Learning Objectives Explain why atomic ratios within a molecule can be applied to moles Calculate the molar mass of a molecule Calculate atoms, molecules, moles, or mass given one of the others Molar Mass Instead of Formula Mass Chemists have to work with molecules, but we can't see them or pick them up. However, we can see and hold groups of molecules. Any measured amount (in grams) can viewed as some amount of a mole, and any amount of a mole can be considered some amount of molecules (Figure \(\PageIndex{1}\). For example, St. Elmo Brady performed many experiments with pyruvic acid (Figure \(\PageIndex{2}\)). Knowing the structure and formula, C 3 H 4 O 3 , let him calculate the mass of one mole of pyruvic acid. He could then use that molar mass with the amount of pyruvid acid he measured out for an experiment to calculate how many moles or how many molecules were present. Definition: Molar Mass The mass, in grams, of a quantity of one mole of a substance. The unit of molar mass is g/mol. Each mole of pyruvic acid contains three moles of carbon, four moles of hydrogen, and three moles of oxygen. We can calculate the total mass for each element and then sum them together to get the molar mass of pyruvic acid. \[\begin{array}{l}3\ mol\ C\times12.01\ \frac{g\ C}{mol\ C}=36.03\ g\ C\\4\ mol\ H\times1.01\ \frac{g\ H}{mol\ H}=4.04\ g\ H\\3\ mol\ O\times16.00\ \frac{g\ O}{mol\ O}=48.00\ g\ O\\\\\stackrel{\underline{\begin{matrix}&36.03\ g\ C&\\&4.04\ g\ H&\\+&48.00\ g\ O&\end{matrix}}}{\ \ \begin{matrix}&88.07\ g&&\end{matrix}}\end{array}\] One mole of pyruvic acid has a mass of 88.07 grams, or pyruvic acid has a molar mass of 88.07 g/mol. St. Elmo Brady (1884-1966) St. Elmo Brady was the first African-American to receive a doctorate in chemistry in 1916 and to be inducted in Phi Lambda Upsilon, the academic chemistry honor society. His research created new ways to prepare and purify types of organic molecules, and included the basis for the modern understanding of the acidity of carboxylic acids. Brady received his undergraduate degree at Fisk University in 1908. He then took a teaching position at Tuskegee Normal and Industrial Institute (now Tuskegee University) before taking a leave of absence to pursue his graduate education at University of Illinois, Urbana-Champagne. Brady's experiences with racism and segregation led him to become a driving force in developing chemistry departments at four of the major historically black college or universities (HBCUs) ; Tougaloo College, Tuskegee University, Fisk University, and Howard University. In 2019, he was honored by the American Chemical Society with a National Historic Chemical Landmark at University of Illinois, Urbana-Champagne; celebrating him as an inspiration for African-Americans in chemistry and a foundational figure in American chemical education. Exercise \(\PageIndex{1}\): Molecular Mass of Freon Calculate the molar mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is \(\ce{CCl_3F}\). Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows: Answer 137.368 g/mol Exercise \(\PageIndex{2}\): Molar mass of Silicon Nitride Calculate the molar mass of \(\ce{Si3N4}\), commonly called silicon nitride. It is an extremely hard and inert material that is used to make cutting tools for machining hard metal alloys. Answer 140.29 g/mol Mass-to-Moles and Moles-to-Mass If St. Elmo Brady measured out 88.07 g of pyruvic acid, then the sample contains one mole of pyruvic acid. If the sample contains one mole of pyruvic acid, then it contains \(6.022\times10^{23}\) molecules of pyruvic acid. Molar mass is a proportion and thus can be used as a conversion factor between the quantity unit, mole, and the mass unit, gram. Figure \(\PageIndex{1}\) is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Examples \(\PageIndex{1}\) and \(\PageIndex{2}\). Example \(\PageIndex{1}\): Combustion of Butane For the combustion of butane (\(\ce{C_4H_{10}}\)) the balanced equation is: \[\ce{2C4H_{10}(l) + 13O2(g) \rightarrow 8CO2(g) + 10H2O(l) }\nonumber \] Calculate the mass of \(\ce{CO_2}\) that is produced in burning 1.00 gram of \(\ce{C_4H_{10}}\). Solution Thus, the overall sequence of steps to solve this problem is: First of all we need to calculate how many moles of butane we have in a 1.00 gram sample: \[ (1.00\; g\; \ce{C_4H_{10}}) \left(\dfrac{1\; mol\; \ce{C_4H_{10}}}{58.0\;g\; \ce{C_4H_{10}}}\right) = 1.72 \times 10^{-2} \; mol\; \ce{C_4H_{10}}\nonumber \] Now, the stoichiometric relationship between \(\ce{C_4H_{10}}\) and \(\ce{CO_2}\) is: \[\left( \dfrac{8\; mol\; \ce{CO_2}}{2\; mol\; \ce{C_4H_{10}}}\right)\nonumber \] Therefore: \[ \left(\dfrac{8\; mol\; \ce{CO_2}}{2\; mol\; \ce{C_4H_{10}}} \right) \times 1.72 \times 10^{-2} \; mol\; \ce{C_4H_{10}} = 6.88 \times 10^{-2} \; mol\; \ce{CO_2} \nonumber \] The question