Sagot :
Answer:
If a substance exists as discrete molecules (as with atoms that are chemically bonded together) then the chemical formula is the molecular formula, and the formula weight is the molecular weight. For example, carbon, hydrogen and oxygen can chemically bond to form a molecule of the sugar glucose with the chemical and molecular formula of C6H12O6. The formula weight and the molecular weight of glucose is thus:If a substance exists as discrete molecules (as with atoms that are chemically bonded together) then the chemical formula is the molecular formula, and the formula weight is the molecular weight. For example, carbon, hydrogen and oxygen can chemically bond to form a molecule of the sugar glucose with the chemical and molecular formula of C6H12O6. The formula weight and the molecular weight of glucose is thus:
6Ă—(12amu)+12Ă—(1.00794amu)+6Ă—(15.9994amu)=180.0amu(3.3.63)
Ionic substances are not chemically 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 (NaCl), for example, has a formula weight of:
23.0amu+35.5amu=58.5amu(3.3.64)
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 chemical 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 ( CH4 ) with a Formula and molecular weight:
1Ă—(12.011amu)+4Ă—(1.008)=16.043amu(3.3.65)
the relative (mass) percentages of carbon and hydrogen are
%C=1Ă—(12.011amu)16.043amu=0.749=74.9%(3.3.66)
%H=4Ă—(1.008amu)16.043amu=0.251=25.1%(3.3.67)
A more complex example is sucrose (table sugar) 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{3.3.1a}
\text {mass of H/mol of sucrose} = 22 \, mol \, H \times {1.008 \, g \, H \over 1 \, mol \, H} = 22.176 \, g \, C \label{3.3.1b}
\text {mass of O/mol of sucrose} = 11 \, mol \, O \times {15.999 \, g \, O \over 1 \, mol \, O} = 175.989 \, g \, O \label{3.3.1c}
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:
mass % C in Sucrose=mass of C/mol sucrosemolar mass of sucroseĂ—100=144.132gC342.297g/molĂ—100=42.12%(3.3.68)
mass % H in Sucrose=mass of H/mol sucrosemolar mass of sucroseĂ—100=22.176gH342.297g/molĂ—100=6.48%(3.3.69)
mass % O in Sucrose=mass of O/mol sucrosemolar mass of sucroseĂ—100=175.989gO342.297g/molĂ—100=51.41%(3.3.70)
This can be checked by verifying that the sum of the percentages of all the elements in the compound is 100%:
42.12%+6.48%+51.41%=100.01%(3.3.71)
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 42.12 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 42.12% carbon, 6.48% hydrogen, and 51.41% oxygen.