8.4: Making Molecules: Mole to Mass (or vice versa) and MasstoMass Conversions
 Page ID
 47503
Skills to Develop
 Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.
We have established that a balanced chemical equation is balanced in terms of moles as well as atoms or molecules. We have used balanced equations to set up ratios, now in terms of moles of materials, that we can use as conversion factors to answer stoichiometric questions, such as how many moles of substance A react with so many moles of reactant B. We can extend this technique even further. Recall that we can relate a molar amount to a mass amount using molar mass. We can use that ability to answer stoichiometry questions in terms of the masses of a particular substance, in addition to moles. We do this using the following sequence:
Collectively, these conversions are called molemass calculations.
As an example, consider the balanced chemical equation
\[Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3 \label{Eq1}\]
If we have 3.59 mol of Fe_{2}O_{3}, how many grams of SO_{3} can react with it? Using the molemass calculation sequence, we can determine the required mass of SO_{3} in two steps. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of SO_{3} needed. Then using the molar mass of SO_{3} as a conversion factor, we determine the mass that this number of moles of SO_{3} has.
The first step resembles the exercises we did in Section 6.4. As usual, we start with the quantity we were given:
\[\mathrm{3.59\: \cancel{ mol\: Fe_2O_3 } \times \left( \dfrac{3\: mol\: SO_3}{1\: \cancel{ mol\: Fe_2O_3}} \right) =10.77\: mol\: SO_3} \label{Eq2}\]
The mol Fe_{2}O_{3} units cancel, leaving mol SO_{3} unit. Now, we take this answer and convert it to grams of SO_{3}, using the molar mass of SO_{3} as the conversion factor:
\[\mathrm{10.77\: \bcancel{mol\: SO_3} \times \left( \dfrac{80.06\: g\: SO_3}{1\: \bcancel{ mol\: SO_3}} \right) =862\: g\: SO_3} \label{Eq3}\]
Our final answer is expressed to three significant figures. Thus, in a twostep process, we find that 862 g of SO_{3} will react with 3.59 mol of Fe_{2}O_{3}. Many problems of this type can be answered in this manner.
The same twostep problem can also be worked out in a single line, rather than as two separate steps, as follows:
\[ 3.59 \cancel{\, mol \, Fe_2O_3} \times \underbrace{\left( \dfrac{ 3 \bcancel{ \, mol\, SO_3}}{ 1 \cancel{\, mol\, Fe_2O_3}} \right)}_{\text{converts to moles of SO}_3} \times \underbrace{ \left( \dfrac{ 80.06 {\, g \, SO_3}}{ 1 \, \bcancel{ mol\, SO_3}} \right)}_{\text{converts to grams of SO}_3} = 862\, g\, SO_3\]
We get exactly the same answer when combining all the math steps together as we do when we calculate one step at a time.
Example \(\PageIndex{1}\): Generation of Aluminum Oxide
How many moles of HCl will be produced when 249 g of AlCl_{3} are reacted according to this chemical equation?
\[2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g) \nonumber\]
Solution
Steps for Problem Solving  Example \(\PageIndex{1}\) 

Identify the "given"information and what the problem is asking you to "find."  Given: 249 g AlCl_{3} Find: moles HCl 
List other known quantities  1 mol AlCl_{3} = 133.33 g/mol 6 mol of HCl to 2 mol AlCl_{3} 
Prepare a concept map and use the proper conversion factor.  
Cancel units and calculate.  \(249\, \cancel{g\, AlCl_{3}}\times \dfrac{1\, \cancel{mol\, AlCl_{3}}}{133.33\, \cancel{g\, AlCl_{3}}}\times \dfrac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}=5.60\, mol\, HCl\) 
Think about your result.  Since 249 g of AlCl_{3} is less than 266.66 g, the mass for 2 moles of AlCl_{3} and the relationship is 6 mol of HCl to 2 mol AlCl_{3 , }the answer should be less than 6 moles of HCl. 
Exercise \(\PageIndex{1}\): Generation of Aluminum Oxide
How many moles of Al_{2}O_{3} will be produced when 23.9 g of H_{2}O are reacted according to this chemical equation?
\[2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g) \nonumber\]
 Answer
 0.442 mol Al_{2}O_{3}
It is a small step from molemass calculations to massmass calculations. If we start with a known mass of one substance in a chemical reaction (instead of a known number of moles), we can calculate the corresponding masses of other substances in the reaction. The first step in this case is to convert the known mass into moles, using the substance’s molar mass as the conversion factor. Then—and only then—we use the balanced chemical equation to construct a conversion factor to convert that quantity to moles of another substance, which in turn can be converted to a corresponding mass. Sequentially, the process is as follows:
This threepart process can be carried out in three discrete steps or combined into a single calculation that contains three conversion factors. The following example illustrates both techniques.
Example \(\PageIndex{2}\): Decomposition of Ammonium Nitrate
Ammonium nitrate decomposes to dinitrogen monoxide and water according to the following equation.
\[\ce{NH_4NO_3} \left( s \right) \rightarrow \ce{N_2O} \left( g \right) + 2 \ce{H_2O} \left( l \right) \nonumber\]
In a certain experiment, \(45.7 \: \text{g}\) of ammonium nitrate is decomposed. Find the mass of each of the products formed.
Steps for Problem Solving  Example \(\PageIndex{2}\) 

Identify the "given"information and what the problem is asking you to "find." 
Given: \(45.7 \: \text{g} \: \ce{NH_4NO_3}\) Mass \(\ce{N_2O} = ? \: \text{g}\) Mass \(\ce{H_2O} = ? \: \text{g}\) 
List other known quantities 
1 mol \(\ce{NH_4NO_3} = 80.06 \: \text{g/mol}\) 1 mol \(\ce{N_2O} = 44.02 \: \text{g/mol}\) 1 mol \(\ce{H_2O} = 18.02 \: \text{g/mol}\) 1 mol NH_{4}NO_{3} to 1 mol N_{2}O to 2 mol H_{2}O 
Prepare two concept maps and use the proper conversion factor. 

Cancel units and calculate. 
\(45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{1 \: \text{mol} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{44.02 \: \text{g} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{N_2O}} = 25.1 \: \text{g} \: \ce{N_2O}\) \(45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{2 \: \ce{H_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{18.02 \: \text{g} \: \ce{H_2O}}{1 \: \text{mol} \: \ce{H_2O}} = 20.6 \: \text{g} \: \ce{H_2O}\) 
Think about your result.  The total mass of the two products is equal to the mass of ammonium nitrate which decomposed, demonstrating the law of conservation of mass. Each answer has three significant figures. 
Exercise \(\PageIndex{2}\): Carbon Tetrachloride
Methane can react with elemental chlorine to make carbon tetrachloride (\(\ce{CCl_4}\)). The balanced chemical equation is as follows.
\[\ce{CH4 (g) + 4 Cl2 (g) → CCl2 (l) + 4 HCl (l) }\]
How many grams of HCl are produced by the reaction of 100.0g of \(\ce{CH4}\)?
 Answer
 908.7g HCl
Contributors
Henry Agnew (UC Davis)