called for the determination of the mass of \(\ce{CO_2}\) produced, thus we have to convert moles of \(\ce{CO_2}\) into grams (by using the molecular weight of \(\ce{CO_2}\)): \[ 6.88 \times 10^{-2} \; mol\; \ce{CO_2} \left( \dfrac{44.0\; g\; \ce{CO_2}}{1\; mol\; \ce{CO_2}} \right) = 3.03\;g \; \ce{CO_2}\nonumber \] Example \(\PageIndex{2}\): Ethylene Glycol For 35.00 g of ethylene glycol (\ce{HOCH2CH2OH}), which is used in inks for ballpoint pens, calculate the number of moles. molecules. Given : mass and molecular formula Asked for : number of moles and number of molecules Strategy : Use the molecular formula of the compound to calculate its molecular mass in grams per mole. Convert from mass to moles by dividing the mass given by the compound’s molar mass. Convert from moles to molecules by multiplying the number of moles by Avogadro’s number. Solution : A The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example \(\PageIndex{1}\): \[ 2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu \nonumber \] \[ 6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu \nonumber \] \[ 2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu \nonumber \] \[ C_2H_6O_2 \text {molecular mass of ethylene glycol} = 62.068 \, amu \nonumber \] The molar mass of ethylene glycol is 62.068 g/mol. B The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole): \[ { \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }\nonumber \] So \[ 35.00 \, g \text {ethylene glycol} \left ( {1 \, mole\, \text {ethylene glycol} \over 62.068 \, g \, \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol} \nonumber \] It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations. C To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number: \[ \begin{align*} \text {molecules of ethylene glycol} &= 0.5639 \, \cancel{mol} \left ( {6.022 \times 10^{23} \, molecules \over 1 \, \cancel{mol} } \right ) \\[4pt] &= 3.396 \times 10^{23} \, molecules \end{align*} \nonumber \] Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 10 23 molecules, which is indeed the case. Exercise \(\PageIndex{3}\): Freon-11 For 75.0 g of CCl 3 F (Freon-11), calculate the number of moles. molecules. Answer a 0.546 mol Answer b 3.29 × 10 23 molecules Example \(\PageIndex{3}\) Calculate the mass of 1.75 mol of each compound. S 2 Cl 2 (common name: sulfur monochloride; systematic name: disulfur dichloride) Ca(ClO) 2 (calcium hypochlorite) Given : number of moles and molecular or empirical formula Asked for : mass Strategy : A Calculate the molecular mass of the compound in grams from its molecular formula (if covalent) or empirical formula (if ionic). B Convert from moles to mass by multiplying the moles of the compound given by its molar mass. Solution : We begin by calculating the molecular mass of S 2 Cl 2 and the formula mass of Ca(ClO) 2 . A The molar mass of S 2 Cl 2 is obtained from its molecular mass as follows: \[ 2S (2 \, atoms)(32.065 \, amu/atom ) = 64.130 \, amu \nonumber \] \[+ 2Cl (2 \, atoms )(35.453 \, amu/atom ) = 70.906 \, amu \nonumber \] \[ S_2 Cl_2 \text {molecular mass of } S_2Cl_2 = 135.036 \, amu \nonumber \] The molar mass of S 2 Cl 2 is 135.036 g/mol. B The mass of 1.75 mol of S 2 Cl 2 is calculated as follows: \[moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g) \nonumber \] \[ 1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2 \nonumber \] A The formula mass of Ca(ClO) 2 is obtained as follows: \[1Ca (1 \, atom)(40.078 \, amu/atom) = 40.078 \, amu \nonumber \] \[2Cl (2 \, atoms)(35.453 \, amu/atom) = 70.906 \, amu \nonumber \] \[+ 2O (2 \, atoms)(15.9994 \, amu/atom) = 31.9988 \, amu \nonumber \] \[ Ca (ClO)_2 \text { formula mass of } Ca (ClO)_2 = 142.983 \, amu\nonumber \] The molar mass of Ca(ClO) 2 142.983 g/mol. B The mass of 1.75 mol of Ca(ClO) 2 is calculated as follows: \[ moles Ca(ClO)_2 \left [{\text {molar mass} Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2} \right ]=mass Ca(ClO)_2\nonumber \] \[ 1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2 \nonumber \] Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is. Exercise \(\PageIndex{4}\) Calculate the mass of 0.0122 mol of each compound. Si 3 N 4 (silicon nitride), used as bearings and rollers (CH 3 ) 3 N (trimethylamine), a corrosion inhibitor Answer a 1.71 g Answer b 0.721 g Conversions Between Grams, Mol, & Atoms: https://youtu.be/rOvErpAnoCg Summary To analyze chemical transformations, it is essential to use a standardized unit of measure called the mole. The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number (6.022 × 10 23 ) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 10 23 atoms, molecules, or formula units of that substance. References St. Elmo Brady . American Chemical Society. (n.d.). Retrieved July 18, 2022, from https://www.acs.org/content/acs/en/e...lmo-brady.html