Below, we see two carbohydrates: glucose and sucrose. Sucrose is almost exactly twice the size of glucose, although their empirical formulas are very similar. Some people can distinguish them on the basis of taste, but it's not a good idea to go around tasting chemicals. The best way to tell glucose and sucrose apart is to determine the molar masses—this approach allows you to easily tell which compound is which. Molecular formulas give the kind and number of atoms of each element present in the molecular compound. In many cases, the molecular formula is the same as the empirical formula. The chemical formula will always be some integer multiple (\(n\)) of the empirical formula (i.e. integer multiples of the subscripts of the empirical formula). \[\text{ Molecular Formula} = n (\text{Empirical formula}) \nonumber \] therefore \[ n = \dfrac{\text{Molecular Formula}}{\text{Empirical Formula}} \nonumber \] The integer multiple, n, can also be obtained by dividing the molar mass, \(MM\), of the compound by the empirical formula mass, \(EFM\) (the molar mass represented by the empirical formula). \[ n = \dfrac{MM ( molar mass)}{EFM (empirical formula molar mass)} \nonumber \] Table \(\PageIndex{1}\) shows the comparison between the empirical and molecular formula of methane, acetic acid, and glucose, and the different values of n. The molecular formula of methane is \(\ce{CH_4}\) and because it contains only one carbon atom, that is also its empirical formula. Sometimes, however, the molecular formula is a simple whole number multiple of the empirical formula. Acetic acid is an organic acid that is the main component of vinegar. Its molecular formula is \(\ce{C_2H_4O_2}\). Glucose is a simple sugar that cells use as a primary source of energy. Its molecular formula is \(\ce{C_6H_{12}O_6}\). The structures of both molecules are shown in Figure \(\PageIndex{2}\). They are very different compounds, yet both have the same empirical formula of \(\ce{CH_2O}\). Empirical formulas can be determined from the percent composition of a compound as discussed in section 6.8. In order to determine its molecular formula, it is necessary to know the molar mass of the compound. Chemists use an instrument called a mass spectrometer to determine the molar mass of compounds. In order to go from the empirical formula to the molecular formula, follow these steps:
The empirical formula of a compound of boron and hydrogen is \(\ce{BH_3}\). Its molar mass is \(27.7 \: \text{g/mol}\). Determine the molecular formula of the compound.
Vitamin C (ascorbic acid) contains 40.92 % C, 4.58 % H, and 54.50 % O, by mass. The experimentally determined molecular mass is 176 amu. What are the empirical and chemical formulas for ascorbic acid? Answer Empirical Formula C3H4O3 Answer Molecular Formula C6H8O6Summary
 LICENSED UNDER We first need to find the empirical formula, which represents the smallest whole number ratio of elements in the compound. After we determine the empirical formula, we determine the empirical molar mass. If the empirical molar mass is the same as the molecular molar mass, then the empirical formula is also the molecular formula. If the empirical molar mass is not the same as the molecular molar mass, we divide the molecular molar mass by the empirical molar mass, which will give us the factor to multiply by the subscripts in the empirical formula. This will give the molecular formula. Empirical Formula Since the masses add to 100, we can start by determining the moles for each element. Moles of each element To determine the moles of each element, divide the given mass for of element by its molar mass to get moles. The molar mass of an element is its atomic weight on the periodic table in g/mol. #"Moles C":# #(40color(red)cancel(color(black)("g C")))/(12.011color(red)cancel(color(black)("g"))/"mol")="3.3 mol"# #"Moles H":# #(6.7color(red)cancel(color(black)("g H")))/(1.008color(red)cancel(color(black)("g"))/"mol")="6.6 mol"# #"Moles O":# #(53.3color(red)cancel(color(black)("g O")))/(15.999color(red)cancel(color(black)("g"))/"mol")="3.33 mol"# Now we need to determine the mole ratio for each element. Mole ratios: subscripts of the empirical formula Divide the number of moles for each element by the least number of moles. #"C":# #(3.3)/(3.3)=1.0# #"H":# #(6.6)/(3.3)=2.0# #"O":# #(3.33)/(3.3)=1.0# The empirical formula is #"CH"_2"O"#. The empirical molar mass is: #(1xx"12.011 g/mol C")+(2xx"1.008 g/mol H")+(1xx"15.999 g/mol O")="30.026 g/mol"# Molecular Formula Divide the molecular molar mass by the empirical molar mass. #(180.18color(red)cancel(color(black)("g"))/color(red)cancel(color(black)("mol")))/(30.026color(red)cancel(color(black)("g"))/color(red)cancel(color(black)("mol")))=6.0008~~6"# Multiply the subscripts in the empirical formula by #6# to get the subscripts of the molecular formula. #"C"_6"H"_12"O"_6"# This is the molecular formula for 6-carbon monosaccharides, such as glucose, fructose, and galactose. |
What is the molecular formula of fructose if its molar mass is 180 g mol 180 g mol and its empirical formula is CH 2 O CH2O?
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