3.9: Gram-to-Gram Stoichiometry
- Page ID
- 423626
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Convert from mass or moles of one substance to mass or moles of another substance in a chemical reaction.
Mole to Mass Conversions
In the previous section, we described how to convert between moles of one substance to moles of another substance using the ratios from the balanced chemical equation. We can combine this concept with molar mass conversions to convert from moles to moles to grams.
As an example, consider the balanced chemical equation
\[\ce{Fe_2O_3 + 3SO_3 \rightarrow Fe_2(SO_4)_3} \label{Eq1} \]
If we have 3.59 mol of \(\ce{Fe2O3}\), how many grams of \(\ce{SO3}\) can react with it? We will need to convert from mol -> mol -> grams. First, we construct the appropriate molar ratio, determined from the balanced chemical equation, to calculate the number of moles of \(\ce{SO3}\) needed. Then, using the molar mass of \(\ce{SO3}\) as a conversion factor, we determine the mass that this number of moles of \(\ce{SO3}\) has.
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 \(\ce{Fe2O3}\) units cancel, leaving mol \(\ce{SO3}\) unit. Now, we take this answer and convert it to grams of \(\ce{SO3}\), using the molar mass of \(\ce{SO3}\) 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 and we find that 862 g of \(\ce{SO3}\) will react with 3.59 mol of \(\ce{Fe2O3}\).
The same two-step problem can also be worked out in a single line, rather than as two separate steps, as follows:
\[ 3.59 \cancel{\, mol \, Fe_2O_3} \times \underbrace{\left( \dfrac{ 3 \bcancel{ \, mol\, SO_3}}{ 1 \cancel{\, mol\, Fe_2O_3}} \right)}_{\text{converts to moles of SO}_3} \times \underbrace{ \left( \dfrac{ 80.06 {\, g \, SO_3}}{ 1 \, \bcancel{ mol\, SO_3}} \right)}_{\text{converts to grams of SO}_3} = 862\, g\, SO_3 \nonumber \]
We get exactly the same answer when combining all math steps together.
How many moles of \(\ce{HCl}\) will be produced when 249 g of \(\ce{AlCl3}\) are reacted according to this chemical equation?
\[\ce{2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g)} \nonumber \]
Solution
| Steps for Problem Solving | Example \(\PageIndex{1}\) |
|---|---|
| Identify the "given" information and what the problem is asking you to "find." | Given: 249 g AlCl3 Find: moles HCl |
| List other known quantities. | 1 mol AlCl3 = 133.33 g AlCl3 6 mol of HCl to 2 mol AlCl3 |
| Prepare a concept map and use the proper conversion factor. |
|
| Cancel units and calculate. | \(249\, \cancel{g\, AlCl_{3}}\times \dfrac{1\, \cancel{mol\, AlCl_{3}}}{133.33\, \cancel{g\, AlCl_{3}}}\times \dfrac{6\, mol\, HCl}{2\, \cancel{mol\, AlCl_{3}}}=5.60\, mol\, HCl\) |
How many moles of \(\ce{Al2O3}\) will be produced when 23.9 g of \(\ce{H2O}\) are reacted according to this chemical equation?
\[\ce{2AlCl_3 + 3H_2O(ℓ) → Al_2O_3 + 6HCl(g)} \nonumber \]
- Answer
- 0.442 mol Al2O3
Mass to Mass Conversions
In the previous examples, we converted from mol -> mol -> grams. If we start with grams of one substance, we can convert to grams of another substance using the following strategy: grams -> mol -> mol -> grams. This calculation strategy is outlined below:
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 N2O that would be formed.
| Steps for Problem Solving | Example \(\PageIndex{2}\) |
|---|---|
| Identify the "given" information and what the problem is asking you to "find." |
Given: \(45.7 \: \text{g} \: \ce{NH_4NO_3}\) |
| List other known quantities. |
1 mol \(\ce{NH_4NO_3} = 80.06 \: \text{g}\) 1 mol \(\ce{N_2O} = 44.02 \: \text{g}\) 1 mol NH4NO3 to 1 mol N2O |
| Prepare concept maps and use the proper conversion factors. |
|
| Cancel units and calculate. |
\(45.7 \: \text{g} \: \ce{NH_4NO_3} \times \dfrac{1 \: \text{mol} \: \ce{NH_4NO_3}}{80.06 \: \text{g} \: \ce{NH_4NO_3}} \times \dfrac{1 \: \text{mol} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{NH_4NO_3}} \times \dfrac{44.02 \: \text{g} \: \ce{N_2O}}{1 \: \text{mol} \: \ce{N_2O}} = 25.1 \: \text{g} \: \ce{N_2O}\) |
Methane can react with elemental chlorine to make carbon tetrachloride (\(\ce{CCl_4}\)). The balanced chemical equation is as follows:
\[\ce{CH4 (g) + 4 Cl2 (g) → CCl2 (l) + 4 HCl (l) } \nonumber \]
How many grams of \(\ce{HCl}\) are produced by the reaction of 100.0 g of \(\ce{CH4}\)?
- Answer
- 908.7g HCl
Summary
- The calculation pathway grams -> mol -> mol -> grams can be used to convert between mass of one substance and mass of another substance in a chemical reaction.