentage Complace from Formulas
In some forms of analyses of it is crucial to know the percent by mass of each kind of element in a compound. The legislation of definite prosections says that a rebab.netical compound constantly includes the very same propercentage of aspects by mass; that is, the percent composition—the percentage of each facet existing in a pure substance—is consistent (although there are exceptions to this law). Take for example methane ((CH_4)) through a Formula and also molecular weight:
<1 imes (12.011 ;amu) + 4 imes (1.008) = 16.043 ;amu>
the family member (mass) percenteras of carbon and also hydrogen are
<\%C = dfrac1 imes (12.011; amu)16.043 amu = 0.749 = 74.9\%>
<\%H = dfrac4 imes (1.008 ;amu)16.043; amu = 0.251 = 25.1\%>
An even more facility example is sucincreased (table sugar), which is 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen by mass. This indicates that 100.00 g of sucincreased always contains 42.11 g of carbon, 6.48 g of hydrogen, and also 51.41 g of oxygen. First the molecular formula of succlimbed (C12H22O11) is supplied to calculate the mass percent of the component elements; the mass portion deserve to then be supplied to determine an empirical formula.
According to its molecular formula, each molecule of sucrose contains 12 carbon atoms, 22 hydrogen atoms, and also 11 oxygen atoms. A mole of succlimbed molecules therefore consists of 12 mol of carbon atoms, 22 mol of hydrogen atoms, and 11 mol of oxygen atoms. This indevelopment have the right to be provided to calculate the mass of each element in 1 mol of sucincreased, which offers the molar mass of sucincreased. These masses have the right to then be used to calculate the percent complace of succlimbed. To 3 decimal locations, the calculations are the following:
< message mass of C/mol of sucrose = 12 , mol , C imes 12.011 , g , C over 1 , mol , C = 144.132 , g , C label3.1.1a>
< ext mass of H/mol of sucrose = 22 , mol , H imes 1.008 , g , H over 1 , mol , H = 22.176 , g , H label3.1.1b>
< message mass of O/mol of sucrose = 11 , mol , O imes 15.999 , g , O over 1 , mol , O = 175.989 , g , O label3.1.1c>
Therefore 1 mol of succlimbed has actually a mass of 342.297 g; note that more than fifty percent of the mass (175.989 g) is oxygen, and also nearly half of the mass (144.132 g) is carbon.
The mass percentage of each facet in sucrose is the mass of the facet current in 1 mol of sucrose separated by the molar mass of sucincreased, multiplied by 100 to offer a portion. The result is displayed to 2 decimal places:
< message mass % C in Sucrose = ext mass of C/mol sucrose over ext molar mass of sucrose imes 100 = 144.132 , g , C over 342.297 , g/mol imes 100 = 42.11 \% >
< message mass % H in Sucrose = message mass of H/mol sucrose over message molar mass of sucrose imes 100 = 22.176 , g , H over 342.297 , g/mol imes 100 = 6.48 \% >
< message mass % O in Sucrose = ext mass of O/mol sucrose over ext molar mass of sucrose imes 100 = 175.989 , g , O over 342.297 , g/mol imes 100 = 51.41 \% >
This have the right to be checked by verifying that the amount of the percentperiods of all the aspects in the compound is 100%:
< 42.11\% + 6.48\% + 51.41\% = 100.00\%>
If the amount is not 100%, an error has actually been made in calculations. (Rounding to the correct number of decimal areas deserve to, however, reason the full to be slightly various from 100%.) Therefore 100.00 g of sucrose has 42.11 g of carbon, 6.48 g of hydrogen, and 51.41 g of oxygen; to 2 decimal places, the percent composition of succlimbed is indeed 42.11% carbon, 6.48% hydrogen, and 51.41% oxygen.
It is also feasible to calculate mass percenteras utilizing atomic masses and also molecular masses, through atomic mass systems. Since the answer is a ratio, expressed as a portion, the devices of mass cancel whether they are grams (utilizing molar masses) or atomic mass units (using atomic and molecular masses).
Example (PageIndex1): NutraSweet
Aspartame is the synthetic sweetener offered as NutraSweet and also Equal. Its molecular formula is (ceC14H18N2O5).
Given: molecular formula and also mass of sample
Asked for: mass percent of all elements and also mass of one element in sample
Strategy:Use atomic masses from the regular table to calculate the molar mass of aspartame. Divide the mass of each facet by the molar mass of aspartame; then multiply by 100 to achieve percentages. To find the mass of an aspect included in a offered mass of aspartame, multiply the mass of aspartame by the mass percentage of that facet, expressed as a decimal.
A We calculate the mass of each facet in 1 mol of aspartame and also the molar mass of aspartame, right here to 3 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>
Hence more than half the mass of 1 mol of aspartame (294.277 g) is carbon (168.154 g).
See more: Negative Dilute Drug Test Results & How Much Water Causes A Diluted Drug Test
B To calculate the mass portion of each element, we divide the mass of each aspect in the compound by the molar mass of aspartame and then multiply by 100 to attain percentperiods, below reported to 2 decimal places:
< mass \% , C = 168.154 , g , C over 294.277 , g , aspartame imes 100 = 57.14 \% C>
< mass \% , H = 18.114 , g , H over 294.277 , g , aspartame imes 100 = 6.16 \% H>
< mass \% , N = 28.014 , g , N over 294.277 , g , aspartame imes 100 = 9.52 \% >
< mass \% , O = 79.995 , g , O over 294.277 , g , aspartame imes 100 = 27.18 \% >
As a inspect, we have the right to include the percentperiods together:
< 57.14\% + 6.16\% + 9.52\% + 27.18\% = 100.00\% >
If you acquire a full that differs from 100% by even more than about ±1%, tbelow have to be an error somewhere in the calculation.